Chapter 13 Playing with Numbers

Let the three-digit number be represented by xyz, where:

'x' is the digit in the hundreds place.

'y' is the digit in the tens place.

'z' is the digit in the units (or ones) place.

To write the generalised form of the number, we multiply each digit by its corresponding place value and sum the results.

Value from hundreds place = $x \times 100$

Value from tens place = $y \times 10$

Value from units place = $z \times 1$

The generalised form of the number xyz is the sum of these values:

$x \times 100 + y \times 10 + z \times 1$

This simplifies to:

$100x + 10y + z$

Now, let's compare this generalised form with the given options:

(a) $x + y + z$ (This is the sum of the digits)

(b) $100x + 10y + z$ (This matches our derived form)

(c) $100z + 10y + x$ (This would be the generalised form of the number 'zyx')

(d) $100y + 10x + z$ (This form does not represent xyz based on place values)

Therefore, the correct generalised form of a three-digit number xyz is $100x + 10y + z$.

The correct option is (b).

We are given the expression in its generalised form:

$100a + b + 10c$

To write this in the usual form of a number, we need to identify the digit corresponding to each place value.

The term $100a$ means that 'a' is the digit in the hundreds place ($100 \times a$).

The term $10c$ means that 'c' is the digit in the tens place ($10 \times c$).

The term $b$ (which is $1 \times b$) means that 'b' is the digit in the units place ($1 \times b$).

Arranging the digits by place value from hundreds to units:

• Hundreds place: a
• Tens place: c
• Units place: b

(Note: Using list here for clarity within thought process, will use p tags in final output)

Hundreds place: a

Tens place: c

Units place: b

When written in the usual form, the number is formed by arranging the digits in the order of their place values (hundreds, tens, units).

The digit in the hundreds place is 'a'.

The digit in the tens place is 'c'.

The digit in the units place is 'b'.

Thus, the usual form of the number is acb.

Let's check the options:

(a) abc (Hundreds: a, Tens: b, Units: c) $\rightarrow 100a + 10b + c$

(b) cab (Hundreds: c, Tens: a, Units: b) $\rightarrow 100c + 10a + b$

(c) bac (Hundreds: b, Tens: a, Units: c) $\rightarrow 100b + 10a + c$

(d) acb (Hundreds: a, Tens: c, Units: b) $\rightarrow 100a + 10c + b$

The expression $100a + b + 10c$ is equivalent to $100a + 10c + b$. This matches the usual form acb.

The correct option is (d).

The given equation is:

$5 \times A = \text{CA}$

Here, A is a single digit (0-9), and CA represents a two-digit number where C is the tens digit and A is the units digit.

The generalised form of the two-digit number CA is:

$\text{CA} = 10 \times C + A$

Substituting this into the given equation, we get:

$5 \times A = 10 \times C + A$

Simplify the equation by subtracting A from both sides:

$5A - A = 10C$

$4A = 10C$

Divide both sides by 2:

$2A = 5C$

Since A and C are single digits (0-9) and C is the tens digit of a two-digit number, C must be a non-zero digit (1-9).

The equation $2A = 5C$ implies that $2A$ must be a multiple of 5. Since 2 and 5 are coprime, A must be a multiple of 5.

Possible single-digit values for A that are multiples of 5 are 0 and 5.

Let's consider these possible values for A:

Case 1: Assume $A = 0$.

Substitute $A=0$ into $2A = 5C$:

$2 \times 0 = 5C$

$0 = 5C$

$C = 0$

If $A=0$ and $C=0$, the number CA would be 00, which is 0. While $5 \times 0 = 0$ is true, CA is typically understood as a two-digit number, requiring $C \neq 0$. So, this case is usually excluded in such puzzles.

Case 2: Assume $A = 5$.

Substitute $A=5$ into $2A = 5C$:

$2 \times 5 = 5C$

$10 = 5C$

Divide by 5:

$C = \frac{10}{5}$

$C = 2$

In this case, $A=5$ and $C=2$. Both are single digits, and $C=2$ is non-zero, making CA a valid two-digit number (25).

Let's verify this solution with the original equation $5 \times A = \text{CA}$:

Substitute $A=5$ and $C=2$:

Left side: $5 \times A = 5 \times 5 = 25$

Right side: $\text{CA} = 25$ (since C=2 and A=5)

Since $25 = 25$, the equation holds true.

Thus, the values of A and C are 5 and 2 respectively.

The values are $A=5$ and $C=2$. The question doesn't provide options, so we state the values directly.

The problem is given as an addition sum involving two-digit numbers:

$\begin{array}{cc} & 5 & A \\ + & 2 & 5 \\ \hline & B & 2 \\ \hline \end{array}$

We can solve this by considering the addition in columns, starting from the units place.

In the units column, we have the addition of A and 5. The units digit of the result is 2.

So, $A + 5$ must result in a number ending in 2. The possible values for A (a single digit from 0 to 9) are:

• If $A=0$, $0+5=5$ (ends in 5)
• If $A=1$, $1+5=6$ (ends in 6)
• If $A=2$, $2+5=7$ (ends in 7)
• If $A=3$, $3+5=8$ (ends in 8)
• If $A=4$, $4+5=9$ (ends in 9)
• If $A=5$, $5+5=10$ (ends in 0)
• If $A=6$, $6+5=11$ (ends in 1)
• If $A=7$, $7+5=12$ (ends in 2)
• If $A=8$, $8+5=13$ (ends in 3)
• If $A=9$, $9+5=14$ (ends in 4)

From this, we see that $A+5$ ends in 2 only when $A=7$.

If $A=7$, then $A+5 = 7+5 = 12$. This means the units digit is 2, and there is a carry-over of 1 to the tens column.

Now, let's consider the tens column.

In the tens column, we add the digits 5 and 2, plus the carry-over from the units column (which is 1).

The sum in the tens column is $5 + 2 + 1$.

The result in the tens column is B.

So, $5 + 2 + 1 = B$.

$8 = B$.

Thus, the values of the digits are $A=7$ and $B=8$.

Let's verify the addition with these values:

$57 + 25 = 82$

This matches the structure B2, where B=8 and the units digit is 2.

The question asks for the value of $A+B$.

$A + B = 7 + 8 = 15$

Comparing this result with the given options:

(a) 15

(b) 10

(c) 8

(d) 7

The value of A + B is 15, which corresponds to option (a).

The correct option is (a).

Let the two-digit number 'ab' be represented in its generalised form. Here, 'a' is the tens digit and 'b' is the units digit.

The generalised form of 'ab' is $10a + b$.

Let the two-digit number 'ba' be represented in its generalised form. Here, 'b' is the tens digit and 'a' is the units digit.

The generalised form of 'ba' is $10b + a$.

We need to find the difference between 'ab' and 'ba':

$ab - ba = (10a + b) - (10b + a)$

Remove the parentheses and rearrange the terms:

$ab - ba = 10a + b - 10b - a$

$ab - ba = (10a - a) + (b - 10b)$

$ab - ba = 9a - 9b$

Factor out the common factor, 9:

$ab - ba = 9(a - b)$

The difference $ab - ba$ is equal to $9$ multiplied by the difference of the digits $(a-b)$.

Since the result is $9(a - b)$, it is always a multiple of 9.

Also, since $a$ and $b$ are digits and $a > b$, $(a-b)$ is a positive integer. The result is also divisible by $(a-b)$.

However, the question asks what it is divisible by in general, regardless of the specific values of $a$ and $b$ (as long as $a>b$). The constant factor is 9.

Thus, the number $ab - ba$ is divisible by 9.

The number ab – ba where a and b are digits and a > b is divisible by 9.

We are given the expression in its generalised form:

$100a + 10c + 9$

To write this in the usual form of a number, we need to identify the digit corresponding to each place value.

The term $100a$ means that 'a' is the digit in the hundreds place ($100 \times a$).

The term $10c$ means that 'c' is the digit in the tens place ($10 \times c$).

The term $9$ (which is $9 \times 1$) means that '9' is the digit in the units place ($9 \times 1$).

Arranging the digits by place value from hundreds to units:

Hundreds place: a

Tens place: c

Units place: 9

When written in the usual form, the number is formed by arranging these digits in order:

The digit in the hundreds place is 'a'.

The digit in the tens place is 'c'.

The digit in the units place is '9'.

Thus, the usual form of the number represented by $100a + 10c + 9$ is ac9.

When written in usual form 100a + 10c + 9 is equal to ac9.

The problem is a multiplication puzzle where A and B represent single digits.

AB represents a two-digit number $10A + B$, where A is the tens digit and B is the units digit. Since AB is a two-digit number, $A \neq 0$.

9B represents a two-digit number $90 + B$, where 9 is the tens digit and B is the units digit. Since 9B is a two-digit number with 9 in the tens place, this is always true if B is a digit. However, if B=0, the number is 90. If B=0, $AB \times B = (10A+0) \times 0 = 0$. $9B = 90+0 = 90$. $0=90$ is false. So B cannot be 0.

The equation is $(10A + B) \times B = 90 + B$.

Let's analyse this using the multiplication in columns:

$\begin{array}{cc} & A & B \\ \times & & B \\ \hline & 9 & B \\ \hline \end{array}$

Consider the units column multiplication: $B \times B$. The units digit of the result is B.

We need to find single digits B (from 1 to 9, since $B \neq 0$) such that $B^2$ ends in B.

• If $B=1$, $1^2 = 1$. Ends in 1. Possible.
• If $B=2$, $2^2 = 4$. Ends in 4. Not 2.
• If $B=3$, $3^2 = 9$. Ends in 9. Not 3.
• If $B=4$, $4^2 = 16$. Ends in 6. Not 4.
• If $B=5$, $5^2 = 25$. Ends in 5. Possible.
• If $B=6$, $6^2 = 36$. Ends in 6. Possible.
• If $B=7$, $7^2 = 49$. Ends in 9. Not 7.
• If $B=8$, $8^2 = 64$. Ends in 4. Not 8.
• If $B=9$, $9^2 = 81$. Ends in 1. Not 9.

So, possible values for B are 1, 5, or 6.

Now consider the tens column. The result of $A \times B$ plus the carry-over from the units column ($c_u$) must result in a number whose units digit is 9 (the tens digit of 9B) and the hundreds digit is 0 (since 9B is a two-digit number).

Case 1: $B = 1$.

Units column: $1 \times 1 = 1$. Units digit is 1, carry-over $c_u = 0$.

Tens column: $A \times 1 + c_u$ must have a units digit of 9 and carry 0.

$A \times 1 + 0 = 9$

$A = 9$

Check: $A=9, B=1$. AB is 91. $91 \times 1 = 91$. 9B is 91. The equation $91 = 91$ is true. This solution works. (A=9, B=1).

Case 2: $B = 5$.

Units column: $5 \times 5 = 25$. Units digit is 5, carry-over $c_u = 2$.

Tens column: $A \times 5 + c_u$ must have a units digit of 9 and carry 0.

$A \times 5 + 2$ must end in 9.

$5A + 2 = 9 \Rightarrow 5A = 7$. No integer A.

$5A + 2 = 19 \Rightarrow 5A = 17$. No integer A.

In general, $5A$ must end in 7. Multiples of 5 end only in 0 or 5. So there is no integer digit A for $B=5$. This case does not work.

Case 3: $B = 6$.

Units column: $6 \times 6 = 36$. Units digit is 6, carry-over $c_u = 3$.

Tens column: $A \times 6 + c_u$ must have a units digit of 9 and carry 0.

$A \times 6 + 3$ must end in 9.

$6A + 3 = 9$

$6A = 6$

$A = 1$

Check the carry-over to the hundreds place: $\lfloor (A \times 6 + 3)/10 \rfloor = \lfloor (1 \times 6 + 3)/10 \rfloor = \lfloor 9/10 \rfloor = 0$. This is correct for a two-digit result 9B.

So, $A=1, B=6$ is a possible solution. Check: AB is 16. $16 \times 6 = 96$. 9B is 96. The equation $96=96$ is true. This solution also works. (A=1, B=6).

Both pairs (A=9, B=1) and (A=1, B=6) satisfy the given equation and constraints (A and B are non-zero digits, AB and 9B are two-digit numbers).

Since the question asks to fill in the blanks, it usually implies a unique answer is expected in this context. We will provide the first solution found.

Thus, A = 9, B = 1.

(Note: Another valid solution is A=1, B=6, but typically one specific answer is expected for fill-in-the-blank questions.)

Let the three-digit numbers formed by the digits a, b, and c be abc, cab, and bca.

The generalised form of the number abc is $100a + 10b + c$.

The generalised form of the number cab is $100c + 10a + b$.

The generalised form of the number bca is $100b + 10c + a$.

We need to find the sum of these three numbers:

Sum = $(100a + 10b + c) + (100c + 10a + b) + (100b + 10c + a)$

Combine the like terms:

Sum = $(100a + 10a + a) + (10b + b + 100b) + (c + 100c + 10c)$

Sum = $111a + 111b + 111c$

Factor out the common factor 111:

Sum = $111(a + b + c)$

Now, let's find the prime factors of 111.

$111 = 3 \times 37$

Substitute this back into the sum expression:

Sum = $3 \times 37 \times (a + b + c)$

Since the sum can be expressed as $37 \times [3 \times (a+b+c)]$, the sum is always a multiple of 37.

Therefore, the sum of the three numbers abc, cab, and bca is always divisible by 37.

The statement is True.

Let the two-digit number 'ab' be represented in its generalised form. Here, 'a' is the tens digit and 'b' is the units digit.

The generalised form of 'ab' is $10a + b$.

Let the number 'ba' (formed by reversing the digits) be represented in its generalised form. Here, 'b' is the tens digit and 'a' is the units digit.

The generalised form of 'ba' is $10b + a$.

We need to find the sum of 'ab' and 'ba':

$ab + ba = (10a + b) + (10b + a)$

Combine the like terms:

$ab + ba = 10a + a + b + 10b$

$ab + ba = 11a + 11b$

Factor out the common factor 11:

$ab + ba = 11(a + b)$

The sum $ab + ba$ is equal to 11 multiplied by the sum of the digits $(a+b)$.

For this sum to be divisible by 9, either 11 must be divisible by 9 (which is false) or the sum of the digits $(a+b)$ must be divisible by 9.

The sum of the digits $(a+b)$ is not always divisible by 9 for any two-digit number 'ab'. For example, if $a=1$ and $b=2$, then $ab=12$ and $ba=21$. Their sum is $12 + 21 = 33$. The sum of the digits $a+b = 1+2 = 3$. The expression $11(a+b) = 11(3) = 33$. 33 is not divisible by 9.

Since $ab + ba$ is not always divisible by 9 for all two-digit numbers ab, the statement is false.

The statement is False.

A number that is divisible by both 2 and 4 is a multiple of both 2 and 4.

The Least Common Multiple (LCM) of 2 and 4 is 4.

So, a number divisible by both 2 and 4 is a multiple of 4.

The statement claims that any multiple of 4 is also a multiple of 8.

Let's consider some multiples of 4:

• 4: Divisible by 2 ($4/2 = 2$), Divisible by 4 ($4/4 = 1$). Is it divisible by 8? $4/8 = 0.5$, No.
• 8: Divisible by 2 ($8/2 = 4$), Divisible by 4 ($8/4 = 2$). Is it divisible by 8? $8/8 = 1$, Yes.
• 12: Divisible by 2 ($12/2 = 6$), Divisible by 4 ($12/4 = 3$). Is it divisible by 8? $12/8 = 1.5$, No.
• 16: Divisible by 2 ($16/2 = 8$), Divisible by 4 ($16/4 = 4$). Is it divisible by 8? $16/8 = 2$, Yes.

From these examples, we can see that numbers like 4 and 12 are divisible by both 2 and 4, but they are not divisible by 8.

Since we found a counterexample (e.g., the number 4), the statement is not always true.

The statement is False.

The given three-digit number is 42x, where x is the digit in the units place.

The divisibility rule for 9 states that a number is divisible by 9 if the sum of its digits is divisible by 9.

The digits of the number 42x are 4, 2, and x.

The sum of the digits is $4 + 2 + x$.

Sum of digits $= 6 + x$.

For the number 42x to be divisible by 9, the sum of its digits, $6 + x$, must be a multiple of 9.

Since x is a digit, its value can be any integer from 0 to 9.

Possible values for $6+x$ are:

If $x=0$, $6+x = 6+0 = 6$

If $x=1$, $6+x = 6+1 = 7$

If $x=2$, $6+x = 6+2 = 8$

If $x=3$, $6+x = 6+3 = 9$

If $x=4$, $6+x = 6+4 = 10$

If $x=5$, $6+x = 6+5 = 11$

If $x=6$, $6+x = 6+6 = 12$

If $x=7$, $6+x = 6+7 = 13$

If $x=8$, $6+x = 6+8 = 14$

If $x=9$, $6+x = 6+9 = 15$

The multiples of 9 are 9, 18, 27, ...

We need to find the value of $6+x$ that is a multiple of 9 and falls within the possible range [6, 15].

The only multiple of 9 in this range is 9.

Set the sum of digits equal to 9:

$6 + x = 9$

Subtract 6 from both sides:

$x = 9 - 6$

$x = 3$

So, the value of x is 3.

The number is 423, and the sum of its digits is $4+2+3 = 9$, which is divisible by 9.

The value of x is 3.

The given problem is a vertical addition:

$\begin{array}{cccc} & 4 & 1 & A \\ + & & B & 4 \\ \hline & 5 & 1 & 2 \\ \hline \end{array}$

Here, A and B represent single digits (0-9).

We solve this column by column, starting from the units place.

Units Column:

The sum of the digits in the units column is A + 4. The units digit of the result is 2.

$A + 4 = \text{something ending in 2}$

For A to be a single digit (0-9), the possible values for A + 4 ending in 2 are 12.

$A + 4 = 12$

$A = 12 - 4$

$A = 8$

So, the value of A is 8. There is a carry-over of 1 to the tens column.

Tens Column:

The sum of the digits in the tens column, plus the carry-over from the units column, is $1 + 1 + B$. The units digit of the result is 1.

$(1 \text{ (carry)} + 1 + B) = \text{something ending in 1}$

$2 + B = \text{something ending in 1}$

For B to be a single digit (0-9), the possible values for $2 + B$ ending in 1 are 11.

$2 + B = 11$

$B = 11 - 2$

$B = 9$

So, the value of B is 9. There is a carry-over of 1 to the hundreds column.

Hundreds Column:

The sum of the digits in the hundreds column, plus the carry-over from the tens column, is $1 + 4$. The result is 5.

$1 \text{ (carry)} + 4 = 5$

This matches the hundreds digit in the result (512).

Thus, the values we found for A and B are correct.

The value of A is 8 and the value of B is 9.

The value of A is 8.

The value of B is 9.

We are given two conditions about the number x based on division and remainders.

Condition 1: The division x ÷ 5 leaves a remainder 4.

According to the division algorithm, this can be written as:

$x = 5q_1 + 4$, where $q_1$ is the quotient.

This means that when x is divided by 5, the remainder is 4. Numbers that satisfy this condition are of the form $5q_1 + 4$.

Let's list some such numbers for $q_1 = 0, 1, 2, 3, ...$:

• $5(0) + 4 = 4$
• $5(1) + 4 = 9$
• $5(2) + 4 = 14$
• $5(3) + 4 = 19$
• $5(4) + 4 = 24$
• $5(5) + 4 = 29$
• ... and so on.

Observe the units digits of these numbers: 4, 9, 4, 9, 4, 9, ...

A number that leaves a remainder of 4 when divided by 5 must have a units digit of either 4 or 9.

Condition 2: The division x ÷ 2 leaves a remainder 1.

According to the division algorithm, this can be written as:

$x = 2q_2 + 1$, where $q_2$ is the quotient.

This means that when x is divided by 2, the remainder is 1. Numbers that satisfy this condition are of the form $2q_2 + 1$.

This is the definition of an odd number.

An odd number always has a units digit that is odd (1, 3, 5, 7, or 9).

Now, we need to find the units digit of x that satisfies both conditions.

From Condition 1, the units digit must be 4 or 9.

From Condition 2, the units digit must be odd (1, 3, 5, 7, 9).

We look for a digit that is in both lists:

List from Condition 1: {4, 9}

List from Condition 2: {1, 3, 5, 7, 9}

The digit that is common to both lists is 9.

Therefore, the ones digit (units digit) of x must be 9.

The ones digit of x is 9.

The given number is the four-digit number 756x, where x is the digit in the units place.

The divisibility rule for 11 states that a number is divisible by 11 if the difference between the sum of the digits at odd places (from the right) and the sum of the digits at even places (from the right) is either 0 or a multiple of 11.

Let's identify the digits at odd and even places starting from the rightmost digit (units place):

Units place (1st place - odd): x

Tens place (2nd place - even): 6

Hundreds place (3rd place - odd): 5

Thousands place (4th place - even): 7

Sum of digits at odd places = (Digit at 1st place) + (Digit at 3rd place)

Sum of digits at odd places = $x + 5$

Sum of digits at even places = (Digit at 2nd place) + (Digit at 4th place)

Sum of digits at even places = $6 + 7 = 13$

Now, find the difference between these two sums:

Difference = (Sum of digits at odd places) - (Sum of digits at even places)

Difference = $(x + 5) - 13$

Difference = $x - 8$

According to the divisibility rule for 11, this difference ($x - 8$) must be 0 or a multiple of 11.

Since x is a single digit (0-9), the possible values for the difference $x - 8$ are:

• If $x=0$, difference = $0 - 8 = -8$
• If $x=1$, difference = $1 - 8 = -7$
• ...
• If $x=8$, difference = $8 - 8 = 0$
• If $x=9$, difference = $9 - 8 = 1$

The possible values for $x-8$ range from $0-8 = -8$ to $9-8 = 1$.

We need to find a value for $x-8$ that is 0 or a multiple of 11 within the range [-8, 1].

The only value in this range that is 0 or a multiple of 11 is 0.

Set the difference equal to 0:

$x - 8 = 0$

Add 8 to both sides:

$x = 8$

So, the value of x is 8.

Let's check the number 7568. Sum of digits at odd places ($8+5=13$). Sum of digits at even places ($6+7=13$). Difference ($13-13=0$). Since the difference is 0, the number 7568 is divisible by 11.

The value of x is 8.

Let the four-digit number be represented by abdc, where:

'a' is the digit in the thousands place.

'b' is the digit in the hundreds place.

'd' is the digit in the tens place.

'c' is the digit in the units place.

To write the generalised form of the number abdc, we multiply each digit by its corresponding place value and sum the results.

Value from thousands place = $a \times 1000$

Value from hundreds place = $b \times 100$

Value from tens place = $d \times 10$

Value from units place = $c \times 1$

The generalised form of the number abdc is the sum of these values:

$a \times 1000 + b \times 100 + d \times 10 + c \times 1$

This simplifies to:

$1000a + 100b + 10d + c$

Now, let's compare this generalised form with the given options:

(a) $1000 a + 100 b + 10 c + d$ (Incorrect, tens digit is d, units digit is c)

(b) $1000 a + 100 c + 10 b + d$ (Incorrect, hundreds digit is b, tens digit is d, units digit is c)

(c) $1000 a + 100 b + 10 d + c$ (This matches our derived form)

(d) $a \times b \times c \times d$ (Incorrect, this is the product of the digits)

Therefore, the correct generalised form of the four-digit number abdc is $1000a + 100b + 10d + c$.

The correct option is (c).

Let the two-digit number be represented by xy, where:

'x' is the digit in the tens place.

'y' is the digit in the units place.

To write the generalised form of the number xy, we multiply each digit by its corresponding place value and sum the results.

Value from tens place = $x \times 10$

Value from units place = $y \times 1$

The generalised form of the number xy is the sum of these values:

$x \times 10 + y \times 1$

This simplifies to:

$10x + y$

Now, let's compare this generalised form with the given options:

(a) $x + y$ (This is the sum of the digits)

(b) $10x + y$ (This matches our derived form)

(c) $10x - y$ (Incorrect)

(d) $10y + x$ (This would be the generalised form of the number 'yx')

Therefore, the correct generalised form of a two-digit number xy is $10x + y$.

The correct option is (b).

We are given the expression in its generalised form:

$1000a + 10b + c$

To write this in the usual form of a number, we need to identify the digit corresponding to each place value.

The term $1000a$ means that 'a' is the digit in the thousands place ($1000 \times a$).

The term $10b$ means that 'b' is the digit in the tens place ($10 \times b$).

The term $c$ (which is $1 \times c$) means that 'c' is the digit in the units place ($1 \times c$).

Notice that there is no term with a coefficient of 100. This means the digit in the hundreds place is 0 ($0 \times 100$).

Arranging the digits by place value from thousands to units:

Thousands place: a

Hundreds place: 0

Tens place: b

Units place: c

When written in the usual form, the number is formed by arranging these digits in order:

The digit in the thousands place is 'a'.

The digit in the hundreds place is '0'.

The digit in the tens place is 'b'.

The digit in the units place is 'c'.

Thus, the usual form of the number represented by $1000a + 10b + c$ is a0bc.

Let's check the options:

(a) abc (Implies $100a + 10b + c$)

(b) abco (Uses 'o' which is not a digit representation)

(c) aobc (Implies a in thousands, 0 in hundreds, b in tens, c in units $\rightarrow 1000a + 100 \times 0 + 10b + c = 1000a + 10b + c$)

(d) aboc (Implies a in thousands, b in hundreds, 0 in tens, c in units $\rightarrow 1000a + 100b + 10 \times 0 + c = 1000a + 100b + c$)

The expression $1000a + 10b + c$ matches the usual form a0bc, represented by option (c).

The correct option is (c).

Let the three-digit number abc be represented in its generalised form. Here, 'a' is the hundreds digit, 'b' is the tens digit, and 'c' is the units digit.

The generalised form of 'abc' is $100a + 10b + c$.

Let the number formed by reversing the digits, cba, be represented in its generalised form. Here, 'c' is the hundreds digit, 'b' is the tens digit, and 'a' is the units digit.

The generalised form of 'cba' is $100c + 10b + a$.

We need to find the difference between 'abc' and 'cba':

$abc - cba = (100a + 10b + c) - (100c + 10b + a)$

Remove the parentheses and rearrange the terms:

$abc - cba = 100a + 10b + c - 100c - 10b - a$

$abc - cba = (100a - a) + (10b - 10b) + (c - 100c)$

$abc - cba = 99a + 0 - 99c$

$abc - cba = 99a - 99c$

Factor out the common factor, 99:

$abc - cba = 99(a - c)$

The difference $abc - cba$ is equal to 99 multiplied by the difference of the first and last digits $(a-c)$.

We know that $99 = 9 \times 11$ and $99 = 3 \times 33$.

So, $abc - cba = (9 \times 11)(a - c) = 9 \times [11(a - c)]$ and $abc - cba = (33 \times 3)(a - c) = 33 \times [3(a - c)]$.

This shows that $abc - cba$ is always divisible by 9, 11, and 33 for any digits a, b, and c.

Now let's check divisibility by 18.

A number is divisible by 18 if it is divisible by both 2 and 9.

We already know that $99(a-c)$ is divisible by 9.

For $99(a-c)$ to be divisible by 2, the expression must be an even number.

Since 99 is an odd number, the product $99(a-c)$ is even if and only if $(a-c)$ is an even number.

The difference $(a-c)$ is even if $a$ and $c$ are both even or both odd.

However, if one digit is even and the other is odd, the difference $(a-c)$ will be odd. In this case, $99(a-c)$ will be an odd number, which is not divisible by 2, and therefore not divisible by 18.

Let's consider an example where $(a-c)$ is odd. Let $a=2$ and $c=1$. abc can be 201, 211, ..., 291. cba would be 102, 112, ..., 192 (assuming c=1). Let's take abc=251 and cba=152. $abc - cba = 251 - 152 = 99$. Here $a-c = 2-1 = 1$. $99(a-c) = 99(1) = 99$.

Is 99 divisible by 18? No, $99 \div 18 = 5.5$.

Since $abc - cba = 99(a-c)$ is not always divisible by 18 (it requires $a-c$ to be even), the statement that $abc - cba$ is *not* divisible by 18 is true for cases where $a-c$ is odd.

The question asks which option it is *not* divisible by (meaning not *always* divisible by).

It is always divisible by 9, 11, and 33.

It is not always divisible by 18.

The number abc – cba is not (always) divisible by 18.

The correct option is (c).

Let the three digits of the number xyz be x, y, and z. The phrasing "the sum of all the numbers formed by the digits x, y and z of the number xyz" typically refers to the sum of the numbers formed by cyclically permuting the digits, which are xyz, yzx, and zxy (similar to Example 8).

The generalised form of the number xyz is $100x + 10y + z$.

The generalised form of the number yzx is $100y + 10z + x$.

The generalised form of the number zxy is $100z + 10x + y$.

The sum of these three numbers is:

Sum = $(100x + 10y + z) + (100y + 10z + x) + (100z + 10x + y)$

Combine the like terms:

Sum = $(100x + x + 10x) + (10y + 100y + y) + (z + 10z + 100z)$

Sum = $111x + 111y + 111z$

Factor out the common factor 111:

Sum = $111(x + y + z)$

Now, let's find the prime factors of 111.

$\begin{array}{c|cc} 3 & 111 \\ \hline 37 & 37 \\ \hline & 1 \end{array}$

So, $111 = 3 \times 37$.

Substitute this back into the sum expression:

Sum = $3 \times 37 \times (x + y + z)$

Since the sum can be expressed as $37 \times [3 \times (x+y+z)]$, the sum is always a multiple of 37.

Let's check the divisibility by the other options:

• Is $111(x+y+z)$ always divisible by 11? No, 111 is not divisible by 11.
• Is $111(x+y+z)$ always divisible by 33? No, 111 is divisible by 3 but not by 11, so it's not divisible by 33.
• Is $111(x+y+z)$ always divisible by 74? No, 111 is an odd number, so it's not divisible by 2, and thus not by 74.

Only divisibility by 37 is guaranteed for any single digits x, y, and z that form a three-digit number (meaning x, y, z are not all zero, and typically x is non-zero for xyz to be a 3-digit number, although the sum logic works even if they are zero). Assuming x,y,z are digits, their sum is an integer.

The sum of the numbers formed by the digits x, y and z of the number xyz (interpreted as cyclic permutations) is always divisible by 37.

The correct option is (c).

The four-digit number is aabb. This means the thousands digit is 'a', the hundreds digit is 'a', the tens digit is 'b', and the units digit is 'b'.

Since it's a four-digit number, the thousands digit 'a' cannot be 0. So, $a \in \{1, 2, ..., 9\}$. The digit 'b' can be any digit from 0 to 9, so $b \in \{0, 1, ..., 9\}$.

The number aabb can be written in generalised form as:

$1000a + 100a + 10b + b$

$1100a + 11b$

The number $1100a + 11b$ is divisible by 55.

Divisibility by 55 means the number must be divisible by both 5 and 11, since $55 = 5 \times 11$ and 5 and 11 are coprime.

Condition 1: Divisibility by 5

A number is divisible by 5 if its units digit is 0 or 5.

The units digit of the number aabb is 'b'.

Therefore, for aabb to be divisible by 5, the value of b must be 0 or 5.

So, possible values for b are 0 and 5.

Condition 2: Divisibility by 11

A number is divisible by 11 if the difference between the sum of the digits at odd places (from the right) and the sum of the digits at even places (from the right) is 0 or a multiple of 11.

The digits of aabb from right to left are b, b, a, a.

Digits at odd places (1st, 3rd): b, a

Digits at even places (2nd, 4th): b, a

Sum of digits at odd places = $b + a$

Sum of digits at even places = $b + a$

Difference = (Sum of digits at odd places) - (Sum of digits at even places)

Difference = $(b + a) - (b + a) = 0$

Since the difference is 0, and 0 is a multiple of 11, the number aabb is always divisible by 11, regardless of the values of a and b (as long as they form a valid four-digit number, i.e., $a \neq 0$).

For the number aabb to be divisible by 55, it must satisfy both conditions:

1. The units digit 'b' must be 0 or 5 (for divisibility by 5).

2. The number is always divisible by 11 (as long as $a \neq 0$).

The possible values of b that satisfy the required condition (divisibility by 5) are 0 and 5. The divisibility by 11 does not impose any further restrictions on b (other than it being a digit).

The possible values of b are 0 and 5.

Comparing with the options:

(a) 0 and 2 (Incorrect, 2 is not possible for divisibility by 5)

(b) 2 and 5 (Incorrect, 2 is not possible for divisibility by 5)

(c) 0 and 5 (Correct)

(d) 7 (Incorrect)

The correct option is (c).

Let the three-digit number abc be represented in its generalised form. Here, 'a' is the hundreds digit, 'b' is the tens digit, and 'c' is the units digit. Since it's a three-digit number, $a \neq 0$.

The generalised form of 'abc' is $100a + 10b + c$.

The number bca is formed by permuting the digits. Its generalised form is $100b + 10c + a$.

The number cab is also formed by permuting the digits. Its generalised form is $100c + 10a + b$.

We need to find the sum of these three numbers:

Sum = $(100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b)$

Combine the like terms:

Sum = $(100a + a + 10a) + (10b + 100b + b) + (c + 10c + 100c)$

Sum = $111a + 111b + 111c$

Factor out the common factor 111:

Sum = $111(a + b + c)$

Now, let's find the prime factors of 111.

$\begin{array}{c|cc} 3 & 111 \\ \hline 37 & 37 \\ \hline & 1 \end{array}$

So, $111 = 3 \times 37$.

Substitute this back into the sum expression:

Sum = $3 \times 37 \times (a + b + c)$

Now, let's check the divisibility of the sum by the given options:

(a) $a + b + c$: The sum is $111 \times (a + b + c)$. Since $a+b+c$ is a factor, the sum is divisible by $a+b+c$ (assuming $a+b+c \neq 0$. As $a \neq 0$, $a+b+c \geq 1$, so this is always true).

(b) 3: The sum is $3 \times 37 \times (a + b + c)$. Since 3 is a factor, the sum is always divisible by 3.

(c) 37: The sum is $3 \times 37 \times (a + b + c)$. Since 37 is a factor, the sum is always divisible by 37.

(d) 9: The sum is $111(a + b + c)$. For this to be divisible by 9, $111(a + b + c)$ must be a multiple of 9.

We know $111 = 3 \times 37$. So, the sum is $3 \times 37 \times (a + b + c)$.

For this to be divisible by $9 = 3 \times 3$, the remaining factor $37 \times (a + b + c)$ must be divisible by 3.

Since 37 is not divisible by 3, $(a + b + c)$ must be divisible by 3.

The sum of the digits $(a+b+c)$ is not always divisible by 3 for any three-digit number abc.

For example, let $a=1, b=0, c=1$. The number is 101. The sum of digits $a+b+c = 1+0+1 = 2$. The sum of numbers is $111(1+0+1) = 111 \times 2 = 222$. Is 222 divisible by 9? The sum of digits of 222 is $2+2+2=6$, which is not divisible by 9. So 222 is not divisible by 9.

This shows that the sum is not always divisible by 9.

The sum abc + bca + cab is not (always) divisible by 9.

The correct option is (d).

The four-digit number is 4ab5. This means the thousands digit is 4, the hundreds digit is 'a', the tens digit is 'b', and the units digit is 5.

'a' and 'b' are single digits (0-9).

The number 4ab5 is divisible by 55. This means the number must be divisible by both 5 and 11, since $55 = 5 \times 11$ and 5 and 11 are coprime.

Condition 1: Divisibility by 5

A number is divisible by 5 if its units digit is 0 or 5.

The units digit of 4ab5 is 5. Since the units digit is 5, the number is always divisible by 5, regardless of the values of a and b.

Condition 2: Divisibility by 11

A number is divisible by 11 if the difference between the sum of the digits at odd places (from the right) and the sum of the digits at even places (from the right) is 0 or a multiple of 11.

The digits of 4ab5 from right to left are 5, b, a, 4.

Digits at odd places (1st, 3rd): 5, a

Digits at even places (2nd, 4th): b, 4

Sum of digits at odd places = $5 + a$

Sum of digits at even places = $b + 4$

Difference = (Sum of digits at odd places) - (Sum of digits at even places)

Difference = $(5 + a) - (b + 4)$

Difference = $5 + a - b - 4$

Difference = $1 + a - b$

According to the divisibility rule for 11, this difference ($1 + a - b$) must be 0 or a multiple of 11.

Since a and b are single digits (0-9), the minimum possible value for $a-b$ is $0-9 = -9$. The maximum possible value for $a-b$ is $9-0 = 9$.

So, the possible values for the difference $1+a-b$ range from $1 + (-9) = -8$ to $1 + 9 = 10$.

We need to find a value for $1+a-b$ that is 0 or a multiple of 11 within the range [-8, 10].

The only value in this range that is 0 or a multiple of 11 is 0.

Set the difference equal to 0:

$1 + a - b = 0$

$a - b = -1$

Multiply by -1:

$b - a = 1$

So, the value of $b - a$ must be 1.

Let's check this. If $b-a = 1$, then $1+a-b = 1-(b-a) = 1-1 = 0$, which confirms divisibility by 11.

Possible pairs of (a, b) where $b-a=1$ are (0,1), (1,2), (2,3), (3,4), (4,5), (5,6), (6,7), (7,8), (8,9).

For example, if a=1, b=2, the number is 4125. Sum odd places ($5+1=6$). Sum even places ($2+4=6$). Difference $6-6=0$. Divisible by 11. Units digit is 5. Divisible by 5. So 4125 is divisible by 55. $b-a = 2-1 = 1$.

The value of b – a is 1.

Comparing with the options:

(a) 0 (Incorrect, $b-a=0 \implies b=a$, difference is 1)

(b) 1 (Correct)

(c) 4 (Incorrect, difference is $1+a-b$, if $b-a=4$, difference is $1-4 = -3$)

(d) 5 (Incorrect, difference is $1+a-b$, if $b-a=5$, difference is $1-5 = -4$)

The correct option is (b).

Let the three-digit number abc be represented in its generalised form:

$abc = 100a + 10b + c$

We are interested in the expression $abc - a - b - c$.

Substitute the generalised form of abc:

$abc - a - b - c = (100a + 10b + c) - a - b - c$

Combine the like terms:

$abc - a - b - c = 100a - a + 10b - b + c - c$

$abc - a - b - c = 99a + 9b + 0$

$abc - a - b - c = 99a + 9b$

Factor out the common factor, 9:

$abc - a - b - c = 9(11a + b)$

Since the expression can be written as 9 multiplied by some integer $(11a+b)$, the number $abc - a - b - c$ is always divisible by 9.

Let's check the other options:

(b) 90: For the number to be divisible by 90, it must be divisible by both 9 and 10. We know it's divisible by 9. For it to be divisible by 10, the units digit of $99a + 9b$ must be 0. The units digit of $99a$ is the units digit of $9a$. The units digit of $9b$ is the units digit of $9b$. The units digit of the sum $99a+9b$ depends on the values of a and b. For example, if a=1 and b=1, $99(1) + 9(1) = 99+9 = 108$. 108 is not divisible by 10. So it's not always divisible by 90.

(c) 10: As seen above, the units digit of $99a+9b$ is not always 0. So it's not always divisible by 10.

(d) 11: For the number to be divisible by 11, $9(11a + b)$ must be a multiple of 11. Since 9 is not divisible by 11, $(11a + b)$ must be divisible by 11. $11a$ is always divisible by 11. So, for $11a+b$ to be divisible by 11, b must be divisible by 11. Since b is a single digit (0-9), the only single digit divisible by 11 is 0. Thus, it is divisible by 11 only if $b=0$. It is not always divisible by 11.

The number abc – a – b – c is always divisible by 9.

The correct option is (a).

Let the three-digit number be abc. The six-digit number formed by repeating this three-digit number is abcabc.

We can write the number abcabc in terms of place values. Let the digits be a, b, c from left to right for the first three digits.

The generalised form of the six-digit number abcabc is:

$100000a + 10000b + 1000c + 100a + 10b + c$

Group the terms with the same digit:

$(100000a + 100a) + (10000b + 10b) + (1000c + c)$

Simplify the terms:

$100100a + 10010b + 1001c$

Factor out the common factor 1001:

$1001(100a + 10b + c)$

The expression inside the parentheses, $100a + 10b + c$, is the generalised form of the original three-digit number abc.

So, the six-digit number abcabc is equal to $1001 \times (abc)$.

Since the six-digit number can be expressed as $1001 \times (\text{the original three-digit number})$, it is always divisible by 1001.

To check the other options, let's find the prime factors of 1001:

$\begin{array}{c|cc} 7 & 1001 \\ \hline 11 & 143 \\ \hline 13 & 13 \\ \hline & 1 \end{array}$

So, $1001 = 7 \times 11 \times 13$.

Since the number abcabc $= 1001 \times (abc) = 7 \times 11 \times 13 \times (abc)$, it is also always divisible by 7, 11, and 13.

Looking at the options:

(a) 7 only - Incorrect, it is also divisible by 11, 13, 1001.

(b) 11 only - Incorrect, it is also divisible by 7, 13, 1001.

(c) 13 only - Incorrect, it is also divisible by 7, 11, 1001.

(d) 1001 - Correct, it is always divisible by 1001.

Any number of this form is divisible by 1001.

The correct option is (d).

The question states that the sum of the digits of a number is divisible by three.

We need to determine which of the given options the number is always divisible by based on this condition.

Let's recall the divisibility rules for the given numbers:

• Divisibility by 2: A number is divisible by 2 if its units digit is 0, 2, 4, 6, or 8 (i.e., the number is even).
• Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3.
• Divisibility by 6: A number is divisible by 6 if it is divisible by both 2 and 3.
• Divisibility by 9: A number is divisible by 9 if the sum of its digits is divisible by 9.

The given condition is "the sum of digits of a number is divisible by three".

According to the divisibility rule for 3, this condition means exactly that the number is divisible by 3.

Let's check if this condition guarantees divisibility by the other options:

• Divisibility by 2: If the sum of digits is divisible by 3, is the number always even? No. Consider the number 15. The sum of its digits is $1 + 5 = 6$, which is divisible by 3. However, 15 is an odd number, so it is not divisible by 2.
• Divisibility by 6: For a number to be divisible by 6, it must be divisible by both 2 and 3. The given condition guarantees divisibility by 3. However, as shown above, it does not guarantee divisibility by 2. Therefore, it does not guarantee divisibility by 6. (Example: 15 is divisible by 3 but not by 6).
• Divisibility by 9: For a number to be divisible by 9, the sum of its digits must be divisible by 9. The given condition only guarantees that the sum of digits is a multiple of 3. It might be a multiple of 3 that is not a multiple of 9 (e.g., 3, 6, 12). Consider the number 12. The sum of its digits is $1 + 2 = 3$, which is divisible by 3. However, 12 is not divisible by 9 ($12 \div 9 \approx 1.33$).

The only option that is always true when the sum of digits of a number is divisible by three is that the number is divisible by 3.

The correct option is (b).

The three-digit number is xyz, where x, y, and z are its digits. This means x, y, and z are integers from 0 to 9. Since it's a three-digit number, $x \neq 0$.

We are given two conditions:

1. The sum of the digits is 6: $x + y + z = 6$.

2. The units digit z is an odd digit: $z \in \{1, 3, 5, 7, 9\}$.

Let's analyse the properties of the number xyz based on these conditions.

Divisibility by 3:

The sum of the digits of the number xyz is $x + y + z$. We are given that $x + y + z = 6$.

Since the sum of the digits (6) is divisible by 3, the number xyz is always divisible by 3, based on the divisibility rule for 3.

So, xyz is a multiple of 3. Now we need to determine if it's an odd or even multiple, and check the other options.

Odd or Even:

A number is odd or even based on its units digit. The units digit of xyz is z.

We are given that z is an odd digit: $z \in \{1, 3, 5, 7, 9\}$.

Therefore, the units digit of xyz is always odd.

A number with an odd units digit is an odd number.

So, the number xyz is an odd number.

Combining the findings:

1. xyz is a multiple of 3.

2. xyz is an odd number.

Therefore, the three-digit number xyz is an odd multiple of 3.

Let's check the options:

(a) an odd multiple of 3 (Matches our finding)

(b) odd multiple of 6 (An odd multiple of 6 would be odd $\times$ even = even. An odd number cannot be an odd multiple of 6. It cannot be a multiple of 6 at all, since it's odd and 6 is even). Incorrect.

(c) even multiple of 3 (We found it is an odd number, so it cannot be an even multiple of 3 unless the multiple itself is 0, which is not the case for a three-digit number). Incorrect.

(d) even multiple of 9 (An even multiple of 9 is even. We found the number is odd). Incorrect.

(Also, divisibility by 9 requires the sum of digits to be divisible by 9. Here, the sum of digits is 6, which is not divisible by 9. So the number is not divisible by 9, unless the number itself is 0, which is not a 3-digit number).

Let's find some examples of xyz satisfying the conditions:

• If $z=1$, $x+y+1=6 \implies x+y=5$. Possible (x,y) pairs with $x \neq 0$: (1,4), (2,3), (3,2), (4,1), (5,0). Numbers are 141, 231, 321, 411, 501. All are odd and their sum of digits is 6. All are divisible by 3. $141 = 3 \times 47$ (47 is odd). $231 = 3 \times 77$ (77 is odd).
• If $z=3$, $x+y+3=6 \implies x+y=3$. Possible (x,y) pairs with $x \neq 0$: (1,2), (2,1), (3,0). Numbers are 123, 213, 303. All are odd and their sum of digits is 6. All are divisible by 3. $123 = 3 \times 41$ (41 is odd). $213 = 3 \times 71$ (71 is odd).
• If $z=5$, $x+y+5=6 \implies x+y=1$. Possible (x,y) pairs with $x \neq 0$: (1,0). Number is 105. Sum of digits 6. Odd. $105 = 3 \times 35$ (35 is odd).
• If $z=7$, $x+y+7=6 \implies x+y = -1$. No non-negative digits possible.
• If $z=9$, $x+y+9=6 \implies x+y = -3$. No non-negative digits possible.

All valid numbers found are odd multiples of 3.

The correct option is (a).

The given problem is a vertical addition:

$\begin{array}{cc} & 5 & A \\ + & B & 3 \\ \hline & 6 & 5 \\ \hline \end{array}$

Here, A and B represent single digits (0-9).

We solve this column by column, starting from the units place.

Units Column:

The sum of the digits in the units column is A + 3. The units digit of the result is 5.

$A + 3 = \text{something ending in 5}$

For A to be a single digit (0-9), the possible values for A + 3 ending in 5 are 5 or 15.

Case 1: $A + 3 = 5$. This gives $A = 5 - 3 = 2$. Carry-over to the tens column is 0.

Case 2: $A + 3 = 15$. This gives $A = 15 - 3 = 12$. This is not a single digit, so this case is not possible.

Thus, the value of A must be 2. There is no carry-over to the tens column from the units column.

Tens Column:

The sum of the digits in the tens column, plus the carry-over from the units column, is $5 + B + 0$. The result is 6.

$5 + B + 0 = 6$

$5 + B = 6$

$B = 6 - 5$

$B = 1$

So, the value of B is 1.

Let's verify the addition with the found values $A=2$ and $B=1$:

The first number is 5A, which is 52.

The second number is B3, which is 13.

The sum is $52 + 13$.

$\begin{array}{cc} & 5 & 2 \\ + & 1 & 3 \\ \hline & 6 & 5 \\ \hline \end{array}$

The result is 65, which matches the given result.

Thus, the values of A and B are 2 and 1 respectively.

A = 2, B = 1.

Comparing with the options:

(a) A = 2, B = 3 (Incorrect, B should be 1)

(b) A = 3, B = 2 (Incorrect, A should be 2, B should be 1)

(c) A = 2, B = 1 (Correct)

(d) A = 1, B = 2 (Incorrect, A should be 2, B should be 1)

The correct option is (c).

The given problem is a vertical addition:

$\begin{array}{cc} & A & 3 \\ + & 8 & B \\ \hline 1 & 5 & 0 \\ \hline \end{array}$

Here, A and B represent single digits (0-9).

A is the tens digit of a two-digit number, so $A \neq 0$. $A \in \{1, 2, ..., 9\}$.

B is a single digit, so $B \in \{0, 1, ..., 9\}$.

We solve this column by column, starting from the units place.

Units Column:

The sum of the digits in the units column is 3 + B. The units digit of the result is 0.

$3 + B = \text{something ending in 0}$

For B to be a single digit (0-9), the possible values for $3 + B$ ending in 0 are 10.

$3 + B = 10$

$B = 10 - 3$

$B = 7$

So, the value of B is 7. There is a carry-over of 1 to the tens column.

Tens Column:

The sum of the digits in the tens column, plus the carry-over from the units column, is $A + 8 + 1$. The result is 15 (since 15 is written below the line). The units digit of this sum is 5, and there is a carry-over of 1 to the hundreds column.

$A + 8 + 1 = \text{something ending in 5, with a carry-over}$

$A + 9 = \text{something ending in 5, with a carry-over}$

For A to be a single digit (1-9), the possible values for $A + 9$ ending in 5 are 15.

$A + 9 = 15$

$A = 15 - 9$

$A = 6$

So, the value of A is 6. The sum is 15, which means the units digit is 5 and there is a carry-over of 1 to the hundreds place. This carry-over of 1 is what appears in the hundreds place of the result (150).

Let's verify the addition with the found values $A=6$ and $B=7$:

The first number is A3, which is 63.

The second number is 8B, which is 87.

The sum is $63 + 87$.

$\begin{array}{cc} & 6 & 3 \\ + & 8 & 7 \\ \hline 1 & 5 & 0 \\ \hline \end{array}$

Units column: $3+7 = 10$. Write 0, carry over 1.

Tens column: $6+8+1 (\text{carry}) = 15$. Write 5, carry over 1.

Hundreds column: $1 (\text{carry}) = 1$. Write 1.

The result is 150, which matches the given result.

Thus, the values of A and B are 6 and 7 respectively.

The question asks for the value of A + B.

$A + B = 6 + 7 = 13$

Comparing with the options:

(a) 13 (Correct)

(b) 12 (Incorrect)

(c) 17 (Incorrect)

(d) 15 (Incorrect)

The correct option is (a).

The given problem is a multiplication puzzle:

$\begin{array}{cc} & 5 & A \\ \times & & A \\ \hline 3 & 9 & 9 \\ \hline \end{array}$

Here, 5A represents the two-digit number $50 + A$, where A is a single digit (0-9). The number must be a two-digit number starting with 5, so A can be any digit from 0 to 9.

The equation is $(50 + A) \times A = 399$.

We can also analyse this using the multiplication in columns:

Consider the units column multiplication: $A \times A$. The units digit of the result is 9.

We need to find single digits A (0-9) such that $A^2$ ends in 9.

• If $A=0$, $0^2 = 0$. Ends in 0. Not 9.
• If $A=1$, $1^2 = 1$. Ends in 1. Not 9.
• If $A=2$, $2^2 = 4$. Ends in 4. Not 9.
• If $A=3$, $3^2 = 9$. Ends in 9. Possible.
• If $A=4$, $4^2 = 16$. Ends in 6. Not 9.
• If $A=5$, $5^2 = 25$. Ends in 5. Not 9.
• If $A=6$, $6^2 = 36$. Ends in 6. Not 9.
• If $A=7$, $7^2 = 49$. Ends in 9. Possible.
• If $A=8$, $8^2 = 64$. Ends in 4. Not 9.
• If $A=9$, $9^2 = 81$. Ends in 1. Not 9.

So, possible values for A are 3 or 7.

Let's test these possible values for A using the original equation $(50 + A) \times A = 399$ or by performing the multiplication:

Case 1: Assume $A = 3$.

The number 5A is 53. We need to check $53 \times 3$.

$\begin{array}{cc} & 5 & 3 \\ \times & & 3 \\ \hline 1 & 5 & 9 \\ \hline \end{array}$

$53 \times 3 = 159$. This is not equal to 399.

Case 2: Assume $A = 7$.

The number 5A is 57. We need to check $57 \times 7$.

$\begin{array}{cc} & 5 & 7 \\ \times & & 7 \\ \hline 3 & 9 & 9 \\ \hline \end{array}$

Units column: $7 \times 7 = 49$. Write 9, carry over 4.

Tens column: $5 \times 7 + 4 (\text{carry}) = 35 + 4 = 39$. Write 39.

The result is 399. This matches the given result.

Thus, the value of A is 7.

Comparing with the options:

(a) 3 (Incorrect)

(b) 6 (Incorrect)

(c) 7 (Correct)

(d) 9 (Incorrect)

The correct option is (c).

The multiplication is $\begin{array}{cc} & 6 & A \\ \times & & B \\ \hline A & 8 & B \\ \hline \end{array}$

Here, A and B are single digits (0-9). Since 6A is a two-digit number, $A$ can be 0-9. Since A8B is a three-digit number, A $\neq$ 0 and B $\neq$ 0 (as B is the multiplier resulting in a 3-digit number).

In the units column, $A \times B$ has a units digit of B.

This implies $A \times B - B$ is a multiple of 10, so $B(A - 1)$ is a multiple of 10.

Possible cases for $B(A-1)$ being a multiple of 10 with $A, B \in \{1..9\}$:

• $A-1 = 0 \implies A=1$. $B(0)=0$, which is $10 \times 0$. This works for any B. The carry $k=0$.
• $A-1 = 5$ and B is even, $A=6$. $5B$ is a multiple of 10 if $B$ is even. Possible B are 2, 4, 6, 8. Carry $k = (A \times B)/10 = 6B/10$.
• B = 5 and $A-1$ is even. $A \in \{1, 3, 5, 7, 9\}$. Carry $k = (A \times B)/10 = 5A/10 = A/2$.
• Other combinations of factors of 10 (e.g., $B=2, A-1=5 \implies A=6$; $B=4, A-1$ ends in 5, not possible).

In the tens column, the sum of $(6 \times B)$'s units digit and the carry $k$ from units results in a number whose units digit is 8, and the carry from this sum is A (the hundreds digit of A8B).

Let $6B + k = 10 \times (\text{hundreds digit}) + (\text{tens digit}) = 10A + 8$.

Test cases satisfying units column rule ($A, B \in \{1..9\}$):

• Case 1: A=1. $k=0$. Equation: $6B + 0 = 10(1) + 8 \implies 6B = 18 \implies B = 3$. Valid digits $A=1, B=3$. Check $61 \times 3 = 183$. This matches A8B form with A=1, B=3. This is a solution.
• Case 2: A=6, B $\in \{2,4,6,8\}$. $k=6B/10$. Equation: $6B + 6B/10 = 10(6) + 8 \implies 6B + 0.6B = 68 \implies 6.6B = 68 \implies B = 68/6.6$. Not an integer. No solution here. (Checking with carry digit: $k=\lfloor 6B/10 \rfloor$. $6B + \lfloor 6B/10 \rfloor = 68$. If $B=2, 12+\lfloor 1.2 \rfloor=13 \neq 68$. If $B=4, 24+\lfloor 2.4 \rfloor=26 \neq 68$. If $B=6, 36+\lfloor 3.6 \rfloor=39 \neq 68$. If $B=8, 48+\lfloor 4.8 \rfloor=52 \neq 68$).
• Case 3: B=5, A $\in \{1,3,5,7,9\}$. $k=\lfloor A/2 \rfloor$. Equation: $6(5) + \lfloor A/2 \rfloor = 10A + 8 \implies 30 + \lfloor A/2 \rfloor = 10A + 8 \implies \lfloor A/2 \rfloor = 10A - 22$. Since $A \geq 1$, $10A - 22 \geq 10-22 = -12$. Also $\lfloor A/2 \rfloor \leq A/2$. $A/2 \geq 10A - 22 \implies 22 \geq 9.5A \implies A \leq 22/9.5 \approx 2.3$. Possible A values are 1. If A=1: $\lfloor 1/2 \rfloor = 0$. $10(1) - 22 = -12$. $0 \neq -12$. No solution for B=5.

The only solution found is $A=1, B=3$.

The value of $A-B = 1-3 = -2$.

The correct option is (a).

For a number to be divisible by 99, it must be divisible by both 9 and 11, since 9 and 11 are coprime factors of 99 ($99 = 9 \times 11$).

Divisibility Rule for 9: A number is divisible by 9 if the sum of its digits is divisible by 9.

Divisibility Rule for 11: A number is divisible by 11 if the difference between the sum of the digits at odd places (from the right) and the sum of the digits at even places (from the right) is either 0 or a multiple of 11.

We will check each option:

Option (a) 913462:

Sum of digits = $9 + 1 + 3 + 4 + 6 + 2 = 25$.

Since 25 is not divisible by 9, the number 913462 is not divisible by 9. Therefore, it is not divisible by 99.

Option (b) 114345:

Sum of digits = $1 + 1 + 4 + 3 + 4 + 5 = 18$.

Since 18 is divisible by 9, the number 114345 is divisible by 9.

Now check for divisibility by 11. Digits from right: 5, 4, 3, 4, 1, 1.

Sum of digits at odd places (1st, 3rd, 5th) = $5 + 3 + 1 = 9$.

Sum of digits at even places (2nd, 4th, 6th) = $4 + 4 + 1 = 9$.

Difference = (Sum of odd places) - (Sum of even places) = $9 - 9 = 0$.

Since the difference is 0, which is a multiple of 11, the number 114345 is divisible by 11.

Since 114345 is divisible by both 9 and 11, it is divisible by 99.

Option (c) 135792:

Sum of digits = $1 + 3 + 5 + 7 + 9 + 2 = 27$.

Since 27 is divisible by 9, the number 135792 is divisible by 9.

Now check for divisibility by 11. Digits from right: 2, 9, 7, 5, 3, 1.

Sum of digits at odd places (1st, 3rd, 5th) = $2 + 7 + 3 = 12$.

Sum of digits at even places (2nd, 4th, 6th) = $9 + 5 + 1 = 15$.

Difference = (Sum of odd places) - (Sum of even places) = $12 - 15 = -3$.

Since -3 is not 0 or a multiple of 11, the number 135792 is not divisible by 11. Therefore, it is not divisible by 99.

Option (d) 3572406:

Sum of digits = $3 + 5 + 7 + 2 + 4 + 0 + 6 = 27$.

Since 27 is divisible by 9, the number 3572406 is divisible by 9.

Now check for divisibility by 11. Digits from right: 6, 0, 4, 2, 7, 5, 3.

Sum of digits at odd places (1st, 3rd, 5th, 7th) = $6 + 4 + 7 + 3 = 20$.

Sum of digits at even places (2nd, 4th, 6th) = $0 + 2 + 5 = 7$.

Difference = (Sum of odd places) - (Sum of even places) = $20 - 7 = 13$.

Since 13 is not 0 or a multiple of 11, the number 3572406 is not divisible by 11. Therefore, it is not divisible by 99.

Only the number 114345 is divisible by both 9 and 11.

The correct option is (b).

To check divisibility by 3, we sum the digits: $3 + 1 + 3 + 4 + 6 + 7 + 3 = 27$. Since 27 is divisible by 3, the number 3134673 is divisible by 3.

We also check for divisibility by 9 by summing the digits. The sum is 27, which is divisible by 9. Therefore, the number 3134673 is also divisible by 9.

The completed statement is: 3134673 is divisible by 3 and 9.

A number is divisible by 3 if the sum of its digits is divisible by 3.

The given number is 20x3. The digits are 2, 0, x, and 3.

The sum of the digits is $2 + 0 + x + 3 = 5 + x$.

For 20x3 to be a multiple of 3, the sum of its digits $(5+x)$ must be a multiple of 3.

Since x is a digit, its possible values are $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$. We check which of these values makes $5+x$ divisible by 3.

- If $x = 0$, $5+0 = 5$ (not divisible by 3)

- If $x = 1$, $5+1 = 6$ (divisible by 3)

- If $x = 2$, $5+2 = 7$ (not divisible by 3)

- If $x = 3$, $5+3 = 8$ (not divisible by 3)

- If $x = 4$, $5+4 = 9$ (divisible by 3)

- If $x = 5$, $5+5 = 10$ (not divisible by 3)

- If $x = 6$, $5+6 = 11$ (not divisible by 3)

- If $x = 7$, $5+7 = 12$ (divisible by 3)

- If $x = 8$, $5+8 = 13$ (not divisible by 3)

- If $x = 9$, $5+9 = 14$ (not divisible by 3)

The values of x for which $5+x$ is divisible by 3 are 1, 4, and 7.

Therefore, 20x3 is a multiple of 3 if the digit x is 1 or 4 or 7.

A number is divisible by 9 if the sum of its digits is divisible by 9.

The given number is 3x5. The digits are 3, x, and 5.

The sum of the digits is $3 + x + 5 = 8 + x$.

For 3x5 to be divisible by 9, the sum of its digits $(8+x)$ must be a multiple of 9.

Since x is a digit, its possible values are $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$. We need to find the value of x such that $8+x$ is a multiple of 9.

The multiples of 9 are $9, 18, 27, ...$

If $8 + x = 9$, then $x = 9 - 8 = 1$. This is a single digit.

If $8 + x = 18$, then $x = 18 - 8 = 10$. This is not a single digit.

Any larger multiple of 9 will result in a value of x greater than 9.

The only single digit value for x that makes the sum $8+x$ a multiple of 9 is 1.

Therefore, 3x5 is divisible by 9 if the digit x is 1.

Let the two-digit number be represented as $10a + b$, where $a$ is the tens digit and $b$ is the units digit.

Since it is a two-digit number, $a$ must be an integer from 1 to 9, and $b$ must be an integer from 0 to 9.

The number obtained by reversing the digits is $10b + a$.

The sum of the original number and the reversed number is:

Sum $= (10a + b) + (10b + a)$

Sum $= 10a + b + 10b + a$

Combine like terms:

Sum $= (10a + a) + (b + 10b)$

Sum $= 11a + 11b$

Factor out the common factor, 11:

Sum $= 11(a + b)$

Since the sum is $11$ multiplied by $(a + b)$, the sum is always a multiple of 11.

Therefore, the sum of a two-digit number and the number obtained by reversing the digits is always divisible by 11.

Let the two-digit number be represented as $10a + b$, where $a$ is the tens digit and $b$ is the units digit.

Here, $a \in \{1, 2, ..., 9\}$ and $b \in \{0, 1, ..., 9\}$.

The number obtained by reversing the digits is $10b + a$.

The difference between the two numbers is $(10a + b) - (10b + a)$ or $(10b + a) - (10a + b)$. Let's consider the absolute difference.

Difference $= (10a + b) - (10b + a)$

Difference $= 10a + b - 10b - a$

Combine like terms:

Difference $= (10a - a) + (b - 10b)$

Difference $= 9a - 9b$

Factor out the common factor, 9:

Difference $= 9(a - b)$

The absolute difference is $|9(a - b)| = 9|a - b|$.

Since $a$ and $b$ are integers, $(a - b)$ is an integer. Thus, $9(a - b)$ is always an integer multiple of 9.

Therefore, the difference of a two–digit number and the number obtained by reversing its digits is always divisible by 9.

Let the three-digit number be represented as $100a + 10b + c$, where $a$ is the hundreds digit, $b$ is the tens digit, and $c$ is the units digit.

Here, $a \in \{1, 2, ..., 9\}$, $b \in \{0, 1, ..., 9\}$, and $c \in \{0, 1, ..., 9\}$.

The number obtained by reversing the digits is $100c + 10b + a$.

The difference between the two numbers is:

Difference $= (100a + 10b + c) - (100c + 10b + a)$

Difference $= 100a + 10b + c - 100c - 10b - a$

Combine like terms:

Difference $= (100a - a) + (10b - 10b) + (c - 100c)$

Difference $= 99a + 0 - 99c$

Difference $= 99a - 99c$

Factor out the common factor, 99:

Difference $= 99(a - c)$

Since the difference is $99(a - c)$, it is a multiple of 99.

We know that $99 = 9 \times 11$.

So, the difference can be written as $9 \times 11 \times (a - c)$.

This shows that the difference is always divisible by 9 and also by 11.

Therefore, the difference of three-digit number and the number obtained by putting the digits in reverse order is always divisible by 9 and 11.

The given problem is an addition of two 2-digit numbers:

The first number is $20 + B$.

The second number is $10A + B$.

The sum is $80 + A$.

The addition can be written as:

$(20 + B) + (10A + B) = 80 + A$

$20 + 10A + 2B = 80 + A$

Alternatively, we can look at the columns:

Units column: $B + B$ results in a units digit of $A$ and a carry-over (let's call it $c_1$) to the tens column.

$2B = A + 10c_1$

Since $B$ is a digit from 0 to 9, $2B$ can range from 0 to 18. Thus, $c_1$ can only be 0 or 1.

Tens column: $2 + A + c_1$ results in a units digit of 8 and a carry-over (let's call it $c_2$) to the hundreds column.

$2 + A + c_1 = 8 + 10c_2$

The sum is a 2-digit number $8A$, which means the hundreds digit is 0. So, $c_2$ must be 0.

The tens column equation becomes: $2 + A + c_1 = 8$

Now we have a system of equations:

(1) $2B = A + 10c_1$

(2) $2 + A + c_1 = 8$

where $A, B \in \{0, 1, ..., 9\}$ and $c_1 \in \{0, 1\}$. Note that $A$ cannot be 0 as it is the first digit of the result 8A.

From equation (2), we can express $A$ in terms of $c_1$:

$A = 8 - 2 - c_1$

$A = 6 - c_1$

Case 1: $c_1 = 0$ (No carry-over from units column)

From $A = 6 - c_1$, we get $A = 6 - 0 = 6$.

From $2B = A + 10c_1$, we get $2B = 6 + 10(0)$

$2B = 6$

$B = \frac{6}{2} = 3$.

This gives $A=6$ and $B=3$. Both are valid digits (A is not 0).

Check this solution:

$\begin{array}{cc} & 2 & 3 \\ + & 6 & 3 \\ \hline & 8 & 6 \\ \hline \end{array}$

This matches the form $\begin{array}{cc} & 2 & B \\ + & A & B \\ \hline & 8 & A \\ \hline \end{array}$ with $A=6$ and $B=3$.

Case 2: $c_1 = 1$ (Carry-over of 1 from units column)

From $A = 6 - c_1$, we get $A = 6 - 1 = 5$.

From $2B = A + 10c_1$, we get $2B = 5 + 10(1)$

$2B = 15$

$B = \frac{15}{2} = 7.5$.

This value of $B$ is not an integer digit, so this case is not possible.

The only valid solution is $A=6$ and $B=3$.

The completed statement is: If $\begin{array}{cccc} & 2 & B \\ + & A & B \\ \hline & 8 & A \\ \hline \end{array}$ then A = 6 and B = 3.

The given problem is a multiplication of a two-digit number AB by a single digit B, resulting in 96.

The number AB can be represented as $10A + B$, where A is the tens digit and B is the units digit. Since it's a two-digit number, A must be a digit from 1 to 9, and B must be a digit from 0 to 9.

The multiplication can be written as the equation:

$(10A + B) \times B = 96$

Let's analyse the multiplication column by column, starting from the units place.

Units column: The product of the units digits is $B \times B$. The units digit of this product must be 6 (the units digit of 96).

We check the possible values for B (digits from 0 to 9) whose square ends in 6:

• $4 \times 4 = 16$ (ends in 6)
• $6 \times 6 = 36$ (ends in 6)

So, B must be either 4 or 6.

Tens column: The product of the tens digit of AB (which is A) by the multiplier B, plus any carry-over from the units column, must result in the tens digit 9.

Let $c_1$ be the carry-over from the units column.
If $B=4$, $B \times B = 4 \times 4 = 16$. The units digit is 6, and the carry-over $c_1 = 1$.

If $B=6$, $B \times B = 6 \times 6 = 36$. The units digit is 6, and the carry-over $c_1 = 3$.

Now consider the tens column calculation: $A \times B + c_1 = 9$.

Case 1: Assume $B=4$.

The carry-over $c_1 = 1$.

The tens column equation is: $A \times 4 + 1 = 9$

$4A + 1 = 9$

$4A = 9 - 1$

$4A = 8$

$A = \frac{8}{4}$

$A = 2$

This gives the solution pair $(A, B) = (2, 4)$. A=2 is a valid non-zero digit.

Check: $24 \times 4 = 96$. This is correct.

Case 2: Assume $B=6$.

The carry-over $c_1 = 3$.

The tens column equation is: $A \times 6 + 3 = 9$

$6A + 3 = 9$

$6A = 9 - 3$

$6A = 6$

$A = \frac{6}{6}$

$A = 1$

This gives the solution pair $(A, B) = (1, 6)$. A=1 is a valid non-zero digit.

Check: $16 \times 6 = 96$. This is correct.

Both pairs $(A, B) = (2, 4)$ and $(A, B) = (1, 6)$ satisfy the given multiplication problem.

Assuming the problem expects a single solution, we can provide either pair.

If $\begin{array}{cccc} A & B \\ \times & B \\ \hline 9 & 6 \\ \hline \end{array}$ then A = 2 and B = 4.

(Note: Another possible answer is A = 1 and B = 6).

The given problem can be written as the multiplication of a two-digit number $(10B + 1)$ by a single digit $B$, which results in the number $(400 + 90 + B)$.

This can be expressed as an equation:

Expand the left side of the equation:

$10B^2 + B = 490 + B$

Subtract $B$ from both sides of the equation:

$10B^2 = 490$

Divide both sides by 10:

$B^2 = \frac{490}{10}$

$B^2 = 49$

Take the square root of both sides:

$B = \sqrt{49}$

$B = \pm 7$

Since $B$ represents a digit in the number system, it must be a non-negative integer from 0 to 9.

The value $B = -7$ is not a valid digit.

The value $B = 7$ is a valid digit (from 0 to 9).

Let's verify the result by performing the multiplication with $B=7$:

The multiplication is $71 \times 7$.

$\begin{array}{cc} & 7 & 1 \\ \times & & 7 \\ \hline 4 & 9 & 7 \\ \hline \end{array}$

The result 497 matches the format 49B with B=7.

Therefore, the value of B is 7.

If $\begin{array}{cccc} & B & 1 \\ & \times & B \\ \hline 4 & 9 & B \\ \hline \end{array}$ then B = 7.

A number is divisible by 9 if the sum of its digits is divisible by 9.

The given number is 1x35, where x is the digit in the hundreds place.

The digits of the number are 1, x, 3, and 5.

The sum of the digits is $1 + x + 3 + 5 = 9 + x$.

For the number 1x35 to be divisible by 9, the sum of its digits $(9+x)$ must be a multiple of 9.

Since x is a single digit, its possible values are $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.

We need to find the value(s) of x such that $9+x$ is a multiple of 9.

If $x = 0$, $9 + 0 = 9$. 9 is divisible by 9.

If $x = 1$, $9 + 1 = 10$. 10 is not divisible by 9.

If $x = 2$, $9 + 2 = 11$. 11 is not divisible by 9.

...

If $x = 9$, $9 + 9 = 18$. 18 is divisible by 9.

The possible single-digit values for x that make $9+x$ a multiple of 9 are 0 and 9.

Since only one blank is provided, we can use either valid digit. By convention, often the smallest digit is listed first if multiple single-digit answers are possible.

Therefore, 1x35 is divisible by 9 if x = 0.

(Note: x = 9 is also a valid answer).

According to the divisibility rule for 11, a number is divisible by 11 if the difference between the sum of its digits at odd places (from the right) and the sum of its digits at even places (from the right) is either 0 or a multiple of 11.

For the number abcd, the sum of digits at odd places (d and b) is $d+b$.

The sum of digits at even places (c and a) is $c+a$.

The difference is $(d+b) - (c+a)$.

For divisibility by 11, $(d+b) - (c+a)$ must be 0 or a multiple of 11.

Case 1: $(d+b) - (c+a) = 0$

This implies $d+b = c+a$.

Case 2: $(d+b) - (c+a) = 11k$ for some integer $k$.

For a four-digit number, the range of $(d+b)-(c+a)$ is limited. The relevant multiples of 11 are usually 11 and -11.

If $(d+b) - (c+a) = 11$, this implies $d+b = c+a + 11$.

If $(d+b) - (c+a) = -11$, this implies $d+b = c+a - 11$.

The question asks for two conditions for $d+b$. Based on common presentations of this rule, the blanks are filled with the conditions derived from the difference being 0 or 11.

Thus, $d + b = c + a$ or $d + b = c + a + 11$.

The completed statement is: A four-digit number abcd is divisible by 11, if d + b = $c + a$ or $c + a + 11$.

The blanks are $c+a$ and $c+a+11$.

The divisibility rule for 11 states that a number is divisible by 11 if the difference between the sum of its digits at odd places (from the right) and the sum of its digits at even places (from the right) is either 0 or a multiple of 11.

Being a "multiple of 11" is equivalent to being "divisible by 11".

Therefore, the blank in the statement should be filled with the number 11.

The completed statement is: A number is divisible by 11 if the differences between the sum of digits at its odd places and that of digits at the even places is either 0 or divisible by 11.

Let the 3-digit number be abc. This represents the number $100a + 10b + c$.

According to the divisibility rule for 11, a number is divisible by 11 if the difference between the sum of its digits at odd places (from the right) and the sum of its digits at even places (from the right) is either 0 or a multiple of 11.

For the number abc:

Digits at odd places (1st and 3rd from right) are c and a. Sum = $c+a$.

Digit at the even place (2nd from right) is b. Sum = $b$.

The difference between the sum of digits at odd places and the sum of digits at even places is $(c+a) - b$ or $(a+c) - b$.

For the number abc to be divisible by 11, this difference must be either 0 or a multiple of 11.

The statement given is "If a 3-digit number abc is divisible by 11, then ______ is either 0 or multiple of 11."

The blank should be filled with the expression representing this difference.

Therefore, the blank is filled with $\mathbf{(a+c) - b}$.

The expression 1A represents a two-digit number where the tens digit is 1 and the units digit is A.

We can write the number 1A in terms of place values as $1 \times 10 + A$.

The given equation is: A $\times$ 3 = 1A.

Substituting the value of 1A, we get:

$3 \times A = 10 + A$

Now, we solve for A:

$3A = 10 + A$

Subtract A from both sides:

$3A - A = 10$

$2A = 10$

Divide both sides by 2:

$A = \frac{10}{2}$

$A = 5$

To verify the solution, substitute A = 5 into the original equation:

A $\times$ 3 = 5 $\times$ 3 = 15.

1A = 15.

Since 15 = 15, the value A = 5 is correct.

Therefore, A = 5.

The equation is given as B $\times$ B = AB.

Here, AB represents a two-digit number, which can be written as $10 \times A + B$. A and B are digits from 0 to 9, and since AB is a two-digit number, A cannot be 0 (A $\in \{1, 2, \dots, 9\}$).

The equation becomes $B^2 = 10A + B$.

Rearranging the terms to isolate 10A:

$B^2 - B = 10A$

$B(B-1) = 10A$

We need to find digits A ($\in \{1, 2, \dots, 9\}$) and B ($\in \{0, 1, \dots, 9\}$) that satisfy this equation.

Since the left side, $B(B-1)$, is equal to $10A$, it must be a multiple of 10. Also, A is a single digit ($1 \le A \le 9$), so $10A$ must be between 10 and 90.

Let's test possible values for B from 0 to 9:

If B = 0, $0(0-1) = 0$. $10A = 0 \implies A = 0$. This gives the number 00, which is not a two-digit number. So B cannot be 0.

If B = 1, $1(1-1) = 0$. $10A = 0 \implies A = 0$. Again, not a two-digit number.

If B = 2, $2(2-1) = 2 \times 1 = 2$. $10A = 2 \implies A = 0.2$. Not an integer digit.

If B = 3, $3(3-1) = 3 \times 2 = 6$. $10A = 6 \implies A = 0.6$. Not an integer digit.

If B = 4, $4(4-1) = 4 \times 3 = 12$. $10A = 12 \implies A = 1.2$. Not an integer digit.

If B = 5, $5(5-1) = 5 \times 4 = 20$. $10A = 20 \implies A = 2$. This is a valid solution: A=2, B=5. Check: $5 \times 5 = 25$, and the number AB is 25. This matches the first possibility given in the question.

If B = 6, $6(6-1) = 6 \times 5 = 30$. $10A = 30 \implies A = 3$. This is a valid solution: A=3, B=6. Check: $6 \times 6 = 36$, and the number AB is 36.

If B = 7, $7(7-1) = 7 \times 6 = 42$. $10A = 42 \implies A = 4.2$. Not an integer digit.

If B = 8, $8(8-1) = 8 \times 7 = 56$. $10A = 56 \implies A = 5.6$. Not an integer digit.

If B = 9, $9(9-1) = 9 \times 8 = 72$. $10A = 72 \implies A = 7.2$. Not an integer digit.

The possible valid pairs (A, B) are (2, 5) and (3, 6).

The question provides A = 2, B = 5 as one option.

The other option is A = 3, B = 6.

Therefore, the blanks are filled with 3 and 6.

The completed statement is: If B × B = AB, then either A = 2, B = 5 or A = 3, B = 6.

Let the 2-digit number be formed by the tens digit t and the ones digit u.

The value of this 2-digit number is $10 \times t + u$.

When the digit 1 is placed after this 2-digit number, the new number will have three digits.

The original 2-digit number ($10t + u$) forms the hundreds and tens place of the new number, and 1 is in the units place.

This means the value of the original 2-digit number gets multiplied by 10, and then the digit 1 is added to the units place.

The new number's value is given by: $10 \times (10t + u) + 1$.

Simplifying the expression:

$10 \times (10t + u) + 1 = 100t + 10u + 1$

The new number is $100t + 10u + 1$.

The blank should be filled with the expression for the new number.

The completed statement is: If the digit 1 is placed after a 2-digit number whose tens is t and ones digit is u, the new number is $\mathbf{100t + 10u + 1}$.

The divisibility rule for the number 2 states that a number is divisible by 2 if its units digit (or last digit) is an even number (0, 2, 4, 6, or 8).

In a two-digit number 'ab', 'a' is the tens digit and 'b' is the units digit.

The statement says that the number 'ab' is divisible by 2 if the units digit 'b' is an even number.

This is exactly what the divisibility rule for 2 states. If the units digit is even, the entire number is divisible by 2.

Therefore, the statement is True.

Let the three-digit number be abc. Here, a, b, and c are digits, where a is the hundreds digit, b is the tens digit, and c is the units digit.

The divisibility rule for the number 5 states that a number is divisible by 5 if its units digit is either 0 or 5.

In the number abc, the units digit is c.

The statement says that the number abc is divisible by 5 if c is an even number.

An even number is a number divisible by 2 (0, 2, 4, 6, 8).

Let's check if this statement is always true:

If c = 0, then c is an even number, and the units digit is 0. According to the rule for 5, a number ending in 0 is divisible by 5 (e.g., 120 is divisible by 5).

If c = 2 (which is an even number), the units digit is 2. According to the rule for 5, a number ending in 2 is NOT divisible by 5 (e.g., 122 is not divisible by 5).

If c = 4, 6, or 8 (which are even numbers), the units digit is not 0 or 5, so the number is not divisible by 5 (e.g., 124, 126, 128 are not divisible by 5).

The condition that the units digit 'c' must be 0 or 5 for divisibility by 5 is not the same as 'c' being an even number. For example, a number ending in 2 is not divisible by 5, even though 2 is an even number.

Therefore, the statement is False.

Let the four-digit number be abcd. This number can be written as $1000a + 100b + 10c + d$.

The number formed by the first two digits is ab, which represents $10a + b$.

The number formed by the last two digits is cd, which represents $10c + d$.

The divisibility rule for 4 states that a number is divisible by 4 if the number formed by its last two digits is divisible by 4.

For the number abcd, this means that the number formed by the digits c and d (i.e., $10c + d$ or cd) must be divisible by 4.

The statement claims that abcd is divisible by 4 if ab ($10a + b$) is divisible by 4.

Let's consider a counterexample:

Let the number be 1234.

Here, ab = 12. 12 is divisible by 4 ($12 \div 4 = 3$). So the condition in the statement is met.

Now, let's check if the number 1234 is divisible by 4.

According to the actual rule for divisibility by 4, we check the last two digits, cd = 34.

Is 34 divisible by 4? $34 \div 4 = 8$ with a remainder of 2. So, 34 is not divisible by 4.

Therefore, 1234 is not divisible by 4.

In this example, ab (12) is divisible by 4, but the number abcd (1234) is not divisible by 4. This contradicts the statement.

The divisibility by 4 depends on the last two digits (cd), not the first two digits (ab).

Therefore, the statement is False.

The divisibility rule for 6 states that a number is divisible by 6 if and only if it is divisible by both 2 and 3.

Let's examine the conditions given in the statement for the three-digit number abc:

Condition 1: c is an even number.

The units digit of the number abc is c. The divisibility rule for 2 states that a number is divisible by 2 if its units digit is even. So, if c is an even number, the number abc is divisible by 2.

Condition 2: a + b + c is a multiple of 3.

The divisibility rule for 3 states that a number is divisible by 3 if the sum of its digits is a multiple of 3. The sum of the digits of abc is a + b + c. So, if a + b + c is a multiple of 3, the number abc is divisible by 3.

The statement says that if both Condition 1 (divisible by 2) and Condition 2 (divisible by 3) are met, then the number abc is divisible by 6.

Since a number is divisible by 6 if and only if it is divisible by both 2 and 3, the given conditions are precisely the requirements for divisibility by 6.

Therefore, the statement is True.

Let the number be represented by M. The form of the number is given as $M = 3N + 2$, where N is an integer (usually considered in such contexts unless otherwise specified).

We are asked about the remainder when M is divided by 3.

According to the division algorithm, for any integer M and a positive integer d, there exist unique integers q (quotient) and r (remainder) such that $M = qd + r$, where $0 \le r < d$.

In this case, the divisor is $d=3$. The given form of the number is $M = 3N + 2$.

Comparing $M = 3N + 2$ with $M = q \times 3 + r$, we can identify $q=N$ and $r=2$.

We must check if the remainder $r=2$ satisfies the condition $0 \le r < d$.

Here, $d=3$. So we check $0 \le 2 < 3$. This inequality is true.

Since the number can be expressed in the form $3N + 2$, where N is the quotient and 2 is the remainder, and the remainder 2 is valid ($0 \le 2 < 3$), the statement is correct.

Therefore, the statement is True.

Let the number be M. The form of the number is given as $M = 7N + 1$.

We want to determine the remainder when M is divided by 7.

According to the division algorithm, when an integer M is divided by a positive integer d, we can write $M = qd + r$, where q is the quotient and r is the remainder, such that $0 \le r < d$.

In this case, the divisor is $d=7$. The given number is $M = 7N + 1$.

Comparing $M = 7N + 1$ with $M = q \times 7 + r$, we can see that $q=N$ and $r=1$.

We check if the remainder $r=1$ is valid for a divisor of 7: $0 \le 1 < 7$. This condition is satisfied.

This means that when any number of the form $7N + 1$ is divided by 7, the quotient is N and the remainder is 1.

Therefore, the statement "Number 7N + 1 will leave remainder 1 when divided by 7" is True.

Let the number be a and the divisor be b.

The statement "a is divisible by b" means that a can be written as a product of b and some integer, say k.

Now, let d be any factor of b. This means that b is divisible by d, or b can be written as a product of d and some integer, say m.

Substitute the expression for b from equation (ii) into equation (i):

a = k $\times$ (m $\times$ d)

Using the associative property of multiplication, we can group the terms:

a = (k $\times$ m) $\times$ d

Since k and m are integers, their product (k $\times$ m) is also an integer. Let's call this integer p.

a = p $\times$ d

The equation a = p $\times$ d shows that the number a can be written as the product of d and some integer p.

By the definition of divisibility, this means that a is divisible by d.

Since d was chosen as any arbitrary factor of b, this holds true for every factor of b.

Therefore, if a number a is divisible by b, it must be divisible by each factor of b.

The statement is True.

The expression AB represents a two-digit number where A is the tens digit and B is the units digit. Its value is $10A + B$.

The given equation is: AB $\times$ 4 = 192.

Substituting the value of AB, we get: $(10A + B) \times 4 = 192$.

To find the value of the two-digit number AB, divide 192 by 4:

$10A + B = \frac{192}{4}$

$10A + B = 48$

The number formed by the digits A and B is 48.

Comparing this with the structure of the number AB, we can identify the digits A and B.

The tens digit A is 4.

The units digit B is 8.

So, we have A = 4 and B = 8.

The statement claims that if AB $\times$ 4 = 192, then A + B = 7.

Using our calculated values for A and B:

A + B = 4 + 8 = 12.

The calculated sum A + B = 12.

The statement claims A + B = 7.

Since $12 \ne 7$, the statement is incorrect.

Therefore, the statement is False.

The expression AB represents a two-digit number with tens digit A and units digit B. Its value is $10A + B$.

The expression 7C represents the product of 7 and the digit C.

The given equation is: AB + 7C = 102.

Substituting the value of AB, we get: $(10A + B) + 7C = 102$.

$10A + B + 7C = 102$

We are given that A, B, and C are digits. Since AB is a two-digit number, A cannot be 0 ($A \in \{1, 2, \dots, 9\}$). We are also given the constraints $B \ne 0$ ($B \in \{1, 2, \dots, 9\}$) and $C \ne 0$ ($C \in \{1, 2, \dots, 9\}$).

We need to find possible values for A, B, and C that satisfy the equation and the constraints, and then check if their sum A + B + C is equal to 14.

Let's rearrange the equation: $10A + B = 102 - 7C$.

Since $10A + B$ is a two-digit number (value from 11 to 99), the right side $102 - 7C$ must fall within this range.

Let's test values for C from $\{1, 2, \dots, 9\}$:

If C = 1, $10A + B = 102 - 7(1) = 95$. This gives A = 9, B = 5. These are valid digits ($A=9 \ne 0$, $B=5 \ne 0$, $C=1 \ne 0$). Sum A + B + C = $9 + 5 + 1 = 15$.

If C = 2, $10A + B = 102 - 7(2) = 102 - 14 = 88$. This gives A = 8, B = 8. These are valid digits ($A=8 \ne 0$, $B=8 \ne 0$, $C=2 \ne 0$). Sum A + B + C = $8 + 8 + 2 = 18$.

If C = 3, $10A + B = 102 - 7(3) = 102 - 21 = 81$. This gives A = 8, B = 1. These are valid digits ($A=8 \ne 0$, $B=1 \ne 0$, $C=3 \ne 0$). Sum A + B + C = $8 + 1 + 3 = 12$.

If C = 4, $10A + B = 102 - 7(4) = 102 - 28 = 74$. This gives A = 7, B = 4. These are valid digits ($A=7 \ne 0$, $B=4 \ne 0$, $C=4 \ne 0$). Sum A + B + C = $7 + 4 + 4 = 15$.

If C = 5, $10A + B = 102 - 7(5) = 102 - 35 = 67$. This gives A = 6, B = 7. These are valid digits ($A=6 \ne 0$, $B=7 \ne 0$, $C=5 \ne 0$). Sum A + B + C = $6 + 7 + 5 = 18$.

If C = 6, $10A + B = 102 - 7(6) = 102 - 42 = 60$. This gives A = 6, B = 0. This is not valid because the constraint $B \ne 0$ is not satisfied.

If C = 7, $10A + B = 102 - 7(7) = 102 - 49 = 53$. This gives A = 5, B = 3. These are valid digits ($A=5 \ne 0$, $B=3 \ne 0$, $C=7 \ne 0$). Sum A + B + C = $5 + 3 + 7 = 15$.

If C = 8, $10A + B = 102 - 7(8) = 102 - 56 = 46$. This gives A = 4, B = 6. These are valid digits ($A=4 \ne 0$, $B=6 \ne 0$, $C=8 \ne 0$). Sum A + B + C = $4 + 6 + 8 = 18$.

If C = 9, $10A + B = 102 - 7(9) = 102 - 63 = 39$. This gives A = 3, B = 9. These are valid digits ($A=3 \ne 0$, $B=9 \ne 0$, $C=9 \ne 0$). Sum A + B + C = $3 + 9 + 9 = 21$.

We found several sets of digits (A, B, C) that satisfy the equation AB + 7C = 102 with the constraints B $\ne$ 0, C $\ne$ 0, and A $\ne$ 0 (since AB is a 2-digit number). Some of these are (9, 5, 1), (8, 8, 2), (8, 1, 3), (7, 4, 4), etc.

For each of these valid solutions, we calculated the sum A + B + C and found the sums to be 15, 18, 12, 15, 18, 15, 18, 21.

Since there exists at least one case (in fact, multiple cases) where AB + 7C = 102 (with the given constraints) but A + B + C is not equal to 14, the statement is false.

Therefore, the statement is False.

Let the number be N = 213x27.

The divisibility rule for 9 states that a number is divisible by 9 if the sum of its digits is divisible by 9.

The digits of the number 213x27 are 2, 1, 3, x, 2, and 7.

The sum of the digits is $2 + 1 + 3 + x + 2 + 7 = 15 + x$.

For the number 213x27 to be divisible by 9, the sum of its digits, $15 + x$, must be divisible by 9.

Since x is a digit, its value must be an integer from 0 to 9 ($x \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$).

We need to find the value(s) of x such that $15 + x$ is a multiple of 9.

Let's check the possible values of $15+x$:

If $x=0$, $15+0 = 15$. 15 is not divisible by 9.

If $x=1$, $15+1 = 16$. 16 is not divisible by 9.

If $x=2$, $15+2 = 17$. 17 is not divisible by 9.

If $x=3$, $15+3 = 18$. 18 is divisible by 9 ($18 = 9 \times 2$). So, $x=3$ is a valid value.

If $x=4$, $15+4 = 19$. 19 is not divisible by 9.

If $x=5$, $15+5 = 20$. 20 is not divisible by 9.

If $x=6$, $15+6 = 21$. 21 is not divisible by 9.

If $x=7$, $15+7 = 22$. 22 is not divisible by 9.

If $x=8$, $15+8 = 23$. 23 is not divisible by 9.

If $x=9$, $15+9 = 24$. 24 is not divisible by 9.

The only digit value for x that makes the sum of digits divisible by 9 is $x=3$.

The statement says that if 213x27 is divisible by 9, then x is 0.

Our analysis shows that if 213x27 is divisible by 9, then x must be 3.

Since $3 \ne 0$, the statement is incorrect.

Therefore, the statement is False.

We are given two conditions about a number N:

Condition 1: N divided by 5 leaves a remainder of 3.

Using the division algorithm, this can be written as:

Condition 2: N divided by 2 leaves a remainder of 0.

This means N is divisible by 2, which implies N is an even number.

From Condition 2, N is even. Therefore, the expression for N from Condition 1, $5k + 3$, must be even.

For $5k + 3$ to be an even number, $5k$ must be an odd number (since Odd + Odd = Even).

For $5k$ to be an odd number, k must be an odd integer.

Since k is an odd integer, it can be written in the form $k = 2j + 1$ for some integer j.

Now, substitute this form of k back into the equation from Condition 1:

$N = 5(2j + 1) + 3$

$N = 10j + 5 + 3$

$N = 10j + 8$

This expression for N is in the form of the division algorithm $N = qd + r$, where $d=10$, $q=j$, and $r=8$.

The remainder r = 8 satisfies the condition $0 \le r < d$, as $0 \le 8 < 10$ is true.

This means that when N is divided by 10, the remainder is 8.

The statement claims that N ÷ 10 leaves remainder 4.

Our calculation shows the remainder is 8, not 4.

Therefore, the statement is False.

To Find:

The least value of the digit 'a' such that the number 91876a2 is divisible by 8.

Solution:

A number is divisible by 8 if the number formed by its last three digits is divisible by 8.

The given number is 91876a2.

The last three digits of the number are 6a2.

For the number 91876a2 to be divisible by 8, the number 6a2 must be divisible by 8.

The number 6a2 can be written as $600 + 10a + 2 = 602 + 10a$.

We need to find the least value of 'a' (which is a digit from 0 to 9) such that $602 + 10a$ is divisible by 8.

Let's test the possible values for 'a' starting from 0:

If $a = 0$, the number is $602 + 10(0) = 602$.

$602 \div 8 = 75$ with a remainder of 2. (Not divisible)

If $a = 1$, the number is $602 + 10(1) = 612$.

$612 \div 8 = 76$ with a remainder of 4. (Not divisible)

If $a = 2$, the number is $602 + 10(2) = 602 + 20 = 622$.

$622 \div 8 = 77$ with a remainder of 6. (Not divisible)

If $a = 3$, the number is $602 + 10(3) = 602 + 30 = 632$.

$632 \div 8 = 79$ with a remainder of 0. (Divisible)

Since we started checking from the least possible value of 'a' (which is 0), the first value for which 6a2 is divisible by 8 is $a = 3$.

Therefore, the least value that must be given to 'a' is 3.

The final answer is $\boxed{3}$.

Given:

The multiplication puzzle:

And the condition: $Q - P = 3$, where P and Q are single digits.

To Find:

The values of P and Q.

Solution:

From the given multiplication, we have the product $(10 \times 1 + P) \times P = 10Q + 6$. This simplifies to $(10 + P) \times P = 10Q + 6$, or $10P + P^2 = 10Q + 6$.

Looking at the units digit of the multiplication, the units digit of $P \times P$ must be 6.

We consider possible single-digit values for P (from 0 to 9) and their squares:

• $0^2 = 0$
• $1^2 = 1$
• $2^2 = 4$
• $3^2 = 9$
• $4^2 = 16$ (Units digit is 6)
• $5^2 = 25$
• $6^2 = 36$ (Units digit is 6)
• $7^2 = 49$
• $8^2 = 64$
• $9^2 = 81$

The possible values for P whose square ends in 6 are 4 and 6.

Now we use the given condition: $Q - P = 3$. So, $Q = P + 3$.

Let's examine the two possible cases for P:

Case 1: P = 4

Using the condition $Q = P + 3$, we get $Q = 4 + 3 = 7$.

Let's check if $P=4$ and $Q=7$ satisfy the original multiplication:

The number $1P$ becomes $14$. The number $Q6$ becomes $76$.

The multiplication is $14 \times 4$.

$14 \times 4 = 56$.

According to the puzzle, the result should be $Q6$, which is $76$.

Since $56 \neq 76$, this case is not valid.

Case 2: P = 6

Using the condition $Q = P + 3$, we get $Q = 6 + 3 = 9$.

Let's check if $P=6$ and $Q=9$ satisfy the original multiplication:

The number $1P$ becomes $16$. The number $Q6$ becomes $96$.

The multiplication is $16 \times 6$.

$16 \times 6 = 96$.

According to the puzzle, the result should be $Q6$, which is $96$.

Since $96 = 96$, this case is valid.

Thus, the values that satisfy both the multiplication and the condition are $P=6$ and $Q=9$.

The final answer is $\boxed{P=6, Q=9}$.

Given:

The addition equation: 1AB + CCA = 697

There is no carry-over in the addition.

A, B, and C are single digits (from 0 to 9).

To Find:

The value of A + B + C.

Solution:

The given addition can be written vertically as:

$\begin{array}{cccc} & 1 & A & B \\ + & C & C & A \\ \hline & 6 & 9 & 7 \\ \hline \end{array}$

We are given that there is no carry-over in any column of the addition.

Considering the columns from right to left:

Units Column:

$B + A = 7$

Since there is no carry-over, $B+A$ must be exactly 7 (not 17, 27, etc.).

Tens Column:

$A + C = 9$

Since there is no carry-over, $A+C$ must be exactly 9 (not 19, 29, etc.).

Hundreds Column:

$1 + C = 6$

Since there is no carry-over from the tens column, $1+C$ must be exactly 6.

From equation (iii), we can solve for C:

$C = 6 - 1$

Now substitute the value of C from equation (iv) into equation (ii):

$A + C = 9$

$A + 5 = 9$

$A = 9 - 5$

Now substitute the value of A from equation (v) into equation (i):

$B + A = 7$

$B + 4 = 7$

$B = 7 - 4$

So, the values are A = 4, B = 3, and C = 5.

Let's check if these values satisfy the original addition:

1AB = 143

CCA = 554

$143 + 554 = 697$.

Units column: $3 + 4 = 7$ (No carry-over)

Tens column: $4 + 5 = 9$ (No carry-over)

Hundreds column: $1 + 5 = 6$ (No carry-over)

The values are correct and the condition of no carry-over is satisfied.

We need to find the value of A + B + C.

$A + B + C = 4 + 3 + 5 = 12$.

The final answer is $\boxed{12}$.

Given:

A five-digit number of the form AABAA is divisible by 33.

A and B are digits.

Since it is a five-digit number, A cannot be 0.

To Find:

All possible numbers of this form.

Solution:

A number is divisible by 33 if and only if it is divisible by both 3 and 11, since $33 = 3 \times 11$ and 3 and 11 are coprime.

The given number is AABAA. We can write this number in expanded form as:

$A \times 10000 + A \times 1000 + B \times 100 + A \times 10 + A \times 1$

$= 10000A + 1000A + 100B + 10A + A = 11011A + 100B$

Divisibility by 11:

For a number to be divisible by 11, the alternating sum of its digits must be divisible by 11.

For the number AABAA, the alternating sum is:

$A - A + B - A + A$

$(A - A) + B + (-A + A) = 0 + B + 0 = B$

So, the digit B must be divisible by 11.

Since B is a single digit (from 0 to 9), the only possible value for B that is divisible by 11 is 0.

Divisibility by 3:

For a number to be divisible by 3, the sum of its digits must be divisible by 3.

The sum of the digits of AABAA is $A + A + B + A + A = 4A + B$.

According to the divisibility rule for 3, $4A + B$ must be divisible by 3.

Substitute the value of B from equation (i) into this condition:

$4A + 0$ must be divisible by 3.

$4A$ must be divisible by 3.

Since 4 is not divisible by 3, for $4A$ to be divisible by 3, A must be divisible by 3.

A is a digit from 1 to 9 (as it is the first digit of a five-digit number). The digits between 1 and 9 that are divisible by 3 are 3, 6, and 9.

So, the possible values for A are 3, 6, and 9.

We have found that B must be 0 and A can be 3, 6, or 9.

The form of the number is AABAA.

If A = 3 and B = 0, the number is 33033.

If A = 6 and B = 0, the number is 66066.

If A = 9 and B = 0, the number is 99099.

Let's verify these numbers:

For 33033:

Sum of digits = $3+3+0+3+3 = 12$. 12 is divisible by 3.

Alternating sum of digits = $3-3+0-3+3 = 0$. 0 is divisible by 11.

Since it is divisible by both 3 and 11, 33033 is divisible by 33.

For 66066:

Sum of digits = $6+6+0+6+6 = 24$. 24 is divisible by 3.

Alternating sum of digits = $6-6+0-6+6 = 0$. 0 is divisible by 11.

Since it is divisible by both 3 and 11, 66066 is divisible by 33.

For 99099:

Sum of digits = $9+9+0+9+9 = 36$. 36 is divisible by 3.

Alternating sum of digits = $9-9+0-9+9 = 0$. 0 is divisible by 11.

Since it is divisible by both 3 and 11, 99099 is divisible by 33.

The numbers of the form AABAA that are divisible by 33 are 33033, 66066, and 99099.

The final answer is $\boxed{33033, 66066, 99099}$.

Given:

The addition puzzle:

Where A, X, and Z are single digits (0-9).

Since A is the leading digit of a two-digit number, $A \neq 0$.

Since X is the leading digit of a three-digit number, $X \neq 0$.

To Find:

The values of the letters A, X, and Z.

Solution:

The given addition can be written in terms of place values.

The number 'AA' represents $10 \times A + A = 11A$.

The number 'XAZ' represents $100 \times X + 10 \times A + Z = 100X + 10A + Z$.

The addition puzzle translates to the equation:

$(10A + A) + (10A + A) = 100X + 10A + Z$

$11A + 11A = 100X + 10A + Z$

$22A = 100X + 10A + Z$

Subtracting $10A$ from both sides:

$12A = 100X + Z$

Let's analyze the addition column by column, starting from the units place.

Units Column:

The sum of the digits in the units column is $A + A = 2A$.

This sum gives the units digit Z and a possible carry-over to the tens column. Let the carry-over be $c_1$.

$2A = Z + 10c_1$

Since A and Z are single digits, $2A$ can range from $2 \times 0 = 0$ to $2 \times 9 = 18$.

If $2A < 10$, $c_1 = 0$ and $Z = 2A$.

If $2A \ge 10$, $c_1 = 1$ and $Z = 2A - 10$.

So, $c_1$ can only be 0 or 1.

Tens Column:

The sum of the digits in the tens column plus the carry-over from the units column is $A + A + c_1 = 2A + c_1$.

This sum gives the tens digit A and a possible carry-over to the hundreds column. Let the carry-over be $c_2$.

$2A + c_1 = A + 10c_2$

Subtracting A from both sides:

$A + c_1 = 10c_2$

Since A is a digit ($0 \le A \le 9$) and $c_1$ is a carry-over (0 or 1), $A+c_1$ can range from $0+0=0$ to $9+1=10$.

Also, $c_2$ is a carry-over (0 or 1).

If $c_2 = 0$, then $A + c_1 = 10 \times 0 = 0$. This is only possible if $A=0$ and $c_1=0$. However, A cannot be 0 as it's the first digit of a two-digit number.

Therefore, $c_2$ must be 1.

Substituting $c_2 = 1$ into equation (ii):

$A + c_1 = 10$

Since $c_1$ can only be 0 or 1:

If $c_1 = 0$, then $A + 0 = 10$, so $A = 10$. This is not possible as A must be a single digit.

If $c_1 = 1$, then $A + 1 = 10$, so $A = 9$. This is a valid single digit.

Thus, we must have $A=9$ and $c_1=1$.

Hundreds Column:

The sum of the digits in the hundreds column plus the carry-over from the tens column is $0 + 0 + c_2$.

This sum gives the hundreds digit X.

$X = c_2$

From our previous derivation, we found $c_2=1$.

Now we can find Z using equation (i) and the values of A and $c_1$ we found:

$2A = Z + 10c_1$

$2(9) = Z + 10(1)$

$18 = Z + 10$

$Z = 18 - 10$

We have found the values: A = 9, X = 1, Z = 8.

Let's check the original addition with these values:

AA = 99

XAZ = 198

The addition becomes $99 + 99 = 198$.

This matches the pattern XAZ where X=1, A=9, and Z=8.

The values of the letters are: A = 9, X = 1, and Z = 8.

The final answer is $\boxed{A=9, X=1, Z=8}$.

Given:

The addition puzzle:

Where A, B, and C are single digits (0-9).

Since B is the leading digit of a three-digit number, $B \neq 0$.

To Find:

The values of the letters A, B, and C.

Solution:

The given addition can be written vertically as:

We analyze the addition column by column, starting from the units place.

Units Column:

The sum of the digits in the units column is $5 + A$. The units digit of the result is 3.

This means that $5 + A$ must result in a number whose units digit is 3. Possible values for $5+A$ are 3, 13, 23, etc.

Since A is a single digit (0-9), the maximum value of $5+A$ is $5+9=14$.

Therefore, the only possibility is $5 + A = 13$.

$5 + A = 13$

$A = 13 - 5$

A = 8

When $5+8=13$, the units digit is 3, and there is a carry-over of 1 to the tens column.

Carry-over to tens column = 1.

Tens Column:

The sum of the digits in the tens column, plus the carry-over from the units column, gives the tens digit C and a possible carry-over to the hundreds column.

The digits in the tens column are 8 and 4. The carry-over from the units column is 1.

Sum = $8 + 4 + 1 = 13$.

The units digit of this sum (13) is the value of C. So, C = 3.

The tens digit of this sum (13) is the carry-over to the hundreds column. So, there is a carry-over of 1 to the hundreds column.

Carry-over to hundreds column = 1.

Hundreds Column:

The sum in the hundreds column consists of the carry-over from the tens column (1) and the implicit 0s from the hundreds place of 85 and 4A.

Sum = $0 + 0 + 1 = 1$.

This sum is the value of B. So, B = 1.

We have found the values: A = 8, B = 1, and C = 3.

Let's check the original addition with these values:

Substitute A=8:

Performing the addition:

$85 + 48 = 133$

The result is 133. This matches the form BC3, where B=1 and C=3. The values are consistent.

The values of the letters are: A = 8, B = 1, and C = 3.

The final answer is $\boxed{A=8, B=1, C=3}$.

Given:

The addition puzzle:

Where A, B, and C are single digits (0-9).

Since 'B6' and '8A' are two-digit numbers, $B \neq 0$ and $A$ is also a digit from 0-9.

Since 'CA2' is a three-digit number, $C \neq 0$.

To Find:

The values of the letters A, B, and C.

Solution:

We analyze the addition column by column, starting from the units place.

Units Column:

Adding the digits in the units column, we have $6 + A$. The units digit of the sum is 2.

This means that $6 + A$ must be a number ending in 2.

Since A is a single digit (0-9), the possible values for $6+A$ are $6, 7, 8, ..., 15$.

The only number in this range that ends in 2 is 12.

So, we must have:

$6 + A = 12$

Solving for A:

$A = 12 - 6$

$A = 6$

When $6 + 6 = 12$, the units digit is 2, and there is a carry-over of 1 to the tens column.

Carry-over to tens column = 1.

Tens Column:

Adding the digits in the tens column (B and 8) and the carry-over from the units column (1), we get the tens digit A and a possible carry-over to the hundreds column.

The sum is $B + 8 + 1 = B + 9$.

The tens digit of the result is A. We already found that $A = 6$.

So, the units digit of $B + 9$ must be 6.

Let the carry-over to the hundreds column be $c$. The sum in the tens column is $B + 9$. This sum equals $A + 10c$.

$B + 9 = 6 + 10c$

$B + 3 = 10c$

Since B is a single digit (0-9), $B+3$ can range from $0+3=3$ to $9+3=12$.

The possible values for $10c$ must be in this range and be a multiple of 10. The only multiple of 10 in the range [3, 12] is 10.

So, we must have:

$10c = 10$

$c = 1$

And

$B + 3 = 10$

$B = 10 - 3$

$B = 7$

The carry-over to the hundreds column is $c = 1$.

Hundreds Column:

The sum in the hundreds column is just the carry-over from the tens column, which is 1.

This sum gives the hundreds digit C.

So, $C = 1$.

We have found the values: A = 6, B = 7, and C = 1.

Let's check these values by performing the addition:

B6 becomes 76.

8A becomes 86.

CA2 becomes 162.

Adding 76 and 86:

$\begin{array}{cc} & 7 & 6 \\ + & 8 & 6 \\ \hline 1 & 6 & 2 \\ \hline \end{array}$

$6 + 6 = 12$ (Write down 2, carry over 1).

$7 + 8 + 1 (\text{carry-over}) = 16$ (Write down 6, carry over 1).

The result is 162, which matches CA2 with C=1, A=6, and the units digit 2. The values are consistent.

The values of the letters are: A = 6, B = 7, and C = 1.

The final answer is $\boxed{A=6, B=7, C=1}$.

Given:

The addition puzzle:

Where A and B are single digits (0-9).

Since 'ABA' is a three-digit number, A must be non-zero ($A \in \{1, 2, ..., 9\}$).

To Find:

The values of the letters A and B.

Solution:

We analyze the addition column by column, from right to left, considering potential carry-overs.

Let $c_1$ be the carry-over from the units column to the tens column.

Let $c_2$ be the carry-over from the tens column to the hundreds column.

Units Column:

Adding the digits in the units column: $A + A$. The units digit of the sum is 2.

$A + A = 2 + 10 \times c_1$

$2A = 2 + 10c_1$

Since A is a single digit (0-9), $2A$ can range from $2 \times 0 = 0$ to $2 \times 9 = 18$.

If $c_1 = 0$, then $2A = 2$, which gives $A=1$.

If $c_1 = 1$, then $2A = 2 + 10 = 12$, which gives $A=6$.

So, based on the units column, A can be 1 (with $c_1=0$) or 6 (with $c_1=1$).

Hundreds Column:

Adding the digits in the hundreds column (1 and A) plus the carry-over from the tens column ($c_2$), the sum is 8.

$1 + A + c_2 = 8$

$A + c_2 = 7$

Since $c_2$ is a carry-over from the tens column where we are adding two single digits and a carry-over from units (max sum $9+9+1=19$), $c_2$ can only be 0 or 1.

If $c_2 = 0$, then $A + 0 = 7$, which gives $A=7$.

If $c_2 = 1$, then $A + 1 = 7$, which gives $A=6$.

So, based on the hundreds column, A can be 7 (with $c_2=0$) or 6 (with $c_2=1$).

Comparing the possible values for A from the units column (1 or 6) and the hundreds column (7 or 6), the only common value is A = 6.

If A = 6, then from the units column analysis, the carry-over $c_1$ must be 1.

If A = 6, then from the hundreds column analysis, the carry-over $c_2$ must be 1.

Tens Column:

Adding the digits in the tens column (B and B) plus the carry-over from the units column ($c_1$), the sum is the tens digit of the result (which is A) plus 10 times the carry-over to the hundreds column ($c_2$).

$B + B + c_1 = A + 10 \times c_2$

$2B + c_1 = A + 10c_2$

Substitute the values we found: A = 6, $c_1 = 1$, and $c_2 = 1$ into this equation.

$2B + 1 = 6 + 10 \times 1$

$2B + 1 = 6 + 10$

$2B + 1 = 16$

$2B = 16 - 1$

$2B = 15$

$B = \frac{15}{2}$

$B = 7.5$

However, B must be a single-digit integer (0-9).

Since $B=7.5$ is not an integer, there is no single-digit integer value for B that satisfies the equation derived from the tens column, given A=6, $c_1=1$, and $c_2=1$. These values of A, $c_1$, and $c_2$ were uniquely determined by the units and hundreds columns based on the problem statement.

This indicates that the problem as stated ($1BA + ABA = 8A2$) has no solution where A and B are single-digit integers with $A \neq 0$. It is likely there is a typo in the original problem.

Based on the problem as given, there are no single-digit integer values for A and B that satisfy the addition.

Conclusion:

The puzzle as written does not have a valid solution for single-digit integers A and B.

Given:

The addition puzzle: CBA + CBA = 1A30.

A, B, C are digits, $C \neq 0$.

To Find:

The values of A, B, and C.

Solution:

From the units column: $A + A$ ends in 0. This means $2A$ is a multiple of 10. Since A is a single digit, $2A$ can be 0, 10, or 20 (not possible as $A \le 9 \implies 2A \le 18$).

If $2A = 0$, then $A = 0$. Carry-over to tens column ($c_1$) = 0.

If $2A = 10$, then $A = 5$. Carry-over to tens column ($c_1$) = 1.

From the hundreds column: $C + C$ plus the carry-over from the tens column ($c_2$) results in a number whose hundreds digit is A and thousands digit is 1 (carry-over $c_3=1$).

$C + C + c_2 = A + 10 \times c_3$

$2C + c_2 = A + 10 \times 1$

Since $C \ge 1$, $2C \ge 2$. Since $c_2$ is 0 or 1, $2C+c_2 \ge 2$. Also, $A \le 9$. So $A+10 \ge 10$. This confirms $c_3$ must be 1, and $c_2$ is 0 or 1.

From the tens column: $B + B$ plus the carry-over from the units column ($c_1$) results in a number whose units digit is 3 and tens digit is the carry-over to the hundreds column ($c_2$).

$B + B + c_1 = 3 + 10 \times c_2$

Let's test the possible values for A from the units column:

Case 1: A = 0 and $c_1 = 0$

Substitute A=0 into equation (i): $2C + c_2 = 0 + 10 = 10$. Since $C \ge 1$, $2C \ge 2$. $c_2$ is 0 or 1. If $c_2=0$, $2C=10 \implies C=5$. If $c_2=1$, $2C=9 \implies C=4.5$ (not a digit).

So, if A=0, then $C=5$ and $c_2=0$.

Now substitute $c_1=0$ and $c_2=0$ into equation (ii): $2B + 0 = 3 + 10(0) \implies 2B = 3$. $B = 1.5$ (not a digit). Case 1 is invalid.

Case 2: A = 5 and $c_1 = 1$

Substitute A=5 into equation (i): $2C + c_2 = 5 + 10 = 15$. Since $C \ge 1$, $2C \ge 2$. $c_2$ is 0 or 1. If $c_2=0$, $2C=15 \implies C=7.5$ (not a digit). If $c_2=1$, $2C=14 \implies C=7$. This is a valid digit, and $C \neq 0$.

So, if A=5, then $C=7$ and $c_2=1$.

Now substitute $c_1=1$ and $c_2=1$ into equation (ii): $2B + 1 = 3 + 10(1) \implies 2B + 1 = 13 \implies 2B = 12 \implies B = 6$. This is a valid digit.

So, A=5, B=6, and C=7 is a potential solution.

Let's check the addition with A=5, B=6, C=7:

CBA = 765

1A30 = 1530

$\begin{array}{cc} & 7 & 6 & 5 \\ + & 7 & 6 & 5 \\ \hline 1 & 5 & 3 & 0 \\ \hline \end{array}$

Units: $5+5=10$ (0, carry 1). Correct.

Tens: $6+6+1=13$ (3, carry 1). Correct.

Hundreds: $7+7+1=15$ (5, carry 1). Correct, and the hundreds digit is A=5.

Thousands: Carry 1. Correct.

The values are consistent.

The values of the letters are: A = 5, B = 6, and C = 7.

The final answer is $\boxed{A=5, B=6, C=7}$.

Given:

The addition puzzle:

Where A and B are single digits (0-9).

Since 'BAA' is a three-digit number, $B \neq 0$.

To Find:

The values of the letters A and B.

Solution:

We analyze the addition column by column, from right to left, considering potential carry-overs.

Let $c_1$ be the carry-over from the units column to the tens column.

Let $c_2$ be the carry-over from the tens column to the hundreds column.

Units Column:

Adding the digits in the units column: $A + A = 2A$. The units digit of the sum is 8.

This means $2A$ must end in 8. Since A is a single digit (0-9), the possible values for $2A$ are 8 or 18.

If $2A = 8$, then $A=4$. The carry-over to the tens column is $c_1 = 0$.

If $2A = 18$, then $A=9$. The carry-over to the tens column is $c_1 = 1$.

Hundreds Column:

Adding the digits in the hundreds column ($B + B$) plus the carry-over from the tens column ($c_2$), the sum is 3, with no carry-over to the thousands column.

$B + B + c_2 = 3$

$2B + c_2 = 3$

Since B is a non-zero single digit ($B \ge 1$), $2B \ge 2$. The carry-over $c_2$ from the tens column can be 0 or 1.

• If $c_2 = 0$, then $2B + 0 = 3 \implies 2B = 3$. This gives $B = 1.5$, which is not a single digit.
• If $c_2 = 1$, then $2B + 1 = 3 \implies 2B = 2$. This gives $B = 1$. This is a valid non-zero single digit.

So, from the hundreds column, we must have B = 1 and $c_2 = 1$.

Tens Column:

Adding the digits in the tens column ($A + A$) plus the carry-over from the units column ($c_1$), the sum gives the tens digit of the result (which is A) plus $10 \times c_2$ (carry-over to the hundreds column).

$A + A + c_1 = A + 10 \times c_2$

$2A + c_1 = A + 10c_2$

Subtracting A from both sides:

Substitute the value of $c_2 = 1$ that we found from the hundreds column:

$A + c_1 = 10 \times 1$

Now we use the possible values for A and $c_1$ from the units column analysis:

• Case 1: A = 4 and $c_1 = 0$. Substitute into equation (i): $4 + 0 = 10 \implies 4 = 10$. This is false.
• Case 2: A = 9 and $c_1 = 1$. Substitute into equation (i): $9 + 1 = 10 \implies 10 = 10$. This is true.

The only consistent values are $A=9$, $c_1=1$, $B=1$, and $c_2=1$.

Thus, the values of the letters are A = 9 and B = 1.

Let's verify the addition with A=9 and B=1:

BAA = 199

3A8 = 398

The addition is $199 + 199$.

$\begin{array}{cc} & 1 & 9 & 9 \\ + & 1 & 9 & 9 \\ \hline & 3 & 9 & 8 \\ \hline \end{array}$

Units: $9 + 9 = 18$. Write 8, carry 1. ($c_1=1$). Correct.

Tens: $9 + 9 + 1 (\text{carry}) = 19$. Write 9, carry 1. ($c_2=1$). The tens digit is A=9. Correct.

Hundreds: $1 + 1 + 1 (\text{carry}) = 3$. Write 3. The hundreds digit is 3. Correct.

The addition $199 + 199 = 398$ is correct and matches the pattern 3A8 with A=9.

The values of the letters are: A = 9 and B = 1.

The final answer is $\boxed{A=9, B=1}$.

Given:

The addition puzzle:

Where A and B are single digits (0-9).

Since 'A01B' is a four-digit number, $A \neq 0$.

Since 'B108' is a four-digit number, $B \neq 0$.

To Find:

The values of the letters A and B.

Solution:

We analyze the addition column by column, from right to left, considering potential carry-overs.

Let $c_1$ be the carry-over from the units column to the tens column.

Let $c_2$ be the carry-over from the tens column to the hundreds column.

Let $c_3$ be the carry-over from the hundreds column to the thousands column.

Units Column:

Adding the digits in the units column: $B + B$. The units digit of the sum is 8.

$B + B = 8 + 10 \times c_1$

Since B is a single digit (0-9), $2B$ can range from $2 \times 0 = 0$ to $2 \times 9 = 18$.

If $c_1 = 0$, then $2B = 8 \implies B = 4$. (Valid, $B \neq 0$)

If $c_1 = 1$, then $2B = 8 + 10 = 18 \implies B = 9$. (Valid, $B \neq 0$)

So, B can be 4 (with $c_1=0$) or 9 (with $c_1=1$).

Hundreds Column:

Adding the digits in the hundreds column ($0 + 0$) plus the carry-over from the tens column ($c_2$), the sum gives the hundreds digit of the result (1) plus $10 \times c_3$ (carry-over to the thousands column).

$0 + 0 + c_2 = 1 + 10 \times c_3$

Since $c_2$ is a carry-over from adding two single digits and a carry (max sum $9+9+1=19$), $c_2$ can only be 0 or 1.

If $c_2 = 0$, equation (ii) becomes $0 = 1 + 10c_3 \implies 10c_3 = -1$, which is impossible for integer $c_3$.

If $c_2 = 1$, equation (ii) becomes $1 = 1 + 10c_3 \implies 10c_3 = 0 \implies c_3 = 0$.

Thus, we must have $c_2 = 1$ and $c_3 = 0$.

Thousands Column:

Adding the digits in the thousands column ($A + 1$) plus the carry-over from the hundreds column ($c_3$), the sum gives the thousands digit of the result (B).

$A + 1 + c_3 = B$

Substitute the value of $c_3 = 0$:

Tens Column:

Adding the digits in the tens column ($1 + A$) plus the carry-over from the units column ($c_1$), the sum gives the tens digit of the result (0) plus $10 \times c_2$ (carry-over to the hundreds column).

$1 + A + c_1 = 0 + 10 \times c_2$

Substitute the value of $c_2 = 1$:

Now we use the possibilities for B and $c_1$ from the units column and the relationship $A+1=B$ from the thousands column.

Case 1: B = 4 and $c_1 = 0$

Using equation (iii), $A + 1 = B \implies A + 1 = 4 \implies A = 3$.

Check if A=3 and $c_1=0$ satisfy equation (v):

$1 + A + c_1 = 1 + 3 + 0 = 4$. Equation (v) requires this to be 10. $4 \neq 10$. So Case 1 is invalid.

Case 2: B = 9 and $c_1 = 1$

Using equation (iii), $A + 1 = B \implies A + 1 = 9 \implies A = 8$.

Check if A=8 and $c_1=1$ satisfy equation (v):

$1 + A + c_1 = 1 + 8 + 1 = 10$. Equation (v) requires this to be 10. $10 = 10$. So Case 2 is valid.

The values found are A = 8 and B = 9. These satisfy $A \neq 0$ and $B \neq 0$.

Let's verify the addition with A=8 and B=9:

A01B becomes 8019.

10AB becomes 1089.

B108 becomes 9108.

Performing the addition:

$\begin{array}{cc} & 8 & 0 & 1 & 9 \\ + & 1 & 0 & 8 & 9 \\ \hline & 9 & 1 & 0 & 8 \\ \hline \end{array}$

Units: $9 + 9 = 18$. Write 8, carry 1. Matches the units digit 8 and $c_1=1$.

Tens: $1 + 8 + 1 (\text{carry}) = 10$. Write 0, carry 1. Matches the tens digit 0 and $c_2=1$.

Hundreds: $0 + 0 + 1 (\text{carry}) = 1$. Write 1, carry 0. Matches the hundreds digit 1 and $c_3=0$.

Thousands: $8 + 1 + 0 (\text{carry}) = 9$. Write 9. Matches the thousands digit B=9.

The values are consistent and the addition is correct.

The values of the letters are: A = 8 and B = 9.

The final answer is $\boxed{A=8, B=9}$.

Given:

The multiplication puzzle:

Where A, B, and C are single digits (0-9).

Since 'AB' is a two-digit number, $A \neq 0$.

Since 'C68' is a three-digit number, $C \neq 0$.

To Find:

The values of the letters A, B, and C.

Solution:

The multiplication can be represented as $(10A + B) \times 6 = 100C + 68$.

We analyze the multiplication column by column, starting from the units place.

Let $c_1$ be the carry-over from the units column to the tens column.

Let $c_2$ be the carry-over from the tens column to the hundreds column.

Units Column:

The product of the units digits, $B \times 6$, must have a units digit of 8.

$B \times 6 = 8 + 10 \times c_1$

We check the single digits B (0-9):

• If $B=0, 0 \times 6 = 0$ (ends in 0)
• If $B=1, 1 \times 6 = 6$ (ends in 6)
• If $B=2, 2 \times 6 = 12$ (ends in 2)
• If $B=3, 3 \times 6 = 18$ (ends in 8). This gives $c_1=1$.
• If $B=4, 4 \times 6 = 24$ (ends in 4)
• If $B=5, 5 \times 6 = 30$ (ends in 0)
• If $B=6, 6 \times 6 = 36$ (ends in 6)
• If $B=7, 7 \times 6 = 42$ (ends in 2)
• If $B=8, 8 \times 6 = 48$ (ends in 8). This gives $c_1=4$.
• If $B=9, 9 \times 6 = 54$ (ends in 4)

The possible values for B are 3 or 8.

Tens Column:

The product of the tens digit (A) and 6, plus the carry-over from the units column ($c_1$), results in a number whose units digit is 6.

$A \times 6 + c_1 = 6 + 10 \times c_2$

This also means that $6A + c_1$ must end in 6.

Hundreds Column:

The carry-over from the tens column ($c_2$) becomes the hundreds digit C.

$C = c_2$

Since C is the first digit of a three-digit number, $C \neq 0$. Therefore, $c_2 \neq 0$, which means $c_2 \ge 1$.

From the tens column equation, $6A + c_1 = 6 + 10c_2$. Since $c_2 \ge 1$, the right side $6 + 10c_2 \ge 6 + 10(1) = 16$.

So, we must have $6A + c_1 \ge 16$.

Let's examine the possibilities for B and $c_1$ from the units column:

Case 1: B = 3 and $c_1 = 1$

Tens column condition: $6A + c_1$ ends in 6 $\implies 6A + 1$ ends in 6 $\implies 6A$ ends in 5.

We check the products of 6 with single digits (A=1 to 9): $6, 12, 18, 24, 30, 36, 42, 48, 54$. None of these end in 5.

So, this case is not possible.

Case 2: B = 8 and $c_1 = 4$

Tens column condition: $6A + c_1$ ends in 6 $\implies 6A + 4$ ends in 6 $\implies 6A$ ends in 2.

We check the products of 6 with single digits (A=1 to 9) that end in 2:

• $6 \times 2 = 12$ (ends in 2). Possible A = 2.
• $6 \times 7 = 42$ (ends in 2). Possible A = 7.

Now, we check the constraint $6A + c_1 \ge 16$ and find $c_2$ using $6A + c_1 = 6 + 10c_2$.

Subcase 2a: A = 2

$6A + c_1 = 6(2) + 4 = 12 + 4 = 16$.

Check constraint: $16 \ge 16$. This is true.

Find $c_2$: $16 = 6 + 10c_2 \implies 10c_2 = 10 \implies c_2 = 1$.

Hundreds column: $C = c_2 = 1$. $C=1$ is a valid non-zero digit.

This gives the solution: A = 2, B = 8, C = 1.

Check: $28 \times 6 = 168$. This matches C68 with C=1, 6, 8.

Subcase 2b: A = 7

$6A + c_1 = 6(7) + 4 = 42 + 4 = 46$.

Check constraint: $46 \ge 16$. This is true.

Find $c_2$: $46 = 6 + 10c_2 \implies 10c_2 = 40 \implies c_2 = 4$.

Hundreds column: $C = c_2 = 4$. $C=4$ is a valid non-zero digit.

This gives a second solution: A = 7, B = 8, C = 4.

Check: $78 \times 6 = 468$. This matches C68 with C=4, 6, 8.

Both solutions satisfy all the conditions derived from the puzzle.

The values of the letters can be A = 2, B = 8, C = 1 or A = 7, B = 8, C = 4.

The final answer is $\boxed{A=2, B=8, C=1 \text{ or } A=7, B=8, C=4}$.

Given:

The multiplication puzzle:

Where A and B are single digits (0-9).

Since 'AB' is a two-digit number, $A \neq 0$.

Since '6AB' is a three-digit number, the hundreds digit is 6.

To Find:

The values of the letters A and B.

Solution:

The two-digit number 'AB' can be written as $10A + B$.

The result '6AB' can be written as $600 + 10A + B$.

The puzzle represents the multiplication $(10A + B) \times (10A + B) = 600 + 10A + B$.

Let $N$ be the number 'AB'. So, $N = 10A + B$.

The equation becomes $N \times N = 600 + N$, or $N^2 = 600 + N$.

Rearrange the equation into a standard quadratic form:

$N^2 - N - 600 = 0$

We can solve this quadratic equation for N using the quadratic formula $N = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1$, $b=-1$, and $c=-600$.

$N = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-600)}}{2(1)}$

$N = \frac{1 \pm \sqrt{1 + 2400}}{2}$

$N = \frac{1 \pm \sqrt{2401}}{2}$

We need to find the square root of 2401.

We know that $40^2 = 1600$ and $50^2 = 2500$. The number 2401 is between 1600 and 2500. Since it ends in 1, its square root must end in 1 or 9.

Let's try $49^2$:

So, $\sqrt{2401} = 49$.

Substitute this value back into the formula for N:

$N = \frac{1 \pm 49}{2}$

This gives two possible values for N:

$N_1 = \frac{1 + 49}{2} = \frac{50}{2} = 25$

$N_2 = \frac{1 - 49}{2} = \frac{-48}{2} = -24$

The number $N = 10A + B$ represents a two-digit number 'AB'. Since A is the tens digit and $A \neq 0$, N must be a positive integer between 10 and 99.

$N_1 = 25$ is a valid value in this range.

$N_2 = -24$ is not a valid value for a two-digit number.

Using $N = 25$, we have $10A + B = 25$. Comparing this with the place value representation, we get:

A = 2

B = 5

These are single digits, and $A=2$ is non-zero, which satisfies the conditions.

Let's verify these values in the original multiplication puzzle:

AB becomes 25.

6AB becomes 625.

The multiplication is $25 \times 25$.

$25 \times 25 = 625$.

The result 625 matches the pattern 6AB with A=2 and B=5. The values are consistent.

The values of the letters are: A = 2 and B = 5.

The final answer is $\boxed{A=2, B=5}$.

Given:

The multiplication puzzle:

Where A, B, and C are single digits (0-9).

Since 'AA' is a two-digit number, $A \neq 0$.

Since 'CAB' is a three-digit number, $C \neq 0$.

To Find:

The values of the letters A, B, and C.

Solution:

The multiplication can be represented as $(10A + A) \times A = 100C + 10A + B$.

$11A \times A = 100C + 10A + B$

$11A^2 = 100C + 10A + B$

We analyze the multiplication column by column, considering potential carry-overs.

Let $c_1$ be the carry-over from the units column to the tens column.

Let $c_2$ be the carry-over from the tens column to the hundreds column.

Units Column:

Multiplying the units digit A by A gives a product whose units digit is B and whose tens digit is $c_1$.

$A \times A = B + 10 \times c_1$

Since A is a non-zero digit (1-9) and B is a digit (0-9), we list the possible values for $A^2$ and the resulting B and $c_1$:

Hundreds Column:

The carry-over from the tens column ($c_2$) becomes the hundreds digit C.

$c_2 = C$

Since C is a non-zero digit, $C \ge 1$. Therefore, $c_2 \ge 1$.

Tens Column:

Multiplying the tens digit A (from 'AA') by A, and adding the carry-over from the units column ($c_1$), the result gives the tens digit A and $10 \times c_2$ (carry-over to the hundreds column).

$A \times A + c_1 = A + 10 \times c_2$

Substitute $c_2 = C$ from equation (ii) into equation (iii):

$A^2 + c_1 = A + 10C$

$A^2 - A + c_1 = 10C$

From equation (iv), $A^2 - A + c_1$ must be a multiple of 10. Also, since $C \ge 1$, $A^2 - A + c_1 \ge 10$.

Let's use the values of A and $c_1$ from the units column analysis and check which ones satisfy $A^2 - A + c_1$ being a multiple of 10 and $c_2 \ge 1$ (which means $C \ge 1$).

• If A = 1, $c_1 = 0$. $A^2 - A + c_1 = 1^2 - 1 + 0 = 0$. $10C = 0 \implies C=0$. Invalid (C must be non-zero).
• If A = 2, $c_1 = 0$. $A^2 - A + c_1 = 2^2 - 2 + 0 = 2$. $10C = 2 \implies C=0.2$. Invalid.
• If A = 3, $c_1 = 0$. $A^2 - A + c_1 = 3^2 - 3 + 0 = 6$. $10C = 6 \implies C=0.6$. Invalid.
• If A = 4, $c_1 = 1$. $A^2 - A + c_1 = 4^2 - 4 + 1 = 16 - 4 + 1 = 13$. $10C = 13 \implies C=1.3$. Invalid.
• If A = 5, $c_1 = 2$. $A^2 - A + c_1 = 5^2 - 5 + 2 = 25 - 5 + 2 = 22$. $10C = 22 \implies C=2.2$. Invalid.
• If A = 6, $c_1 = 3$. $A^2 - A + c_1 = 6^2 - 6 + 3 = 36 - 6 + 3 = 33$. $10C = 33 \implies C=3.3$. Invalid.
• If A = 7, $c_1 = 4$. $A^2 - A + c_1 = 7^2 - 7 + 4 = 49 - 7 + 4 = 46$. $10C = 46 \implies C=4.6$. Invalid.
• If A = 8, $c_1 = 6$. $A^2 - A + c_1 = 8^2 - 8 + 6 = 64 - 8 + 6 = 62$. $10C = 62 \implies C=6.2$. Invalid.
• If A = 9, $c_1 = 8$. $A^2 - A + c_1 = 9^2 - 9 + 8 = 81 - 9 + 8 = 72 + 8 = 80$. $10C = 80 \implies C=8$. This is a valid non-zero digit.

The only possibility that yields a valid integer for C is A = 9.

If A = 9, then from the units column analysis, $B=1$ and $c_1=8$.

From the hundred column analysis, $C = (A^2 - A + c_1)/10 = (9^2 - 9 + 8)/10 = 80/10 = 8$. So $C=8$. Also $c_2 = C = 8$.

Let's verify these values (A=9, B=1, C=8) in the original puzzle:

AA = 99

A = 9

CAB = 891

We perform the multiplication $99 \times 9$:

$9 \times 9 = 81$. Write down 1 (B=1), carry 8 ($c_1=8$).

$9 \times 9 + \text{carry } 8 = 81 + 8 = 89$. Write down 9 (A=9), carry 8 ($c_2=8$).

The carry 8 becomes the hundreds digit C. So C=8.

The result is 891, which matches CAB with C=8, A=9, B=1. The values are consistent.

The values of the letters are: A = 9, B = 1, and C = 8.

The final answer is $\boxed{A=9, B=1, C=8}$.

Given:

The subtraction puzzle:

Where A and B are single digits (0-9).

Since 'AB' is a two-digit number, $A \neq 0$.

Since 'B7' is a two-digit number, $B \neq 0$.

To Find:

The values of the letters A and B.

Solution:

We analyze the subtraction column by column, from right to left, considering borrowing.

Units Column:

We have the subtraction $B - 7$. The units digit of the result is 5.

If we directly subtract B - 7, it would result in 5. For example, if B=12, $12-7=5$. Since B is a single digit, $B-7=5$ implies $B=12$, which is not a single digit.

This means we must have borrowed from the tens column of the top number.

So, the calculation in the units column is $(10 + B) - 7 = 5$.

Solve equation (i) for B:

$3 + B = 5$

$B = 5 - 3$

$B = 2$

This is a valid single digit and satisfies the condition $B \neq 0$.

Since we performed $(10 + B) - 7$, we borrowed 1 from the tens column of the top number (AB).

Tens Column:

The original tens digit in the top number was A. However, we borrowed 1 from A for the units column. So, the effective digit in the tens column of the top number is $A - 1$.

The subtraction in the tens column is $(A - 1) - B$. The result in the tens column is 4.

$(A - 1) - B = 4$

Substitute the value of B = 2 into equation (ii):

$(A - 1) - 2 = 4$

$A - 3 = 4$

$A = 4 + 3$

$A = 7$

This is a valid single digit and satisfies the condition $A \neq 0$.

We have found the values: A = 7 and B = 2.

Let's verify these values in the original subtraction:

AB becomes 72.

B7 becomes 27.

The subtraction is $72 - 27$.

$\begin{array}{cc} & 7 & 2 \\ - & 2 & 7 \\ \hline & 4 & 5 \\ \hline \end{array}$

Units column: $2 - 7$. We borrow from the tens place. 12 - 7 = 5. Correct.

Tens column: The 7 in the tens place becomes $7-1=6$ after borrowing. $6 - 2 = 4$. Correct.

The result 45 matches the puzzle. The values are consistent.

The values of the letters are: A = 7 and B = 2.

The final answer is $\boxed{A=7, B=2}$.

Given:

The subtraction puzzle:

Where A, B, C, and D are single digits (0-9).

Since 'ABC5' is a four-digit number, $A \neq 0$.

Since 'D488' is a four-digit number, $D \neq 0$.

To Find:

The values of the letters A, B, C, and D.

Solution:

We analyze the subtraction column by column, from right to left, considering borrowing.

Units Column:

In the units column, we have $C - 5$ resulting in a units digit of 8. This is only possible if we borrow from the tens column of the top number. The calculation is $(10 + C) - 5 = 8$.

$10 + C - 5 = 8$

$5 + C = 8$

$C = 8 - 5$

$C = 3$

A borrow of 1 was taken from the tens digit (B) of the top number.

Tens Column:

In the tens column, we originally had B, but 1 was borrowed for the units column, so we have $B - 1$. The subtraction is $(B - 1) - C$. This results in a units digit of 8.

Substitute $C = 3$: $(B - 1) - 3 = B - 4$. The units digit of $B - 4$ is 8. This is only possible if we borrow from the hundreds column of the top number.

The calculation is $(10 + (B - 1)) - 3 = 8$.

$10 + B - 1 - 3 = 8$

$B + 6 = 8$

$B = 8 - 6$

$B = 2$

A borrow of 1 was taken from the hundreds digit (A) of the top number.

Hundreds Column:

In the hundreds column, we originally had A, but 1 was borrowed for the tens column, so we have $A - 1$. The subtraction is $(A - 1) - B$. This results in a hundreds digit of 4.

Substitute $B = 2$: $(A - 1) - 2 = 4$.

$A - 3 = 4$

$A = 4 + 3$

$A = 7$

Since $A=7$, the calculation in the hundreds column is $(7 - 1) - 2 = 6 - 2 = 4$. This result is 4, which matches the hundreds digit of the difference. No borrow is needed from the thousands column (8) of the top number.

The value $A=7$ is a valid non-zero digit.

Thousands Column:

In the thousands column, we have 8 (from the top number) minus A (from the bottom number). No borrow was taken from the 8.

The calculation is $8 - A$. This results in the thousands digit D.

Substitute $A = 7$: $8 - 7 = D$.

$D = 1$

The value $D=1$ is a valid non-zero digit.

We have found the values: A = 7, B = 2, C = 3, and D = 1.

Let's verify these values in the original subtraction puzzle:

8ABC becomes 8723.

ABC5 becomes 7235.

D488 becomes 1488.

Perform the subtraction $8723 - 7235$:

Units: $3 - 5$. Borrow 1 from 2 (making it 1). $13 - 5 = 8$. Correct.

Tens: $1 - 3$. Borrow 1 from 7 (making it 6). $11 - 3 = 8$. Correct.

Hundreds: $6 - 2 = 4$. Correct.

Thousands: $8 - 7 = 1$. Correct.

The result 1488 matches D488 with D=1. All values are consistent and satisfy the conditions.

The values of the letters are: A = 7, B = 2, C = 3, and D = 1.

The final answer is $\boxed{A=7, B=2, C=3, D=1}$.

Given:

The equation: 2A7 $\div$ A = 33

Another equation: B – A = 1

A and B are single digits.

From the context of 2A7 being a three-digit number, A is a digit in the tens place. From the division by A, A cannot be 0.

To Find:

The value of the letter A.

Solution:

The number 2A7 is a three-digit number where the digit A is in the tens place. This number can be written in expanded form as $2 \times 100 + A \times 10 + 7 = 200 + 10A + 7$.

The given equation is:

$\frac{200 + 10A + 7}{A} = 33$

$\frac{207 + 10A}{A} = 33$

Multiply both sides by A (since A $\neq$ 0):

$207 + 10A = 33 \times A$

$207 + 10A = 33A$

Subtract $10A$ from both sides of the equation:

$207 = 33A - 10A$

$207 = 23A$

To find A, divide 207 by 23:

$A = \frac{207}{23}$

Performing the division:

$207 \div 23 = 9$

So, A = 9.

The value A=9 is a single digit (0-9) and is not 0, which fits the context of the puzzle.

The second equation B – A = 1 is not needed to find the value of A. If we were asked for B, using A=9 would give B – 9 = 1, so B = 10, which is not a single digit. However, the question specifically asks only for the value of A.

The value of the letter A is 9.

The final answer is $\boxed{9}$.

Given:

The number 212x5 is a multiple of both 3 and 11.

x is a single digit (0-9).

To Find:

The value of the digit x.

Solution:

Since the number 212x5 is a multiple of both 3 and 11, it must be divisible by both 3 and 11.

Divisibility by 3:

A number is divisible by 3 if the sum of its digits is divisible by 3.

The digits of the number 212x5 are 2, 1, 2, x, and 5.

The sum of the digits is $2 + 1 + 2 + x + 5 = 10 + x$.

For the number to be divisible by 3, $10 + x$ must be divisible by 3.

Since x is a single digit (0-9), the possible values for $10+x$ are $10, 11, 12, 13, 14, 15, 16, 17, 18, 19$.

The values in this list that are divisible by 3 are 12 and 15 and 18.

• If $10 + x = 12$, then $x = 2$.
• If $10 + x = 15$, then $x = 5$.
• If $10 + x = 18$, then $x = 8$.

So, the possible values for x based on divisibility by 3 are 2, 5, or 8.

Divisibility by 11:

A number is divisible by 11 if the alternating sum of its digits is divisible by 11.

For the number 212x5, the alternating sum of the digits (starting from the rightmost digit with a positive sign) is:

$5 - x + 2 - 1 + 2$

$(5 + 2 - 1 + 2) - x = 8 - x$

For the number to be divisible by 11, $8 - x$ must be divisible by 11.

Since x is a single digit (0-9), $8-x$ can range from $8-9 = -1$ to $8-0 = 8$.

The only number in the range $[-1, 8]$ that is divisible by 11 is 0.

So, we must have $8 - x = 0$.

$x = 8$

We need to find the value of x that satisfies both conditions.

From divisibility by 3, x can be 2, 5, or 8.

From divisibility by 11, x must be 8.

The only value common to both sets of possibilities is 8.

Thus, the value of x is 8.

Let's check the number with x=8: 21285.

Sum of digits: $2+1+2+8+5 = 18$. 18 is divisible by 3.

Alternating sum of digits: $5-8+2-1+2 = (5+2+2) - (8+1) = 9 - 9 = 0$. 0 is divisible by 11.

Since 21285 is divisible by both 3 and 11, it is divisible by 33. The value x=8 is correct.

The final answer is $\boxed{8}$.

Given:

The number 31k2 is divisible by 6.

k is a single digit (0-9).

To Find:

The value of the digit k.

Solution:

A number is divisible by 6 if and only if it is divisible by both 2 and 3, since $6 = 2 \times 3$ and 2 and 3 are coprime.

The given number is 31k2.

Divisibility by 2:

A number is divisible by 2 if its units digit is even.

The units digit of 31k2 is 2.

Since 2 is an even number, the number 31k2 is always divisible by 2, regardless of the value of k.

Divisibility by 3:

A number is divisible by 3 if the sum of its digits is divisible by 3.

The digits of the number 31k2 are 3, 1, k, and 2.

The sum of the digits is $3 + 1 + k + 2 = 6 + k$.

For the number to be divisible by 3, $6 + k$ must be divisible by 3.

Since 6 is divisible by 3, for $6+k$ to be divisible by 3, k must also be divisible by 3.

Since k is a single digit (0-9), the possible values for k that are divisible by 3 are 0, 3, 6, or 9.

For the number 31k2 to be divisible by 6, it must satisfy both the divisibility rule for 2 and the divisibility rule for 3.

The divisibility rule for 2 is always satisfied by the number 31k2.

The divisibility rule for 3 is satisfied when k is 0, 3, 6, or 9.

Therefore, the possible values for k are 0, 3, 6, and 9.

The question asks for "the value of k" (singular). This implies there might be only one possible value in the context of the problem, or the question is implicitly asking for all possible values. Assuming it means "all possible values", we list them.

The values of k for which 31k2 is divisible by 6 are 0, 3, 6, and 9.

The final answer is $\boxed{0, 3, 6, 9}$.

Given:

The number 1y3y6 is divisible by 11.

y is a single digit (0-9).

To Find:

The value of the digit y.

Solution:

A number is divisible by 11 if the alternating sum of its digits is divisible by 11.

The digits of the number 1y3y6 are 1, y, 3, y, and 6.

We calculate the alternating sum of the digits, starting from the rightmost digit (6) with a positive sign:

$6 - y + 3 - y + 1$

$(6 + 3 + 1) - (y + y)$

$10 - 2y$

For the number 1y3y6 to be divisible by 11, the alternating sum $10 - 2y$ must be divisible by 11.

Since y is a single digit (0-9), we find the possible values for $10 - 2y$:

• If y = 0, $10 - 2(0) = 10$.
• If y = 1, $10 - 2(1) = 8$.
• If y = 2, $10 - 2(2) = 6$.
• If y = 3, $10 - 2(3) = 4$.
• If y = 4, $10 - 2(4) = 2$.
• If y = 5, $10 - 2(5) = 0$.
• If y = 6, $10 - 2(6) = -2$.
• If y = 7, $10 - 2(7) = -4$.
• If y = 8, $10 - 2(8) = -6$.
• If y = 9, $10 - 2(9) = -8$.

The range of possible values for $10 - 2y$ is $[-8, 10]$.

The only value in this range that is divisible by 11 is 0.

So, we must have $10 - 2y = 0$.

$10 = 2y$

$y = \frac{10}{2}$

$y = 5$

The value y=5 is a single digit, which is valid.

Let's check the number with y=5: 15356.

Alternating sum of digits: $6 - 5 + 3 - 5 + 1 = (6+3+1) - (5+5) = 10 - 10 = 0$.

Since the alternating sum is 0, the number 15356 is divisible by 11. The value y=5 is correct.

The value of y is 5.

The final answer is $\boxed{5}$.

Given:

The number 756x is a multiple of 11.

x is a single digit (0-9).

To Find:

The value of the digit x.

Solution:

A number is divisible by 11 if the alternating sum of its digits is divisible by 11.

The digits of the number 756x are 7, 5, 6, and x.

We calculate the alternating sum of the digits, starting from the rightmost digit (x) with a positive sign:

$x - 6 + 5 - 7$

$x + (5 - 6 - 7)$

$x + (5 - 13)$

$x - 8$

For the number 756x to be divisible by 11, the alternating sum $x - 8$ must be divisible by 11.

Since x is a single digit (0-9), we find the possible values for $x - 8$:

• If x = 0, $0 - 8 = -8$.
• If x = 1, $1 - 8 = -7$.
• If x = 2, $2 - 8 = -6$.
• If x = 3, $3 - 8 = -5$.
• If x = 4, $4 - 8 = -4$.
• If x = 5, $5 - 8 = -3$.
• If x = 6, $6 - 8 = -2$.
• If x = 7, $7 - 8 = -1$.
• If x = 8, $8 - 8 = 0$.
• If x = 9, $9 - 8 = 1$.

The range of possible values for $x - 8$ is $[-8, 1]$.

The only value in this range that is divisible by 11 is 0.

So, we must have $x - 8 = 0$.

$x = 8$

The value x=8 is a single digit, which is valid.

Let's check the number with x=8: 7568.

Alternating sum of digits: $8 - 6 + 5 - 7 = (8+5) - (6+7) = 13 - 13 = 0$.

Since the alternating sum is 0, the number 7568 is divisible by 11. The value x=8 is correct.

The value of x is 8.

The final answer is $\boxed{8}$.

Given:

The sum of two numbers: $2a3 + 326 = 5b9$

The resulting number 5b9 is divisible by 9.

a and b are single digits (0-9).

To Find:

The value of b – a.

Solution:

First, let's perform the addition $2a3 + 326 = 5b9$. We can write this vertically:

We analyze the addition column by column, from right to left, considering potential carry-overs.

Let $c_1$ be the carry-over from the units column to the tens column.

Let $c_2$ be the carry-over from the tens column to the hundreds column.

Units Column:

$3 + 6 = 9$.

The units digit of the result is 9, which matches. The carry-over to the tens column is $c_1 = 0$.

Hundreds Column:

The sum of the digits in the hundreds column plus the carry-over from the tens column ($c_2$) results in 5.

$2 + 3 + c_2 = 5$

$5 + c_2 = 5$

$c_2 = 5 - 5$

$c_2 = 0$

The carry-over to the hundreds column is $c_2 = 0$. This matches the thousands digit of the result (which is 0, as it's a three-digit number).

Tens Column:

The sum of the digits in the tens column (a and 2) plus the carry-over from the units column ($c_1$) gives the tens digit of the result (b) plus $10 \times c_2$ (carry-over to the hundreds column).

$a + 2 + c_1 = b + 10 \times c_2$

Substitute the values $c_1 = 0$ and $c_2 = 0$:

$a + 2 + 0 = b + 10 \times 0$

$a + 2 = b$

$b - a = 2$

Now we use the condition that the resulting number 5b9 is divisible by 9.

A number is divisible by 9 if the sum of its digits is divisible by 9.

The digits of 5b9 are 5, b, and 9.

The sum of the digits is $5 + b + 9 = 14 + b$.

For 5b9 to be divisible by 9, $14 + b$ must be divisible by 9.

Since b is a single digit (0-9), the possible values for $14+b$ range from $14+0=14$ to $14+9=23$.

The multiples of 9 in this range are 18.

So, we must have $14 + b = 18$.

$b = 18 - 14$

$b = 4$

The value b=4 is a single digit, which is valid.

Now that we have the value of b, we can find the value of a using equation (i):

$b - a = 2$

$4 - a = 2$

$a = 4 - 2$

$a = 2$

The value a=2 is a single digit, which is valid. The number 2a3 becomes 223.

Let's check the addition with a=2 and b=4:

2a3 becomes 223.

5b9 becomes 549.

The addition is $223 + 326 = 549$. This matches $2a3 + 326 = 5b9$ with a=2 and b=4.

Now check if 549 is divisible by 9. The sum of digits of 549 is $5+4+9 = 18$. 18 is divisible by 9, so 549 is divisible by 9. This condition is also satisfied.

We are asked to find the value of b – a.

b – a = 4 – 2 = 2.

The final answer is $\boxed{2}$.

Given:

The addition puzzle:

$\begin{array}{cccccc} & & B & A & S & E \\ + & & B & A & L & L \\ \hline & G & A & M & E & S \\ \hline \end{array}$

Given values: E = 3, B = 7, A = 4.

A, B, E, S, L, G, M represent unique digits from 0 to 9.

To Find:

The values of the other digits S, L, G, and M.

Solution:

Substitute the given values into the puzzle:

$\begin{array}{cccccc} & & 7 & 4 & S & 3 \\ + & & 7 & 4 & L & L \\ \hline & G & 4 & M & 3 & S \\ \hline \end{array}$

We analyze the addition column by column, from right to left, considering carry-overs.

Let $c_1, c_2, c_3, c_4$ be the carry-overs from the units, tens, hundreds, and thousands columns respectively.

Units Column:

$E + L = S + 10 \times c_1$

$3 + L = S + 10c_1$

Since L and S are digits, $3+L$ can be at most $3+9=12$. So $c_1$ can be 0 or 1.

Tens Column:

$S + L + c_1 = E + 10 \times c_2$

$S + L + c_1 = 3 + 10c_2$

Since S, L, $c_1$ are small, $S+L+c_1$ can be at most $9+9+1=19$. So $c_2$ can be 0 or 1.

Hundreds Column:

$A + A + c_2 = M + 10 \times c_3$

$4 + 4 + c_2 = M + 10c_3$

$8 + c_2 = M + 10c_3$

Since $c_2$ is 0 or 1, $8+c_2$ is 8 or 9. Since M is a digit, $10c_3$ must be 0. So $c_3 = 0$.

Substitute $c_3=0$ into the hundreds column equation:

$8 + c_2 = M + 10(0)$

$8 + c_2 = M$

If $c_2=0$, M=8. If $c_2=1$, M=9.

Thousands Column:

$B + B + c_3 = A + 10 \times c_4$

$7 + 7 + c_3 = 4 + 10c_4$

$14 + c_3 = 4 + 10c_4$

Substitute $c_3 = 0$:

$14 + 0 = 4 + 10c_4$

$14 = 4 + 10c_4$

$10 = 10c_4$

$c_4 = 1$

Ten Thousands Column:

The carry-over $c_4$ becomes the digit G.

$G = c_4$

Since $c_4 = 1$, G = 1.

We know E=3, B=7, A=4, G=1. These are 3, 7, 4, 1. They are unique so far.

Now we need to find S, L, M using equations (i) and (ii) and the possible values for $c_1$ and $c_2$.

From $8+c_2=M$, we know that if $c_2=0$, M=8, and if $c_2=1$, M=9.

From equation (ii): $S + L + c_1 = 3 + 10c_2$.

Case 1: $c_2 = 0$ (implies M=8)

Equation (ii) becomes $S + L + c_1 = 3 + 10(0) = 3$.

Possible values for $c_1$ (0 or 1):

If $c_1=0$: $S + L = 3$. From equation (i): $3 + L = S + 10(0) \implies S = L+3$. Substitute $S$ into $S+L=3$: $(L+3) + L = 3 \implies 2L+3=3 \implies 2L=0 \implies L=0$. Then $S=0+3=3$. But $S$ cannot be equal to E=3 (digits must be unique). So this possibility is invalid.

If $c_1=1$: $S + L + 1 = 3 \implies S + L = 2$. From equation (i): $3 + L = S + 10(1) \implies S = L+3-10 \implies S = L-7$. Substitute $S$ into $S+L=2$: $(L-7) + L = 2 \implies 2L-7=2 \implies 2L=9 \implies L=4.5$. This is not an integer digit. So this possibility is invalid.

Case 1 ($c_2=0$, M=8) leads to no valid integer solutions for L and S.

Case 2: $c_2 = 1$ (implies M=9)

Equation (ii) becomes $S + L + c_1 = 3 + 10(1) = 13$.

Possible values for $c_1$ (0 or 1):

If $c_1=0$: $S + L = 13$. From equation (i): $3 + L = S + 10(0) \implies S = L+3$. Substitute $S$ into $S+L=13$: $(L+3) + L = 13 \implies 2L+3=13 \implies 2L=10 \implies L=5$. Then $S=5+3=8$.

Let's check if these values are unique and valid: A=4, B=7, E=3, G=1. New values are L=5, S=8, M=9. The set of digits is {1, 3, 4, 5, 7, 8, 9}. All are unique and valid single digits.

This combination seems promising: A=4, B=7, E=3, G=1, L=5, S=8, M=9.

If $c_1=1$: $S + L + 1 = 13 \implies S + L = 12$. From equation (i): $3 + L = S + 10(1) \implies S = L+3-10 \implies S = L-7$. Substitute $S$ into $S+L=12$: $(L-7) + L = 12 \implies 2L-7=12 \implies 2L=19 \implies L=9.5$. This is not an integer digit. So this possibility is invalid.

Case 2 ($c_2=1$, M=9) yields a unique valid integer solution when $c_1=0$: L=5, S=8.

The values derived are A=4, B=7, E=3, G=1, L=5, S=8, M=9.

Let's verify the original sum with these values:

BASE = 7483

BALL = 7455

GAMES = 14938

Adding 7483 and 7455:

$\begin{array}{cccccc} & & 7 & 4 & 8 & 3 \\ + & & 7 & 4 & 5 & 5 \\ \hline & 1 & 4 & 9 & 3 & 8 \\ \hline \end{array}$

Units: $3 + 5 = 8$. Matches S=8. Carry $c_1=0$. Correct.

Tens: $8 + 5 + c_1 = 8 + 5 + 0 = 13$. Units digit is 3 (matches E=3). Carry $c_2=1$. Correct.

Hundreds: $4 + 4 + c_2 = 4 + 4 + 1 = 9$. Units digit is 9 (matches M=9). Carry $c_3=0$. Correct.

Thousands: $7 + 7 + c_3 = 7 + 7 + 0 = 14$. Units digit is 4 (matches A=4). Carry $c_4=1$. Correct.

Ten Thousands: $c_4 = 1$. Digit is 1 (matches G=1). Correct.

The sum is 14938, which matches GAMES with G=1, A=4, M=9, E=3, S=8. The values are consistent.

The given digits are E=3, B=7, A=4.

The other digits are S, L, G, and M.

We found S=8, L=5, G=1, and M=9.

The final answer is $\boxed{S=8, L=5, G=1, M=9}$.

Given:

The addition puzzle:

$\begin{array}{@{}c@{\,}c@{}c@{}c@{}c} & & M & A & D \\ & + & & A & S \\ & + & & & A \\ \hline & B & U & L & L \\ \hline \end{array}$

Given values: D = 3, L = 7, A = 8.

M, A, D, S, B, U, L represent unique digits from 0 to 9.

To Find:

The values of the other digits M, S, B, and U.

Solution:

Substitute the given values into the puzzle:

$\begin{array}{@{}c@{\,}c@{}c@{}c@{}c} & & M & 8 & 3 \\ & + & & 8 & S \\ & + & & & 8 \\ \hline & B & U & 7 & 7 \\ \hline \end{array}$

We analyze the addition column by column, from right to left, considering carry-overs.

Let $c_1, c_2, c_3$ be the carry-overs from the units, tens, and hundreds columns respectively.

Units Column:

$D + S + A = L + 10 \times c_1$

$3 + S + 8 = 7 + 10c_1$

$11 + S = 7 + 10c_1$

$4 + S = 10c_1$

Since S is a single digit (0-9), $4+S$ can range from $4+0=4$ to $4+9=13$.

The only multiple of 10 in this range is 10.

So, we must have $4 + S = 10$.

$S = 10 - 4$

$S = 6$

If $4+S=10$, then $10c_1=10$, so $c_1=1$.

We know D=3, L=7, A=8, S=6. These are {3, 7, 8, 6}. They are unique so far.

Tens Column:

The sum of the digits in the tens column ($A + A$) plus the carry-over from the units column ($c_1$) gives the tens digit of the result (L) plus $10 \times c_2$ (carry-over to the hundreds column).

$A + A + c_1 = L + 10 \times c_2$

$8 + 8 + c_1 = 7 + 10c_2$

$16 + c_1 = 7 + 10c_2$

Substitute $c_1 = 1$:

$16 + 1 = 7 + 10c_2$

$17 = 7 + 10c_2$

$10 = 10c_2$

$c_2 = 1$

Hundreds Column:

The sum of the digits in the hundreds column (M and implicit 0s) plus the carry-over from the tens column ($c_2$) gives the hundreds digit of the result (U) plus $10 \times c_3$ (carry-over to the thousands column).

$M + 0 + 0 + c_2 = U + 10 \times c_3$

$M + c_2 = U + 10c_3$

Substitute $c_2 = 1$:

$M + 1 = U + 10c_3$

Since M and U are single digits, $M+1$ can range from $0+1=1$ to $9+1=10$.

If $c_3 = 0$, then $M + 1 = U$. This implies $U$ is one greater than M. Possible pairs (M, U) are (0,1), (1,2), (2,3), ..., (8,9).

If $c_3 = 1$, then $M + 1 = U + 10$. Since $M \le 9$ and $U \ge 0$, $M+1 \le 10$ and $U+10 \ge 10$. This equation requires $M+1 \ge 10$, so $M=9$ and $10=U+10 \implies U=0$. Possible pair (M, U) is (9,0).

Thousands Column:

The carry-over from the hundreds column ($c_3$) becomes the thousands digit B.

$c_3 = B$

Since B is the first digit of the result (BULL), $B \neq 0$. Thus, $c_3 \neq 0$, so $c_3=1$.

Therefore, B = 1.

From the hundreds column equation $M+1 = U + 10c_3$, substitute $c_3=1$: $M+1 = U + 10$. As derived before, this requires $M=9$ and $U=0$.

So, M = 9 and U = 0.

We have found the values: D=3, L=7, A=8, S=6, B=1, U=0, M=9.

Let's check if these digits are unique: {3, 7, 8, 6, 1, 0, 9}. Yes, they are all unique single digits.

Let's verify the addition with these values:

MAD = 983

AS = 86

A = 8

BULL = 1077

Adding 983 + 86 + 8:

$\begin{array}{cc} & 983 \\ + & \phantom{9}86 \\ + & \phantom{99}8 \\ \hline 1077 \\ \hline \end{array}$

Units: $3 + 6 + 8 = 17$. Units digit is 7 (matches L=7). Carry $c_1=1$. Correct.

Tens: $8 + 8 + c_1 = 8 + 8 + 1 = 17$. Units digit is 7 (matches L=7). Carry $c_2=1$. Correct.

Hundreds: $9 + 0 + 0 + c_2 = 9 + 1 = 10$. Units digit is 0 (matches U=0). Carry $c_3=1$. Correct.

Thousands: $c_3 = 1$. Digit is 1 (matches B=1). Correct.

The sum is 1077, which matches BULL with B=1, U=0, L=7, L=7. The values are consistent.

The given digits are D=3, L=7, A=8.

The other digits are M, S, B, and U.

We found M=9, S=6, B=1, and U=0.

The final answer is $\boxed{M=9, S=6, B=1, U=0}$.

Problem:

Find the count and list of all two-digit numbers such that the difference between the number and the number formed by reversing its digits is a perfect cube.

Solution:

Let the two-digit number be $N = 10a + b$, where $a$ is the tens digit and $b$ is the units digit.

Since it is a two-digit number, $a$ must be a non-zero digit, so $a \in \{1, 2, \dots, 9\}$.

The units digit $b$ can be any digit from 0 to 9, so $b \in \{0, 1, \dots, 9\}$.

The number formed by reversing the digits is $N_{rev} = 10b + a$.

The difference between the original number and the reversed number is $N - N_{rev}$.

$N - N_{rev} = (10a + b) - (10b + a)$

$N - N_{rev} = 10a + b - 10b - a$

$N - N_{rev} = 9a - 9b = 9(a - b)$.

We are given that this difference is a perfect cube. A perfect cube is an integer $k^3$ where $k$ is an integer.

Let $9(a - b) = k^3$.

Let $d = a - b$. The difference $d$ between the digits $a$ and $b$ can range from $1 - 9 = -8$ to $9 - 0 = 9$.

So, we have the condition $9d = k^3$, where $d \in \{-8, -7, \dots, 8, 9\}$.

For $9d = 3^2 d$ to be a perfect cube $k^3$, the prime factorization of $d$ must include $3^1$ and possibly other factors that, when combined with $3^1$, form a perfect cube. This means $d$ must be of the form $3m^3$ for some integer $m$.

We check possible integer values for $m$ such that $d = 3m^3$ falls within the range $[-8, 9]$:

• If $m = 0$, $d = 3 \times 0^3 = 0$. This is in the range $[-8, 9]$. The perfect cube is $9 \times 0 = 0 = 0^3$.
• If $m = 1$, $d = 3 \times 1^3 = 3$. This is in the range $[-8, 9]$. The perfect cube is $9 \times 3 = 27 = 3^3$.
• If $m = 2$, $d = 3 \times 2^3 = 3 \times 8 = 24$. This is outside the range $[-8, 9]$.
• If $m = -1$, $d = 3 \times (-1)^3 = -3$. This is in the range $[-8, 9]$. The perfect cube is $9 \times (-3) = -27 = (-3)^3$.
• If $m = -2$, $d = 3 \times (-2)^3 = 3 \times (-8) = -24$. This is outside the range $[-8, 9]$.

The possible values for the difference between the digits, $d = a - b$, are 0, 3, and -3.

We now find the two-digit numbers for each possible value of $a - b$, where $a \in \{1, \dots, 9\}$ and $b \in \{0, \dots, 9\}$.

Case 1: $a - b = 0 \implies a = b$

The digits must be equal. Since $a \neq 0$, $a$ can be any digit from 1 to 9. $b$ will be the same digit.

Possible pairs (a, b): (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7), (8, 8), (9, 9).

The corresponding numbers are: 11, 22, 33, 44, 55, 66, 77, 88, 99.

There are 9 such numbers.

Case 2: $a - b = 3 \implies a = b + 3$

We list the pairs (a, b) where $a = b + 3$, $a \in \{1, \dots, 9\}$, $b \in \{0, \dots, 9\}$:

• If $b=0$, $a=3$. Number: 30.
• If $b=1$, $a=4$. Number: 41.
• If $b=2$, $a=5$. Number: 52.
• If $b=3$, $a=6$. Number: 63.
• If $b=4$, $a=7$. Number: 74.
• If $b=5$, $a=8$. Number: 85.
• If $b=6$, $a=9$. Number: 96.
• If $b \ge 7$, $a \ge 10$, which is not a single digit.

The corresponding numbers are: 30, 41, 52, 63, 74, 85, 96.

There are 7 such numbers.

Case 3: $a - b = -3 \implies b = a + 3$

We list the pairs (a, b) where $b = a + 3$, $a \in \{1, \dots, 9\}$, $b \in \{0, \dots, 9\}$:

• If $a=1$, $b=4$. Number: 14.
• If $a=2$, $b=5$. Number: 25.
• If $a=3$, $b=6$. Number: 36.
• If $a=4$, $b=7$. Number: 47.
• If $a=5$, $b=8$. Number: 58.
• If $a=6$, $b=9$. Number: 69.
• If $a \ge 7$, $b \ge 10$, which is not a single digit.

The corresponding numbers are: 14, 25, 36, 47, 58, 69.

There are 6 such numbers.

Combining the numbers from all possible cases, the total number of such two-digit numbers is the sum of the counts from each case.

Total number of numbers = (Count from Case 1) + (Count from Case 2) + (Count from Case 3)

Total number of numbers = $9 + 7 + 6 = 22$.

The list of all such numbers is the collection of numbers found in each case:

From Case 1: 11, 22, 33, 44, 55, 66, 77, 88, 99

From Case 2: 30, 41, 52, 63, 74, 85, 96

From Case 3: 14, 25, 36, 47, 58, 69

Combining and listing them in increasing order (optional, but good practice):

11, 14, 22, 25, 30, 33, 36, 41, 44, 47, 52, 55, 58, 63, 66, 69, 74, 77, 85, 88, 96, 99.

There are 22 such numbers.

The final answer is $\boxed{22 \text{ numbers}}$, and the numbers are 11, 14, 22, 25, 30, 33, 36, 41, 44, 47, 52, 55, 58, 63, 66, 69, 74, 77, 85, 88, 96, 99.

Given:

The multiplication puzzle: 12345679 $\times$ 9.

Calculation:

We perform the multiplication:

$\begin{array}{@{}c@{\,}c@{}c@{}c@{}c@{}c@{}c@{}c@{}c} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 9 \\ & & & & & & \times & 9 \\ \hline 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}$

The result of the multiplication is $12345679 \times 9 = 111111111$ (nine 1s).

Using the result to answer the questions:

Let $N = 12345679$. We found that $N \times 9 = 111111111$.

(a) What will be 12345679 × 45?

We can write 45 as $5 \times 9$.

$12345679 \times 45 = 12345679 \times (5 \times 9)$

Using the associative property of multiplication:

$= (12345679 \times 9) \times 5$

Substitute the result of the initial multiplication:

$= 111111111 \times 5$

Multiply 111111111 by 5:

$111111111 \times 5 = 555555555$

The result is a number consisting of nine 5s.

(b) What will be 12345679 × 63?

We can write 63 as $7 \times 9$.

$12345679 \times 63 = 12345679 \times (7 \times 9)$

Using the associative property of multiplication:

$= (12345679 \times 9) \times 7$

Substitute the result of the initial multiplication:

$= 111111111 \times 7$

Multiply 111111111 by 7:

$111111111 \times 7 = 777777777$

The result is a number consisting of nine 7s.

(c) By what number should 12345679 be multiplied to get 888888888?

Let the unknown number be $x$. We want to solve for $x$ in the equation:

$12345679 \times x = 888888888$

We know that $111111111 \times 8 = 888888888$.

We also know that $12345679 \times 9 = 111111111$.

Substitute the second equation into the first one:

$(12345679 \times 9) \times 8 = 888888888$

$12345679 \times (9 \times 8) = 888888888$

$12345679 \times 72 = 888888888$

Comparing this with the original equation $12345679 \times x = 888888888$, we see that $x = 72$.

So, 12345679 should be multiplied by 72 to get 888888888.

(d) By what number should 12345679 be multiplied to get 999999999?

Let the unknown number be $y$. We want to solve for $y$ in the equation:

$12345679 \times y = 999999999$

We know that $111111111 \times 9 = 999999999$.

We also know that $12345679 \times 9 = 111111111$.

Substitute the second equation into the first one:

$(12345679 \times 9) \times 9 = 999999999$

$12345679 \times (9 \times 9) = 999999999$

$12345679 \times 81 = 999999999$

Comparing this with the original equation $12345679 \times y = 999999999$, we see that $y = 81$.

So, 12345679 should be multiplied by 81 to get 999999999.

Answers:

(a) $12345679 \times 45 = 555555555$

(b) $12345679 \times 63 = 777777777$

(c) The number is 72.

(d) The number is 81.

The final answers are $\boxed{(a) 555555555, (b) 777777777, (c) 72, (d) 81}$.

Given:

Two puzzles involving letter digits.

To Find:

The values of the letters in each puzzle.

(i) Multiplication Puzzle:

$\begin{array}{cccc} & P & Q \\ & \times & 6 \\ \hline Q & Q & Q \\ \hline \end{array}$

Where P and Q are single digits (0-9).

Since 'PQ' is a two-digit number, $P \neq 0$.

Since 'QQQ' is a three-digit number, $Q \neq 0$.

The multiplication can be represented as $(10P + Q) \times 6 = 100Q + 10Q + Q = 111Q$.

$60P + 6Q = 111Q$

$60P = 111Q - 6Q$

$60P = 105Q$

Divide both sides by 15:

$\frac{60P}{15} = \frac{105Q}{15}$

$4P = 7Q$

Since P and Q are single digits, and P $\neq$ 0, Q $\neq$ 0, we look for pairs of non-zero single digits (P, Q) that satisfy this equation.

Equation (i) implies that 7Q must be a multiple of 4. Since 7 is not a multiple of 4, Q must be a multiple of 4.

Possible non-zero single digits for Q that are multiples of 4 are 4 and 8.

• If Q = 4: $4P = 7 \times 4 \implies 4P = 28 \implies P = \frac{28}{4} = 7$. This is a valid non-zero single digit. (P=7, Q=4)
• If Q = 8: $4P = 7 \times 8 \implies 4P = 56 \implies P = \frac{56}{4} = 14$. This is not a single digit.

The only solution for non-zero single digits is P = 7 and Q = 4.

Let's check the multiplication: $74 \times 6$.

$74 \times 6 = 444$.

This matches the pattern QQQ with Q=4. The values are consistent.

The values of the letters are P = 7 and Q = 4.

(ii) Addition Puzzle:

$\begin{array}{cccc} & 2 & L & M \\ + & L & M & 1 \\ \hline & M & 1 & 8 \\ \hline \end{array}$

Where L and M are single digits (0-9).

Since 2LM and LM1 are three-digit numbers, L and M must be digits. The result M18 is also a three-digit number, so M $\neq$ 0.

We analyze the addition column by column, considering carry-overs.

Let $c_1$ be the carry-over from the units column to the tens column.

Let $c_2$ be the carry-over from the tens column to the hundreds column.

Units Column:

$M + 1 = 8 + 10 \times c_1$

Since M is a single digit (0-9), M+1 can range from $0+1=1$ to $9+1=10$.

If $c_1 = 0$, then $M + 1 = 8 \implies M = 7$. (Valid, $M \neq 0$)

If $c_1 = 1$, then $M + 1 = 8 + 10 = 18 \implies M = 17$. (Not a single digit)

So, from the units column, we must have M = 7 and $c_1 = 0$.

Hundreds Column:

The sum of the digits in the hundreds column (2 and L) plus the carry-over from the tens column ($c_2$) gives the hundreds digit of the result (M).

$2 + L + c_2 = M$

Substitute M = 7 into equation (iii):

$2 + L + c_2 = 7$

$L + c_2 = 5$

Since $c_2$ is a carry-over from the tens column (sum of L, M, and $c_1$), $c_2$ can be 0 or 1.

• If $c_2 = 0$, then $L + 0 = 5 \implies L = 5$. (Valid single digit)
• If $c_2 = 1$, then $L + 1 = 5 \implies L = 4$. (Valid single digit)

So, L can be 5 (with $c_2=0$) or 4 (with $c_2=1$).

Tens Column:

The sum of the digits in the tens column (L and M) plus the carry-over from the units column ($c_1$) gives the tens digit of the result (1) plus $10 \times c_2$ (carry-over to the hundreds column).

$L + M + c_1 = 1 + 10 \times c_2$

Substitute M = 7 and $c_1 = 0$ into equation (v):

$L + 7 + 0 = 1 + 10c_2$

$L + 7 = 1 + 10c_2$

$L + 6 = 10c_2$

Now we use the possible values for L and $c_2$ from the hundreds column analysis:

• Case 1: L = 5 and $c_2 = 0$. Substitute into equation (vi): $5 + 6 = 10(0) \implies 11 = 0$. This is false.
• Case 2: L = 4 and $c_2 = 1$. Substitute into equation (vi): $4 + 6 = 10(1) \implies 10 = 10$. This is true.

The only consistent values are L = 4, M = 7, $c_1 = 0$, and $c_2 = 1$.

The values of the letters are L = 4 and M = 7.

Let's verify the addition with L=4 and M=7:

2LM becomes 247.

LM1 becomes 471.

M18 becomes 718.

Adding 247 and 471:

$\begin{array}{cc} & 247 \\ + & 471 \\ \hline & 718 \\ \hline \end{array}$

Units: $7 + 1 = 8$. Matches 8. Carry $c_1=0$. Correct.

Tens: $4 + 7 + c_1 = 4 + 7 + 0 = 11$. Units digit is 1. Carry $c_2=1$. Correct.

Hundreds: $2 + 4 + c_2 = 2 + 4 + 1 = 7$. Units digit is 7. Matches M=7. Correct.

The sum is 718, which matches M18 with M=7. The values are consistent.

The values of the letters are L = 4 and M = 7.

The final answers are $\boxed{(i) P=7, Q=4, (ii) L=4, M=7}$.

Given:

The number 148101B095 is divisible by 33.

B is a single digit (0-9).

To Find:

The value of the digit B.

Solution:

A number is divisible by 33 if and only if it is divisible by both 3 and 11, since $33 = 3 \times 11$ and 3 and 11 are coprime.

The given number is 148101B095.

Divisibility by 3:

A number is divisible by 3 if the sum of its digits is divisible by 3.

The digits of the number 148101B095 are 1, 4, 8, 1, 0, 1, B, 0, 9, 5.

The sum of the digits is $1 + 4 + 8 + 1 + 0 + 1 + B + 0 + 9 + 5$

Sum $= (1 + 4 + 8 + 1 + 0 + 1 + 0 + 9 + 5) + B$

Sum $= 29 + B$

For the number to be divisible by 3, $29 + B$ must be divisible by 3.

We can check the divisibility of 29 by 3: $29 = 3 \times 9 + 2$. So $29 \equiv 2 \pmod{3}$.

Thus, $29 + B \equiv 2 + B \pmod{3}$.

For $29 + B$ to be divisible by 3, $2 + B$ must be divisible by 3.

Since B is a single digit (0-9), we find values of B such that $2+B$ is a multiple of 3:

• If B = 1, $2+1=3$. (Divisible by 3)
• If B = 4, $2+4=6$. (Divisible by 3)
• If B = 7, $2+7=9$. (Divisible by 3)
• If B = 0, $2+0=2$. (Not divisible by 3)
• If B = 2, $2+2=4$. (Not divisible by 3)
• If B = 3, $2+3=5$. (Not divisible by 3)
• If B = 5, $2+5=7$. (Not divisible by 3)
• If B = 6, $2+6=8$. (Not divisible by 3)
• If B = 8, $2+8=10$. (Not divisible by 3)
• If B = 9, $2+9=11$. (Not divisible by 3)

So, the possible values for B based on divisibility by 3 are 1, 4, or 7.

Divisibility by 11:

A number is divisible by 11 if the alternating sum of its digits is divisible by 11.

The digits of the number 148101B095 are 1, 4, 8, 1, 0, 1, B, 0, 9, 5.

We calculate the alternating sum of the digits, starting from the rightmost digit (5) with a positive sign:

$5 - 9 + 0 - B + 1 - 0 + 1 - 8 + 4 - 1$

$(5 + 0 + 1 + 1 + 4) - (9 + B + 0 + 8 + 1)$

$(11) - (18 + B)$

$11 - 18 - B$

$-7 - B$

For the number to be divisible by 11, the alternating sum $-7 - B$ must be divisible by 11.

Since B is a single digit (0-9), we find the possible values for $-7 - B$:

• If B = 0, $-7 - 0 = -7$.
• If B = 1, $-7 - 1 = -8$.
• If B = 2, $-7 - 2 = -9$.
• If B = 3, $-7 - 3 = -10$.
• If B = 4, $-7 - 4 = -11$.
• If B = 5, $-7 - 5 = -12$.
• If B = 6, $-7 - 6 = -13$.
• If B = 7, $-7 - 7 = -14$.
• If B = 8, $-7 - 8 = -15$.
• If B = 9, $-7 - 9 = -16$.

The range of possible values for $-7 - B$ is $[-16, -7]$.

The only value in this range that is divisible by 11 is -11.

So, we must have $-7 - B = -11$.

$-B = -11 + 7$

$-B = -4$

$B = 4$

We need to find the value of B that satisfies both divisibility by 3 and divisibility by 11.

From divisibility by 3, B can be 1, 4, or 7.

From divisibility by 11, B must be 4.

The only value common to both sets of possibilities is 4.

Thus, the value of B is 4.

Let's check the number with B=4: 1481014095.

Sum of digits: $1+4+8+1+0+1+4+0+9+5 = 33$. 33 is divisible by 3.

Alternating sum of digits: $5-9+0-4+1-0+1-8+4-1 = (5+0+1+1+4) - (9+4+0+8+1) = 11 - 22 = -11$. -11 is divisible by 11.

Since the number is divisible by both 3 and 11, it is divisible by 33. The value B=4 is correct.

The value of B is 4.

The final answer is $\boxed{4}$.

Given:

The number 123123A4 is divisible by 11.

A is a single digit (0-9).

To Find:

The value of the digit A.

Solution:

A number is divisible by 11 if the alternating sum of its digits is divisible by 11.

The digits of the number 123123A4 are 1, 2, 3, 1, 2, 3, A, 4.

We calculate the alternating sum of the digits, starting from the rightmost digit (4) with a positive sign:

$4 - A + 3 - 2 + 1 - 3 + 2 - 1$

$(4 + 3 + 1 + 2) - (A + 2 + 3 + 1)$

$(10) - (A + 6)$

$10 - A - 6$

$4 - A$

For the number 123123A4 to be divisible by 11, the alternating sum $4 - A$ must be divisible by 11.

Since A is a single digit (0-9), we find the possible values for $4 - A$:

• If A = 0, $4 - 0 = 4$.
• If A = 1, $4 - 1 = 3$.
• If A = 2, $4 - 2 = 2$.
• If A = 3, $4 - 3 = 1$.
• If A = 4, $4 - 4 = 0$.
• If A = 5, $4 - 5 = -1$.
• If A = 6, $4 - 6 = -2$.
• If A = 7, $4 - 7 = -3$.
• If A = 8, $4 - 8 = -4$.
• If A = 9, $4 - 9 = -5$.

The range of possible values for $4 - A$ is $[-5, 4]$.

The only value in this range that is divisible by 11 is 0.

So, we must have $4 - A = 0$.

$A = 4$

The value A=4 is a single digit, which is valid.

Let's check the number with A=4: 12312344.

Alternating sum of digits: $4 - 4 + 3 - 2 + 1 - 3 + 2 - 1 = (4+3+1+2) - (4+2+3+1) = 10 - 10 = 0$.

Since the alternating sum is 0, the number 12312344 is divisible by 11. The value A=4 is correct.

The value of A is 4.

The final answer is $\boxed{4}$.

Given:

The number 56x32y is divisible by 18.

x and y are single digits (0-9).

To Find:

The least value of the digit y.

Solution:

A number is divisible by 18 if and only if it is divisible by both 2 and 9, since $18 = 2 \times 9$ and 2 and 9 are coprime.

The given number is 56x32y.

Divisibility by 2:

A number is divisible by 2 if its units digit is an even number (0, 2, 4, 6, or 8).

The units digit of 56x32y is y.

For the number to be divisible by 2, y must be an even digit.

Possible values for y: 0, 2, 4, 6, 8.

We are looking for the least value of y, so we will check these values starting from the smallest (0).

Divisibility by 9:

A number is divisible by 9 if the sum of its digits is divisible by 9.

The digits of the number 56x32y are 5, 6, x, 3, 2, y.

The sum of the digits is $5 + 6 + x + 3 + 2 + y = 16 + x + y$.

For the number to be divisible by 9, $16 + x + y$ must be divisible by 9.

We can check the divisibility of 16 by 9: $16 = 9 \times 1 + 7$. So $16 \equiv 7 \pmod{9}$.

Thus, $16 + x + y \equiv 7 + x + y \pmod{9}$.

For $16 + x + y$ to be divisible by 9, $7 + x + y$ must be divisible by 9.

So, $7 + x + y = 9n$ for some integer n.

Since x and y are single digits (0-9), the maximum value of $x+y$ is $9+9=18$.

The minimum value of $x+y$ is $0+0=0$.

So, $7 + x + y$ can range from $7+0=7$ to $7+18=25$.

The multiples of 9 in the range [7, 25] are 9 and 18.

Case A: $7 + x + y = 9 \implies x + y = 2$.

Case B: $7 + x + y = 18 \implies x + y = 11$.

We are looking for the least value of y from the set {0, 2, 4, 6, 8} such that there exists a digit x (0-9) that satisfies either $x+y=2$ or $x+y=11$.

Let's check the possible values of y in increasing order:

• If y = 0:
  • Check Case A: $x + 0 = 2 \implies x = 2$. Is $x=2$ a single digit (0-9)? Yes.
  • Since we found a valid value for x (which is 2) when y=0, the number 562320 is divisible by 9.

We found that y=0 satisfies the condition for divisibility by 2 (since 0 is even) and there exists a digit x=2 such that the number is divisible by 9.

Since we started checking y values from the least possible value (0) and found a valid case, the least value of y is 0.

To be complete, we can check the other possible values of y, but it is not necessary to find the least value.

• If y = 2:
  • Check Case A: $x + 2 = 2 \implies x = 0$. Valid digit.
  • Check Case B: $x + 2 = 11 \implies x = 9$. Valid digit.

  If y=2, x can be 0 or 9.

• If y = 4:
  • Check Case A: $x + 4 = 2 \implies x = -2$. Invalid digit.
  • Check Case B: $x + 4 = 11 \implies x = 7$. Valid digit.

  If y=4, x can be 7.

• If y = 6:
  • Check Case A: $x + 6 = 2 \implies x = -4$. Invalid digit.
  • Check Case B: $x + 6 = 11 \implies x = 5$. Valid digit.

  If y=6, x can be 5.

• If y = 8:
  • Check Case A: $x + 8 = 2 \implies x = -6$. Invalid digit.
  • Check Case B: $x + 8 = 11 \implies x = 3$. Valid digit.

  If y=8, x can be 3.

All the possible even values for y (0, 2, 4, 6, 8) allow for at least one valid digit x such that the number is divisible by 9. However, the question asks for the least value of y.

The least value of y that satisfies the divisibility rule for 2 is 0. For y=0, we found a valid x=2 such that the divisibility rule for 9 is satisfied ($16+2+0 = 18$, which is divisible by 9). Thus, the number 562320 is divisible by 18.

Since we checked y values in increasing order and the first one (y=0) worked, the least value of y is 0.

The final answer is $\boxed{0}$.