Chapter 13 Playing with Numbers (Class 8 - Maths NCERT Exemplar Solutions)

Given:

A three-digit number $xyz$.

To Find:

The generalised form of the number.

Solution:

In a three-digit number, the digits represent place values of hundreds, tens, and units.

For the number $xyz$:

Digit at hundreds place = $x$

Digit at tens place = $y$

Digit at units place = $z$

The generalised form is calculated by multiplying each digit by its respective place value and adding them together.

$\text{Generalised form} = (100 \times x) + (10 \times y) + (1 \times z)$

$\text{Generalised form} = 100x + 10y + z$

Final Answer:

The correct option is (b) 100x + 10y + z.

Given:

The generalised form: $100a + b + 10c$.

To Find:

The usual form of the given expression.

Solution:

To find the usual form, we first arrange the terms according to descending powers of $10$ (place values):

$100a + 10c + b$

Here:

The digit at the hundreds place is $a$.

The digit at the tens place is $c$.

The digit at the units place is $b$.

Combining these digits, we get the number as $acb$.

Final Answer:

The correct option is (d) acb.

Given:

$5 \times A = CA$

To Find:

The values of digits $A$ and $C$.

Solution:

The product $5 \times A$ ends in the digit $A$.

Let's check the multiples of $5$:

$5 \times 0 = 0$ (If $A=0$, then $CA=0$, but $C$ must be a digit to form a two-digit number $CA$. So $A \neq 0$.)

$5 \times 1 = 5$ (Units digit is $5$, but $A$ was $1$. Not matching.)

$5 \times 2 = 10$ (Units digit is $0$. Not matching.)

$5 \times 3 = 15$ (Units digit is $5$. Not matching.)

$5 \times 4 = 20$ (Units digit is $0$. Not matching.)

$5 \times 5 = 25$

In the case of $A = 5$, the units digit of the product is also $5$.

Comparing $5 \times 5 = 25$ with $CA$:

$A = 5$

$C = 2$

Final Answer:

The correct values are $A = 5$ and $C = 2$. Therefore, the correct option is (c).

Given:

$5A + 25 = B2$

To Find:

The value of $A + B$.

Solution:

Let us write the addition in vertical form:

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

Looking at the units place column:

$A + 5$ must end in $2$.

This is possible if $A + 5 = 12$.

$A = 12 - 5 = 7$

Since $A+5 = 12$, we carry over $1$ to the tens place column.

Now looking at the tens place column:

$\text{Carry} + 5 + 2 = B$

$1 + 5 + 2 = B$

$B = 8$

Now, we need to find the value of $A + B$:

$A + B = 7 + 8 = 15$

Final Answer:

The value of $A + B$ is $15$. The correct option is (a).

Solution:

Let the two-digit number $ab$ be written in generalized form:

Similarly, the number $ba$ is:

Subtracting (ii) from (i):

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

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

$ab - ba = 9a - 9b$

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

Since the result is a multiple of $9$, it is divisible by $9$.

Final Answer:

The number $ab - ba$ is divisible by $9$.

Solution:

The given expression is $100a + 10c + 9$.

Digit at hundreds place = $a$

Digit at tens place = $c$

Digit at units place = $9$

To write the usual form, we place the digits side by side according to their place values.

The usual form is $ac9$.

Final Answer:

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

Solution:

Given the multiplication: $AB \times B = 9B$.

This means a two-digit number $AB$ multiplied by the single digit $B$ results in a two-digit number $9B$.

From the units place: $B \times B$ must result in a number ending with $B$.

The digits that satisfy this property are $0, 1, 5,$ and $6$.

1. If $B = 0$, then $A0 \times 0 = 00$, but the result must be $90$. Not possible.

2. If $B = 1$, then $A1 \times 1 = 91$. This gives $A = 9$.

3. If $B = 5$, then $A5 \times 5 = 95$. Let's check: $15 \times 5 = 75$, $25 \times 5 = 125$ (not possible as it is 3-digit). So $B \neq 5$.

4. If $B = 6$, then $A6 \times 6 = 96$. We know $16 \times 6 = 96$. This gives $A = 1$.

Both $(A=9, B=1)$ and $(A=1, B=6)$ satisfy the equation.

Final Answer:

One set of values is $A = 9$, $B = 1$. (Alternatively $A = 1, B = 6$ is also correct.)

Solution:

Let the three numbers be written in their generalized forms:

$abc = 100a + 10b + c$

$cab = 100c + 10a + b$

$bca = 100b + 10c + a$

Now, let us find the sum of these three numbers:

$\text{Sum} = (100a + 10b + c) + (100c + 10a + b) + (100b + 10c + a)$

$\text{Sum} = 111a + 111b + 111c$

$\text{Sum} = 111(a + b + c)$

We know that $111 = 3 \times 37$. Therefore, the expression can be written as:

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

Since the sum is a multiple of $37$, it is always divisible by $37$.

Final Answer:

The statement is True (T).

Solution:

Let the two-digit number $ab$ be $10a + b$ and $ba$ be $10b + a$.

Let us find their sum:

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

$ab + ba = 11a + 11b$

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

The sum $ab + ba$ is always a multiple of $11$, not $9$. For example, if $a=1, b=2$, then $12 + 21 = 33$, which is not divisible by $9$.

Final Answer:

The statement is False (F).

Solution:

To check if a number divisible by both $2$ and $4$ is always divisible by $8$, let us consider a counter-example.

Consider the number $12$.

1. $12$ is divisible by $2$ ($12 \div 2 = 6$).

2. $12$ is divisible by $4$ ($12 \div 4 = 3$).

However, $12$ is not divisible by $8$ ($12 \div 8 = 1.5$).

Divisibility by $2$ and $4$ only ensures divisibility by the Least Common Multiple (LCM) of $2$ and $4$, which is $4$. It does not guarantee divisibility by their product ($8$) unless the numbers are co-prime.

Final Answer:

The statement is False (F).

Given:

The number $42x$ is divisible by $9$.

To Find:

The value of the digit $x$.

Solution:

According to the divisibility rule of $9$, the sum of the digits of the number must be divisible by $9$.

Sum of digits = $4 + 2 + x = 6 + x$

Since $x$ is a digit ($0$ to $9$), the sum $6 + x$ can be $9$ or $18$.

Case 1: $6 + x = 9$

$x = 9 - 6 = 3$

Case 2: $6 + x = 18$

$x = 18 - 6 = 12$

Since $x$ must be a single digit, $x = 12$ is not possible.

Final Answer:

The value of $x$ is $3$.

Given:

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

To Find:

The values of digits $A$ and $B$.

Solution:

1. From the units column: $A + 4 = 2$ or $12$.

Since $A$ is a digit, $A + 4$ cannot be $2$. Thus, $A + 4 = 12$.

$A = 12 - 4 = 8$

There is a carry of $1$ to the tens column.

2. From the tens column: $1 (\text{carry}) + 1 + B = 1$ or $11$.

$2 + B = 11$ (as $2+B=1$ is not possible for a digit $B$).

$B = 11 - 2 = 9$

There is a carry of $1$ to the hundreds column.

3. Checking the hundreds column: $1 (\text{carry}) + 4 = 5$. This matches the result.

Final Answer:

The values are $A = 8$ and $B = 9$.

Given:

1. $x \div 5$ leaves remainder $4$.

2. $x \div 2$ leaves remainder $1$.

To Find:

The ones digit of $x$.

Solution:

From the first condition ($x \div 5$ leaves remainder $4$):

Any number divided by $5$ has a remainder $4$ if its ones digit is either $4$ or $9$.

From the second condition ($x \div 2$ leaves remainder $1$):

Any number divided by $2$ has a remainder $1$ if it is an odd number. Therefore, its ones digit must be odd.

The only digit that satisfies both conditions (i) and (ii) is $9$.

Final Answer:

The ones digit of $x$ is $9$.

Given:

The number $756x$ is divisible by $11$.

To Find:

The value of digit $x$.

Solution:

A number is divisible by $11$ if the difference between the sum of digits at odd places and the sum of digits at even places is either $0$ or a multiple of $11$.

Digits at odd places (from right): $x$ and $5$.

$\text{Sum of odd places} = x + 5$

Digits at even places (from right): $6$ and $7$.

$\text{Sum of even places} = 6 + 7 = 13$

The difference is:

$\text{Difference} = (x + 5) - 13 = x - 8$

For the number to be divisible by $11$, $x - 8$ must be $0$ (since $x$ is a digit, it cannot result in $11$ or $-11$).

$x - 8 = 0$

$x = 8$

Final Answer:

The value of $x$ is $8$.

Given:

A four-digit number $abdc$.

Solution:

In the four-digit number $abdc$, the digits occupy the following place values:

Digit $a$ is at the thousands place.

Digit $b$ is at the hundreds place.

Digit $d$ is at the tens place.

Digit $c$ is at the units (ones) place.

The generalised form is the sum of the products of each digit and its place value:

$\text{Generalised form} = (1000 \times a) + (100 \times b) + (10 \times d) + (1 \times c)$

$\text{Generalised form} = 1000a + 100b + 10d + c$

Final Answer:

The correct option is (c) 1000 a + 100 b + 10 d + c.

Solution:

In the two-digit number $xy$:

Digit $x$ is at the tens place.

Digit $y$ is at the units place.

To write the generalised form, we multiply the tens digit by $10$ and the units digit by $1$.

$\text{Generalised form} = 10x + y$

Final Answer:

The correct option is (b) 10x + y.

Given:

The generalised form $1000a + 10b + c$.

Solution:

Let us identify the digits at each place value:

Digit at thousands place ($1000 \times \dots$) = $a$

Digit at hundreds place ($100 \times \dots$) = $0$ (since no $100$ term is present)

Digit at tens place ($10 \times \dots$) = $b$

Digit at units place ($1 \times \dots$) = $c$

Combining these in order, we get the number: $a0bc$.

Final Answer:

The correct option is (c) aobc.

Solution:

Let us expand the numbers in their generalised form:

Subtracting (ii) from (i):

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

$abc - cba = 99a - 99c$

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

Now, we factorise $99$: $99 = 9 \times 11 = 3 \times 33$.

The expression $99(a-c)$ is always divisible by $9$, $11$, and $33$. It is not necessarily divisible by $18$.

Final Answer:

The correct option is (c) 18.

Solution:

The digits are $x$, $y$, and $z$. The six possible three-digit numbers are:

$xyz, xzy, yxz, yzx, zxy, zyx$.

The sum of these numbers is:

$\text{Sum} = 111(x+y+z) + 111(x+y+z)$ (considering the permutations)

$\text{Sum} = 222(x + y + z)$

We can factorise $222$ as $6 \times 37$.

$\text{Sum} = 37 \times [6(x + y + z)]$

Since the sum is a multiple of $37$, it is always divisible by $37$.

Final Answer:

The correct option is (c) 37.

Given:

The number $aabb$ is divisible by $55$.

Solution:

If a number is divisible by $55$, it must be divisible by both $5$ and $11$ ($5 \times 11 = 55$).

1. Divisibility by 5: A number is divisible by $5$ if its last digit is either $0$ or $5$.

In $aabb$, the last digit is $b$. Therefore, $b$ must be $0$ or $5$.

2. Divisibility by 11: The difference between the sum of digits at odd places and even places must be $0$ or a multiple of $11$.

Sum of odd places ($b + a$) - Sum of even places ($b + a$) = $0$. This is always true for any value of $a$ and $b$.

Thus, the only requirement is divisibility by $5$.

Final Answer:

The possible values of $b$ are $0$ and $5$. The correct option is (c).

Solution:

We find the sum using the generalised form:

$abc + bca + cab = (100a+10b+c) + (100b+10c+a) $$ + (100c+10a+b)$

$abc + bca + cab = 111a + 111b + 111c$

$abc + bca + cab = 111(a + b + c)$

Factorising $111$:

$abc + bca + cab = 3 \times 37 \times (a + b + c)$

The sum is divisible by $3$, $37$, and $(a + b + c)$. It is not necessarily divisible by $9$.

Final Answer:

The correct option is (d) 9.

Given:

The number $4ab5$ is divisible by $55$.

Solution:

For $4ab5$ to be divisible by $55$, it must be divisible by $11$ and $5$.

The last digit is $5$, so it is already divisible by $5$.

For divisibility by $11$:

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

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

Difference = $(5 + a) - (b + 4) = 1 + a - b$

This difference must be $0$ or a multiple of $11$.

$1 + a - b = 0$

$b - a = 1$

Final Answer:

The value of $b - a$ is $1$. The correct option is (b).

Given:

A three-digit number $abc$.

To Find:

The divisor of the expression $abc - a - b - c$.

Solution:

The generalised form of a three-digit number $abc$ is:

$abc = 100a + 10b + c$

Now, let us subtract the sum of its digits from the number:

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

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

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

Taking $9$ as a common factor:

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

Since the result is a multiple of $9$, the number is always divisible by $9$.

Final Answer:

The correct option is (a) 9.

Given:

A six-digit number formed by repeating a three-digit number (e.g., $256256$).

Solution:

Let the three-digit number be $x$. When it is repeated to form a six-digit number, it can be written as:

$\text{Number} = x \times 1000 + x$

$\text{Number} = 1001x$

We can find the prime factors of $1001$:

$1001 = 7 \times 11 \times 13$

Since the number is a multiple of $1001$, it is divisible by $7$, $11$, and $13$. Thus, any such number is divisible by $1001$.

Final Answer:

The correct option is (d) 1001.

Solution:

According to the divisibility rule of 3, if the sum of the digits of a number is divisible by $3$, then the number itself is divisible by $3$.

For example, consider the number $123$. The sum of digits is $1 + 2 + 3 = 6$, which is divisible by $3$. Therefore, $123$ is divisible by $3$.

Final Answer:

The correct option is (b) 3.

Given:

A three-digit number $xyz$ where $x + y + z = 6$ and $z$ is an odd digit.

Solution:

1. Since the sum of the digits is $6$, and $6$ is divisible by $3$, the number $xyz$ must be a multiple of 3.

2. The digit $z$ is the ones (units) digit of the number $xyz$. It is given that $z$ is odd ($1, 3, 5, 7,$ or $9$).

3. If the ones digit of a number is odd, the entire number is odd.

Therefore, $xyz$ is an odd multiple of $3$.

Final Answer:

The correct option is (a) an odd multiple of 3.

Given:

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

Solution:

Looking at the units place column:

$A = 5 - 3$

$A = 2$

Now, looking at the tens place column (there is no carry from the units place):

$B = 6 - 5$

$B = 1$

Final Answer:

The values are $A = 2$ and $B = 1$. The correct option is (c).

Given:

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

Solution:

1. Units column: $3 + B$ ends in $0$. This implies $3 + B = 10$.

$B = 10 - 3 = 7$

There is a carry of $1$ to the tens place.

2. Tens column: $1 (\text{carry}) + A + 8 = 15$.

$A + 9 = 15$

$A = 15 - 9 = 6$

3. Now we find the value of $A + B$:

$A + B = 6 + 7 = 13$

Final Answer:

The correct option is (a) 13.

Given:

$5A \times A = 399$

Solution:

The product $A \times A$ must end in $9$. The possible digits for $A$ are $3$ (since $3 \times 3 = 9$) or $7$ (since $7 \times 7 = 49$).

Case 1: If $A = 3$, then $53 \times 3$:

$53 \times 3 = 159$ (Not equal to $399$)

Case 2: If $A = 7$, then $57 \times 7$:

$57 \times 7 = 399$ (Matches the given result)

Final Answer:

The value of $A$ is 7. The correct option is (c).

Given:

$6A \times B = A8B$

Solution:

The result of the multiplication is a three-digit number $A8B$.

Looking at the units place: $A \times B$ must end in $B$. This happens if $B=0$ (not possible here) or if $A=1$.

Let's test $A = 1$:

$61 \times B = 18B$

If $B = 3$, then:

$61 \times 3 = 183$

This matches the form $A8B$ where $A=1$ and $B=3$.

Now, calculate $A - B$:

$A - B = 1 - 3 = -2$

Final Answer:

The value of $A - B$ is $-2$. The correct option is (a).

Solution:

A number is divisible by $99$ if it is divisible by both 9 and 11.

Let's check option (b) $114345$:

1. Divisibility by 9: Sum of digits $= 1+1+4+3+4+5 = 18$. Since $18$ is divisible by $9$, the number is divisible by $9$.

2. Divisibility by 11: Difference of sum of digits at odd and even places:

$\text{Sum at odd places} = 5 + 3 + 1 = 9$

$\text{Sum at even places} = 4 + 4 + 1 = 9$

$\text{Difference} = 9 - 9 = 0$. Since the difference is $0$, the number is divisible by $11$.

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

Final Answer:

The correct option is (b) 114345.

Given:

The number is $3134673$.

Solution:

To check divisibility by $3$ and other numbers, we first find the sum of the digits of the given number.

$\text{Sum of digits} = 3 + 1 + 3 + 4 + 6 + 7 + 3 = 27$

Since the sum of digits ($27$) is divisible by $3$, the number $3134673$ is divisible by $3$.

Similarly, according to the divisibility rule of $9$, if the sum of the digits is divisible by $9$, the number is divisible by $9$.

Since $27$ is divisible by $9$ ($27 = 9 \times 3$), the number $3134673$ is also divisible by $9$.

Final Answer:

3134673 is divisible by 3 and 9.

Given:

The number $20x3$ is a multiple of $3$.

Solution:

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

$\text{Sum of digits} = 2 + 0 + x + 3 = 5 + x$

For $5 + x$ to be divisible by $3$, it must be $6, 9, 12, 15, \dots$

1. If $5 + x = 6 \Rightarrow x = 1$

2. If $5 + x = 9 \Rightarrow x = 4$

3. If $5 + x = 12 \Rightarrow x = 7$

4. If $5 + x = 15 \Rightarrow x = 10$ (Not possible as $x$ must be a single digit)

Final Answer:

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

Given:

The number $3x5$ is divisible by $9$.

Solution:

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

$\text{Sum of digits} = 3 + x + 5 = 8 + x$

For $8 + x$ to be divisible by $9$, the smallest possible multiple of $9$ is $9$ itself.

$8 + x = 9$

$x = 9 - 8 = 1$

Final Answer:

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

Solution:

Let the two-digit number be $ab$, which can be written in generalised form as:

The number obtained by reversing the digits is $ba$, which is:

Adding equations (i) and (ii):

$\text{Sum} = (10a + b) + (10b + a)$

$\text{Sum} = 11a + 11b$

$\text{Sum} = 11(a + b)$

Since the sum is a multiple of $11$, it is always divisible by $11$.

Final Answer:

The sum is always divisible by 11.

Solution:

Let the two-digit number be $10a + b$.

The reversed number is $10b + a$.

Assuming $a > b$, their difference is:

$\text{Difference} = (10a + b) - (10b + a)$

$\text{Difference} = 10a - a + b - 10b$

$\text{Difference} = 9a - 9b$

$\text{Difference} = 9(a - b)$

Since the difference is a multiple of $9$, it is always divisible by $9$.

Final Answer:

The difference is always divisible by 9.

Solution:

Let the three-digit number be $abc$, which is $100a + 10b + c$.

The reversed number is $cba$, which is $100c + 10b + a$.

Subtracting the numbers:

$\text{Difference} = (100a + 10b + c) - (100c + 10b + a)$

$\text{Difference} = 99a - 99c$

$\text{Difference} = 99(a - c)$

Since $99 = 9 \times 11$, the difference is divisible by both $9$ and $11$.

Final Answer:

The difference is always divisible by 9 and 11.

Given:

$\begin{array}{cc} & 2 & B \\ + & A & B \\ \hline & 8 & A \\ \hline \end{array}$

Solution:

From the units column:

$B + B = A$ or $B + B = 10 + A$

From the tens column:

$2 + A = 8$ (assuming no carry from units column)

$A = 8 - 2 = 6$

Substituting $A = 6$ in equation (i):

$2B = 6 \Rightarrow B = 3$

Checking the values:

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

This matches the given condition.

Final Answer:

A = 6 and B = 3.

Given:

$\begin{array}{cc} & A & B \\ \times & & B \\ \hline & 9 & 6 \\ \hline \end{array}$

Solution:

The product of digit $B$ with itself must end in $6$. The possible digits for $B$ are $4$ ($4 \times 4 = 16$) or $6$ ($6 \times 6 = 36$).

Case 1: If $B = 4$

The equation becomes $A4 \times 4 = 96$.

$A4 = 96 \div 4 = 24$.

This gives $A = 2$.

Case 2: If $B = 6$

The equation becomes $A6 \times 6 = 96$.

$A6 = 96 \div 6 = 16$.

This gives $A = 1$.

Final Answer:

A = 2 and B = 4 (or A = 1 and B = 6).

Given:

$\begin{array}{cccc} & B & 1 \\ \times & & B \\ \hline 4 & 9 & B \\ \hline \end{array}$

Solution:

From the units column, we have $1 \times B = B$, which is true for any digit $B$.

From the tens and hundreds column, we have $B \times B = 49$.

Since $B$ is a digit, we take the square root of $49$:

$B = 7$

Let us check: $71 \times 7 = 497$, which matches the form $49B$.

Final Answer:

B = 7.

Given:

The number $1x35$ is divisible by $9$.

Solution:

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

$\text{Sum of digits} = 1 + x + 3 + 5 = 9 + x$

For $9 + x$ to be divisible by $9$, it must be a multiple of $9$ such as $9, 18, \dots$

Case 1: $9 + x = 9 \Rightarrow x = 0$

Case 2: $9 + x = 18 \Rightarrow x = 9$

Final Answer:

x = 0 or 9.

Solution:

For the four-digit number $abcd$ to be divisible by $11$, the difference between the sum of digits at odd places and even places must be $0$ or a multiple of $11$.

Sum of digits at odd places (from right) $= d + b$

Sum of digits at even places (from right) $= c + a$

Thus, $(d + b) - (c + a) = 0 \Rightarrow d + b = c + a$

Or $(d + b) - (c + a) = 11 \Rightarrow d + b = a + c + 11$

Final Answer:

d + b = a + c or a + c + 11.

Final Answer:

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.

Solution:

For a 3-digit number $abc$, the digits at odd places are $c$ and $a$. The digit at the even place is $b$.

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

Final Answer:

If a 3-digit number abc is divisible by 11, then (a + c) - b is either 0 or multiple of 11.

Given:

$A \times 3 = 1A$

Solution:

Writing the equation in generalised form:

$3A - A = 10$

$2A = 10$

$A = 5$

Check: $5 \times 3 = 15$, which matches $1A$.

Final Answer:

A = 5.

Given:

$B \times B = AB$

Solution:

The product of digit $B$ with itself must result in a two-digit number ending in $B$.

Checking squares of digits:

$5 \times 5 = 25 \Rightarrow A = 2, B = 5$ (Given)

$6 \times 6 = 36 \Rightarrow A = 3, B = 6$

Final Answer:

A = 3, B = 6.

Given:

Tens digit $= t$, Ones digit $= u$.

Solution:

The original two-digit number is $10t + u$.

If digit $1$ is placed after this number, the original digits shift one place to the left.

Digit $t$ moves to the hundreds place.

Digit $u$ moves to the tens place.

Digit $1$ occupies the units place.

The new number $= 100t + 10u + 1$

Final Answer:

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

According to the divisibility rule of 2, a number is divisible by 2 if its units digit is even ($0, 2, 4, 6,$ or $8$).

In the number $ab$, $b$ is the units digit. If $b$ is an even number, the whole number $ab$ is divisible by 2.

Final Answer: True (T)

According to the divisibility rule of 5, a number is divisible by 5 if its units digit is either 0 or 5.

The statement says the number is divisible by 5 if $c$ is even. While $c=0$ (even) works, $c=2, 4, 6,$ or $8$ (even) does not make the number divisible by 5. Also, $c=5$ (odd) makes it divisible by 5 but is not mentioned.

Final Answer: False (F)

According to the divisibility rule of 4, a number is divisible by 4 if the number formed by its last two digits is divisible by 4.

In $abcd$, the last two digits form the number $cd$. Divisibility depends on $cd$, not on $ab$ (the first two digits).

Final Answer: False (F)

A number is divisible by 6 if it is divisible by both 2 and 3.

1. If the unit digit $c$ is even, the number is divisible by 2.

2. If the sum of digits $a+b+c$ is a multiple of 3, the number is divisible by 3.

Since both conditions are met, the number is divisible by 6.

Final Answer: True (T)

Using Euclid's Division Lemma, $a = bq + r$, where $r$ is the remainder and $0 \leq r < b$.

In the expression $3N + 2$, $3$ is the divisor, $N$ is the quotient, and $2$ is the remainder.

Final Answer: True (T)

In the expression $7N + 1$, the term $7N$ is exactly divisible by 7. The additional 1 represents the remainder when the total number is divided by 7.

Final Answer: True (T)

If $b$ divides $a$, and $k$ is a factor of $b$, then $k$ must also divide $a$. For example, if a number is divisible by 10, it is also divisible by the factors of 10 (2 and 5).

Final Answer: True (T)

Given: $AB \times 4 = 192$

$\text{To find AB: } AB = \frac{192}{4}$

$AB = 48$

Comparing digits: $A = 4$ and $B = 8$.

$\text{Calculating sum: } A + B = 4 + 8 = 12$.

The statement says $A + B = 7$, which is incorrect.

Final Answer: False (F)

Given: $AB + 7C = 102$, where $B, C \neq 0$.

Looking at the units place: $B + C$ must end in 2. Since $B, C$ are non-zero digits, $B + C$ can be 2 or 12.

Case 1: $B + C = 2$. Since $B, C \neq 0$, the only possibility is $B = 1, C = 1$. Tens place: $A + 7 = 10 \Rightarrow A = 3$. Sum $A+B+C = 3+1+1 = 5$.

Case 2: $B + C = 12$. There is a carry of 1. Tens place: $A + 7 + 1 = 10 \Rightarrow A = 2$. Sum $A+B+C = 2+B+C = 2+12 = 14$.

Since the statement $A+B+C=14$ is not true for all valid cases (it fails for $A=3, B=1, C=1$), the statement is False.

Final Answer: False (F)

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

$\text{Sum of digits} = 2 + 1 + 3 + x + 2 + 7 = 15 + x$

For $15 + x$ to be divisible by 9, $x$ must be 3 ($15 + 3 = 18$).

The statement says $x = 0$, which is incorrect.

Final Answer: False (F)

Given:

1. $N \div 5$ leaves remainder 3. This means the last digit of $N$ is 3 or 8.

2. $N \div 2$ leaves remainder 0. This means $N$ is even.

From these two, the last digit must be 8.

When a number is divided by 10, the remainder is equal to its last digit. Thus, $N \div 10$ leaves remainder 8, not 4.

Final Answer: False (F)

Given:

The number is $91876a2$.

To Find:

The least value of digit $a$ such that the number is divisible by $8$.

Solution:

According to the divisibility rule of 8, a number is divisible by $8$ if the number formed by its last three digits is divisible by $8$.

The last three digits of $91876a2$ form the number $6a2$.

We need to find the smallest digit $a$ ($0$ to $9$) such that $6a2$ is divisible by $8$.

1. If $a = 0$, number is $602$. $602 \div 8 = 75$ with remainder $2$. (Not divisible)

2. If $a = 1$, number is $612$. $612 \div 8 = 76$ with remainder $4$. (Not divisible)

3. If $a = 2$, number is $622$. $622 \div 8 = 77$ with remainder $6$. (Not divisible)

4. If $a = 3$, number is $632$. $632 \div 8 = 79$ with remainder $0$. (Divisible)

The least value of $a$ is $3$.

Final Answer:

The least value that must be given to $a$ is $3$.

Given:

$\begin{array}{cc} & 1 & P \\ \times & & P \\ \hline & Q & 6 \\ \hline \end{array}$ and $Q - P = 3$.

To Find:

The values of digits $P$ and $Q$.

Solution:

Looking at the units place of the multiplication, $P \times P$ results in a number ending with $6$.

The digits whose squares end in $6$ are $4$ ($4 \times 4 = 16$) and $6$ ($6 \times 6 = 36$).

Case 1: If $P = 4$

The multiplication is $14 \times 4$.

$14 \times 4 = 56$

Comparing this with $Q6$, we get $Q = 5$.

Now, check the condition $Q - P = 3$:

$5 - 4 = 1 \neq 3$. (This case is rejected)

Case 2: If $P = 6$

The multiplication is $16 \times 6$.

$16 \times 6 = 96$

Comparing this with $Q6$, we get $Q = 9$.

Now, check the condition $Q - P = 3$:

Final Answer:

The values are $P = 6$ and $Q = 9$.

Given:

$1AB + CCA = 697$

Condition: No carry-over in addition.

To Find:

The value of $A + B + C$.

Solution:

Let us write the addition in vertical form:

$\begin{array}{ccc} & 1 & A & B \\ + & C & C & A \\ \hline & 6 & 9 & 7 \\ \hline \end{array}$

Since there is no carry-over, we can solve column by column:

1. Hundreds column: $1 + C = 6 \Rightarrow C = 5$

2. Tens column: $A + C = 9$

Substituting $C = 5$:

$A + 5 = 9 \Rightarrow A = 4$

3. Units column: $B + A = 7$

Substituting $A = 4$:

$B + 4 = 7 \Rightarrow B = 3$

Now, we find the sum of $A$, $B$, and $C$:

$A + B + C = 4 + 3 + 5 = 12$

Final Answer:

The value of $A + B + C$ is $12$.

Given:

A five-digit number $AABAA$ is divisible by $33$.

To Find:

All possible numbers of this form.

Solution:

If a number is divisible by $33$, it must be divisible by both its co-prime factors 3 and 11.

1. Divisibility by 11:

The difference between the sum of digits at odd places and even places must be $0$ or a multiple of $11$.

$\text{Sum of odd places (1st, 3rd, 5th)} = A + B + A = 2A + B$

$\text{Sum of even places (2nd, 4th)} = A + A = 2A$

$\text{Difference} = (2A + B) - (2A) = B$

For the number to be divisible by $11$, the difference $B$ must be $0$ or a multiple of $11$. Since $B$ is a single digit, $B = 0$.

2. Divisibility by 3:

The sum of all digits must be divisible by $3$.

$\text{Sum of digits} = A + A + B + A + A = 4A + B$

Since we found $B = 0$:

$\text{Sum of digits} = 4A + 0 = 4A$

For $4A$ to be divisible by $3$, $A$ must be a multiple of $3$ (as $4$ is not divisible by $3$).

Possible values for the non-zero digit $A$: $3, 6, 9$.

1. If $A = 3$, the number is $33033$.

2. If $A = 6$, the number is $66066$.

3. If $A = 9$, the number is $99099$.

Final Answer:

The numbers of this form are $33033$, $66066$, and $99099$.

Given:

$\begin{array}{cc} & A & A \\ + & A & A \\ \hline X & A & Z \\ \hline \end{array}$

Solution:

Looking at the addition in the tens column, we see that $A + A + \text{carry} = XA$. Since $A$ is a digit, the maximum value of $A + A$ is $18$. Thus, $X$ must be 1.

From the units column:

$A + A = Z$ (with no carry) or $A + A = 10 + Z$ (with carry 1).

From the tens column:

$A + A + \text{carry} = 10 + A$ (since $X=1$)

$2A + \text{carry} = 10 + A$

$A + \text{carry} = 10$

If carry is 0, $A = 10$ (not possible). Thus, carry must be 1.

$A + 1 = 10 \Rightarrow A = 9$

Substituting $A = 9$ in the units column with carry 1:

$9 + 9 = 18 \Rightarrow Z = 8$

Checking: $99 + 99 = 198$. Here $X=1, A=9, Z=8$.

Final Answer: $A = 9, X = 1, Z = 8$.

Given:

$\begin{array}{cc} & 8 & 5 \\ + & 4 & A \\ \hline B & C & 3 \\ \hline \end{array}$

Solution:

1. From the units column: $5 + A$ ends in $3$. Since $A$ is a digit, $5 + A = 13$.

$A = 13 - 5 = 8$

There is a carry of 1 to the tens column.

2. From the tens column: $1 (\text{carry}) + 8 + 4 = BC$

$13 = BC$

Comparing the digits, we get $B = 1$ and $C = 3$.

Final Answer: $A = 8, B = 1, C = 3$.

Given:

$\begin{array}{cc} & B & 6 \\ + & 8 & A \\ \hline C & A & 2 \\ \hline \end{array}$

Solution:

1. From the units column: $6 + A$ ends in $2$. Thus, $6 + A = 12$.

$A = 12 - 6 = 6$

There is a carry of 1 to the tens column.

2. From the tens column: $1 (\text{carry}) + B + 8 = CA$

$9 + B = 10C + A$

Substitute $A = 6$:

$9 + B = 10C + 6$

$B + 3 = 10C$

Since $B$ is a digit and $10C$ is a multiple of 10, $B + 3$ must be 10. Thus, $B = 7$ and $C = 1$.

Final Answer: $A = 6, B = 7, C = 1$.

Given:

The following addition of two three-digit numbers:

$\begin{array}{cccc} & 1 & B & A \\ + & A & B & A \\ \hline & 8 & B & 2 \\ \hline \end{array}$

To Find:

The values of the digits A and B.

Solution:

We solve the cryptarithmetic addition by analyzing the place value columns from right to left.

Step 1: Units Column

From the units column, we have $A + A$ ending in $2$. This gives two possibilities for $A$:

1. $2A = 2 \Rightarrow A = 1$ (no carry-over)

2. $2A = 12 \Rightarrow A = 6$ (carry-over of $1$ to the tens place)

Step 2: Hundreds Column

Looking at the hundreds column, we have $1 + A = 8$.

If $A = 1$, the sum is $2$, which does not match the result $8$.

If $A = 6$, then $1 + 6 = 7$. To get $8$ as the result, there must be a carry-over of 1 from the tens place.

This confirms that $A = 6$ and that a carry-over is required from the tens place.

Step 3: Tens Column

Since $A = 6$, the units column ($6 + 6 = 12$) provides a carry-over of $1$ to the tens column. The sum in the tens column ends in $B$ and must provide a carry-over of $1$ to the hundreds column.

Solving for $B$:

$1 + 2B = 10 + B$

$2B - B = 10 - 1$

$B = 9$

Verification:

Substitute $A = 6$ and $B = 9$ into the addition:

$\begin{array}{cc} & 1 & 9 & 6 \\ + & 6 & 9 & 6 \\ \hline & 8 & 9 & 2 \\ \hline \end{array}$

The addition is mathematically correct: $196 + 696 = 892$.

Final Answer:

The required values are A = 6 and B = 9.

Given:

$\begin{array}{ccc} & C & B & A \\ + & C & B & A \\ \hline 1 & A & 3 & 0 \\ \hline \end{array}$

Solution:

1. Units column: $A + A = 0$ or $10$. If $A=0$, then $B+B=3$ (not possible). Thus $A = 5$. Carry = 1.

2. Tens column: $1 (\text{carry}) + B + B$ ends in 3. $2B + 1 = 13 \Rightarrow 2B = 12 \Rightarrow B = 6$. Carry = 1.

3. Hundreds column: $1 (\text{carry}) + C + C = 10 + A$.

$1 + 2C = 10 + 5 \Rightarrow 1 + 2C = 15 \Rightarrow 2C = 14 \Rightarrow C = 7$.

Checking: $765 + 765 = 1530$. Matches.

Final Answer: $A = 5, B = 6, C = 7$.

Given:

$\begin{array}{ccc} & B & A & A \\ + & B & A & A \\ \hline & 3 & A & 8 \\ \hline \end{array}$

Solution:

1. Units column: $A + A = 8$ or $18$.

Case 1: $A = 4$. Tens: $A + A = 4 + 4 = 8$. But the tens digit in the result is $A=4$. Not matching.

Case 2: $A = 9$. Units: $9 + 9 = 18$ (Carry 1). Tens: $1 (\text{carry}) + 9 + 9 = 19$ (Matches $A=9$, carry 1). Hundreds: $1 (\text{carry}) + B + B = 3$.

$1 + 2B = 3 \Rightarrow 2B = 2 \Rightarrow B = 1$.

Checking: $199 + 199 = 398$. Matches.

Final Answer: $A = 9, B = 1$.

Given:

$\begin{array}{cccc} & A & 0 & 1 & B \\ + & 1 & 0 & A & B \\ \hline & B & 1 & 0 & 8 \\ \hline \end{array}$

Solution:

1. Units column: $B + B = 8$ or $18$. If $B=4$, then thousands $A+1=4 \Rightarrow A=3$. Tens: $1+A=0 \Rightarrow 1+3=4 \neq 0$.

2. If $B=9$: Units $9+9=18$ (Carry 1). Tens: $1 (\text{carry}) + 1 + A = 10 \Rightarrow A = 8$ (Carry 1). Hundreds: $1 (\text{carry}) + 0 + 0 = 1$ (Matches). Thousands: $A + 1 = 8 + 1 = 9$ (Matches $B=9$).

Checking: $8019 + 1089 = 9108$. Matches.

Final Answer: $A = 8, B = 9$.

Given:

$\begin{array}{cc} & A & B \\ \times & & 6 \\ \hline C & 6 & 8 \\ \hline \end{array}$

Solution:

1. Units place: $B \times 6$ must end with the digit $8$. The possible values for $B$ are $3$ (since $3 \times 6 = 18$) and $8$ (since $8 \times 6 = 48$).

2. Case 1: If $B = 3$

$\text{Carry} = 1$. For the tens place: $A \times 6 + 1$ must end in $6$.

$6A + 1 = \dots 6 \Rightarrow 6A = \dots 5$. Since $6A$ is always even, no digit $A$ satisfies this.

3. Case 2: If $B = 8$

$\text{Carry} = 4$. For the tens place: $A \times 6 + 4$ must end in $6$.

$6A + 4 = \dots 6 \Rightarrow 6A = \dots 2$.

This is possible if $A = 2$ ($6 \times 2 = 12$) or $A = 7$ ($6 \times 7 = 42$).

If $A = 2$: $28 \times 6 = 168$. Here, $C = 1$. This matches the form $C68$.

If $A = 7$: $78 \times 6 = 468$. Here, $C = 4$. This also matches the form $C68$.

Final Answer:

The possible values are A = 2, B = 8, C = 1 or A = 7, B = 8, C = 4.

Given:

$\begin{array}{cc} & A & B \\ \times & A & B \\ \hline 6 & A & B \\ \hline \end{array}$

Solution:

The problem implies that a two-digit number $AB$ squared results in a three-digit number $6AB$.

Let the two-digit number be $x$.

$x^2 - x - 600 = 0$

$(x - 25)(x + 24) = 0$

Since $x$ must be a positive number, $x = 25$.

Comparing $AB$ with $25$, we get $A = 2$ and $B = 5$.

Checking: $25 \times 25 = 625$. Here $A=2$ and $B=5$ are correctly placed.

Final Answer:

The values are A = 2 and B = 5.

Given:

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

Solution:

The product of a two-digit number $AA$ and a single digit $A$ is a three-digit number $CAB$.

The generalised form is $(11A) \times A = 100C + 10A + B$, which simplifies to $11A^2 = 100C + 10A + B$.

Let us test values of $A$ where $11A^2$ results in a three-digit number ending with digit $A$ at the tens place:

1. If $A = 4$: $44 \times 4 = 176$ (Tens digit is 7, not 4).

2. If $A = 5$: $55 \times 5 = 275$ (Tens digit is 7, not 5).

3. If $A = 9$: $99 \times 9 = 891$.

In this case, the tens digit of the result is $9$, which matches $A$.

Comparing $891$ with $CAB$:

$C = 8, A = 9, B = 1$.

Final Answer:

The values are A = 9, B = 1, C = 8.

Given:

$\begin{array}{cc} & A & B \\ - & B & 7 \\ \hline & 4 & 5 \\ \hline \end{array}$

Solution:

1. Units column: $B - 7 = 5$. Since $B$ is a digit, we must have borrowed from the tens place.

$(10 + B) - 7 = 5$

$B + 3 = 5 \Rightarrow B = 2$

2. Tens column: Having given a borrow, the digit $A$ becomes $A - 1$.

Substituting $B = 2$:

$A - 1 - 2 = 4$

$A - 3 = 4 \Rightarrow A = 7$

Checking: $72 - 27 = 45$. Correct.

Final Answer:

The values are A = 7 and B = 2.

Given:

$\begin{array}{cccc} & 8 & A & B & C \\ - & A & B & C & 5 \\ \hline & D & 4 & 8 & 8 \\ \hline \end{array}$

Solution:

We solve the subtraction column by column from right to left:

1. Units place: $C - 5 = 8$. Borrowing is required: $10 + C - 5 = 8 \Rightarrow C = 3$.

2. Tens place: $(B - 1) - C = 8$. Substituting $C = 3$:

$B - 4 = 8$. Borrowing is required: $10 + B - 4 = 8 \Rightarrow B = 2$.

3. Hundreds place: $(A - 1) - B = 4$. Substituting $B = 2$:

$A - 3 = 4 \Rightarrow A = 7$. (No borrowing from thousands needed as $7-3=4$).

4. Thousands place: $8 - A = D$. Substituting $A = 7$:

$8 - 7 = 1 \Rightarrow D = 1$.

Final Answer:

The values are A = 7, B = 2, C = 3, D = 1.

Given:

The division $2A7 \div A = 33$.

The condition $B - A = 1$.

To Find:

The value of digit $A$.

Solution:

We are given that a three-digit number $2A7$ divided by a single digit $A$ equals $33$. This can be written as:

In generalised form, the three-digit number $2A7$ is $200 + 10A + 7$. Substituting this into equation (i):

$207 + 10A = 33A$

$207 = 33A - 10A$

$207 = 23A$

$A = \frac{207}{23}$

$A = 9$

Substituting $A = 9$ into the second condition $B - A = 1$:

$B - 9 = 1 \Rightarrow B = 10$.

Final Answer:

The value of $A$ is $9$.

Given:

The number $212x5$ is a multiple of $3$ and $11$.

To Find:

The value of the digit $x$.

Solution:

Step 1: Divisibility by 11

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

Sum of digits at odd places (from right) $= 5 + 2 + 2 = 9$

Sum of digits at even places (from right) $= x + 1$

$8 - x = 0$

$x = 8$

Step 2: Check Divisibility by 3

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

$\text{Sum of digits} = 2 + 1 + 2 + x + 5 = 10 + x$

Substituting $x = 8$:

$\text{Sum of digits} = 10 + 8 = 18$

Since $18$ is divisible by $3$, $x = 8$ is correct.

Final Answer:

The value of $x$ is $8$.

Given:

The number $31k2$ is divisible by $6$.

To Find:

All possible values of digit $k$.

Solution:

A number is divisible by $6$ if it is divisible by both 2 and 3.

1. Divisibility by 2: The last digit of $31k2$ is $2$, which is even. So, the number is always divisible by $2$ regardless of $k$.

2. Divisibility by 3: The sum of the digits must be divisible by $3$.

$\text{Sum of digits} = 3 + 1 + k + 2 = 6 + k$

For $6 + k$ to be divisible by $3$, $k$ must be a multiple of $3$.

The possible values for digit $k$ are $0, 3, 6, \text{ and } 9$.

Final Answer:

The possible values of $k$ are 0, 3, 6, or 9.

Given:

The number $1y3y6$ is divisible by $11$.

To Find:

The value of digit $y$.

Solution:

For divisibility by $11$, the difference between the sum of digits at odd places and even places must be $0$ or divisible by $11$.

$\text{Sum of digits at odd places (from right)} = 6 + 3 + 1 = 10$

$\text{Sum of digits at even places (from right)} = y + y = 2y$

Setting the difference to $0$:

$2y = 10$

$y = 5$

Final Answer:

The value of $y$ is $5$.

Given:

The number $756x$ is a multiple of $11$.

To Find:

The value of the digit $x$.

Solution:

According to the divisibility rule of 11, the difference between the sum of digits at odd places and the sum of digits at even places must be either $0$ or a multiple of $11$.

Let's identify the digits at different places (from right to left):

$\text{Digit at 1st place (Odd)} = x$

$\text{Digit at 2nd place (Even)} = 6$

$\text{Digit at 3rd place (Odd)} = 5$

$\text{Digit at 4th place (Even)} = 7$

Now, we find the sums:

$\text{Sum of digits at odd places} = x + 5$

$\text{Sum of digits at even places} = 6 + 7 = 13$

The difference is given by:

$\text{Difference} = 13 - (x + 5)$

$\text{Difference} = 8 - x$

For the number to be divisible by $11$, this difference must be $0$.

$8 - x = 0$

$x = 8$

Final Answer:

The value of $x$ is $8$.

Given:

1. The addition: $2a3 + 326 = 5b9$.

2. $5b9$ is divisible by $9$.

To Find:

The value of $b - a$.

Solution:

First, we find the value of $b$ using the divisibility rule of $9$. A number is divisible by $9$ if the sum of its digits is a multiple of $9$.

$\text{Sum of digits of } 5b9 = 5 + b + 9 = 14 + b$

The smallest multiple of $9$ greater than $14$ is $18$.

$14 + b = 18$

$b = 4$

Now, we use the addition: $2a3 + 326 = 549$ (since $b = 4$).

Let's check the columns:

Units column: $3 + 6 = 9$ (Satisfied)

Tens column: $a + 2 = b \Rightarrow a + 2 = 4$

$a = 2$

Hundreds column: $2 + 3 = 5$ (Satisfied)

Finally, we calculate $b - a$:

$b - a = 4 - 2 = 2$

Final Answer:

The value of $b - a$ is $2$.

Given:

$E = 3$, $B = 7$, and $A = 4$.

$\begin{array}{cccccc} & & B & A & S & E \\ + & & B & A & L & L \\ \hline & G & A & M & E & S \\ \hline \end{array}$

Solution:

Substituting the given values into the addition:

$\begin{array}{cccccc} & & 7 & 4 & S & 3 \\ + & & 7 & 4 & L & L \\ \hline & G & 4 & M & 3 & S \\ \hline \end{array}$

1. Units column: $3 + L = S$ (Assuming no carry to tens) or $3 + L = 10 + S$ (Carry 1).

2. Thousands column: $7 + 7 = 14$. The result shows $G4$. This implies $G = 1$ and there is no carry from the hundreds column ($14 + 0 = 14$).

3. Hundreds column: $4 + 4 + \text{carry}_2 = M$. Since there is no carry to thousands, $M$ can be $8$ or $9$.

4. Tens column: $S + L + \text{carry}_1 = 3$ or $13$. Since $M$ is $8$ or $9$, the units column likely gives a carry. Let's try $3 + L = S$.

Substitute $S = L + 3$ into $S + L = 13$ (assuming carry 1 to hundreds):

$(L + 3) + L = 13 \Rightarrow 2L = 10 \Rightarrow L = 5$

Then, $S = 5 + 3 = 8$.

Now, checking hundreds with carry 1: $1 (\text{carry}) + 4 + 4 = 9$. Thus, $M = 9$.

Final Answer:

The digits are G = 1, M = 9, S = 8, L = 5.

Given:

$D = 3$, $L = 7$, and $A = 8$.

$\begin{array}{ccccc} & & M & A & D \\ + & & & A & S \\ + & & & & A \\ \hline & B & U & L & L \\ \hline \end{array}$

Solution:

Substituting the given values:

$\begin{array}{ccccc} & & M & 8 & 3 \\ + & & & 8 & S \\ + & & & & 8 \\ \hline & B & U & 7 & 7 \\ \hline \end{array}$

1. Units column: $3 + S + 8 = 11 + S$. The result ends in $7$.

$11 + S = 17 \Rightarrow S = 6$. (Carry of $1$ to tens).

2. Tens column: $1 (\text{carry}) + 8 + 8 = 17$. The result ends in $7$. (Carry of $1$ to hundreds).

3. Hundreds/Thousands: $1 (\text{carry}) + M = BU$.

Since $BU$ is a two-digit number and $M$ is a single digit, $M$ must be $9$.

$1 + 9 = 10$.

Comparing $10$ with $BU$, we get $B = 1$ and $U = 0$.

Final Answer:

The other digits are M = 9, S = 6, B = 1, U = 0.

Given:

A two-digit number $ab$. The reversed number is $ba$. The difference $(ab - ba)$ is a perfect cube.

Solution:

Let the two-digit number be $10a + b$. The reversed number is $10b + a$.

$\text{Difference} = (10a + b) - (10b + a) = 9a - 9b = 9(a - b)$

For $9(a - b)$ to be a perfect cube, let's test possible values of $a - b$ (which range from $1$ to $9$):

1. If $a - b = 1 \Rightarrow 9(1) = 9$ (Not a cube)

2. If $a - b = 2 \Rightarrow 9(2) = 18$ (Not a cube)

3. If $a - b = 3 \Rightarrow 9(3) = 27 = 3^3$ (Perfect cube)

Any higher value of $a - b$ like $4, 5, \dots, 9$ will not result in a perfect cube when multiplied by $9$ (as $64$ and $125$ are not multiples of $9$).

So, we find pairs $(a, b)$ such that $a - b = 3$:

$(3, 0) \rightarrow 30$

$(4, 1) \rightarrow 41$

$(5, 2) \rightarrow 52$

$(6, 3) \rightarrow 63$

$(7, 4) \rightarrow 74$

$(8, 5) \rightarrow 85$

$(9, 6) \rightarrow 96$

Similarly, if we consider $ba - ab$ to be a positive cube, then $b - a = 3$:

$(1, 4) \rightarrow 14$, $(2, 5) \rightarrow 25$, $(3, 6) \rightarrow 36$, $(4, 7) \rightarrow 47$, $(5, 8) \rightarrow 58$, $(6, 9) \rightarrow 69$.

Final Answer:

There are 13 such numbers (if considering absolute difference): 30, 41, 52, 63, 74, 85, 96, 14, 25, 36, 47, 58, 69.

Given:

The multiplication $12345679 \times 9$.

Solution:

First, we perform the basic multiplication:

$\begin{array}{cc}& & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 9 \\ \times & & & & & & & & & 9 \\ \hline & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}$

The result is $111,111,111$. We use this result to answer the sub-questions.

(a) $12345679 \times 45$:

$45$ can be written as $9 \times 5$.

$12345679 \times 9 \times 5 = 111,111,111 \times 5 = 555,555,555$.

(b) $12345679 \times 63$:

$63$ can be written as $9 \times 7$.

$12345679 \times 9 \times 7 = 111,111,111 \times 7 = 777,777,777$.

(c) Multiplication to get $888,888,888$:

$888,888,888 = 111,111,111 \times 8$

Since $111,111,111 = 12345679 \times 9$,

$888,888,888 = (12345679 \times 9) \times 8 = 12345679 \times 72$.

The number is 72.

(d) Multiplication to get $999,999,999$:

$999,999,999 = 111,111,111 \times 9$

$999,999,999 = (12345679 \times 9) \times 9 = 12345679 \times 81$.

The number is 81.

(i) Given:

$\begin{array}{ccc} & P & Q \\ \times & & 6 \\ \hline Q & Q & Q \\ \hline \end{array}$

The product $Q \times 6$ must end in $Q$. Possible values for $Q$ are $2, 4, 6,$ and $8$.

If $Q = 4$:

$4 \times 6 = 24$. So, the unit digit is $4$ and carry is $2$.

Tens place: $6P + 2 = QQ \Rightarrow 6P + 2 = 44$.

$6P = 42 \Rightarrow P = 7$.

Check: $74 \times 6 = 444$. Correct.

Final Answer (i): P = 7, Q = 4

(ii) Given:

$\begin{array}{cccc} & 2 & L & M \\ + & L & M & 1 \\ \hline & M & 1 & 8 \\ \hline \end{array}$

1. Units column: $M + 1 = 8 \Rightarrow M = 7$.

2. Tens column: $L + M = 11$ (ends in 1). Since $M = 7$, $L + 7 = 11 \Rightarrow L = 4$. (Carry 1).

3. Hundreds column: $2 + L + \text{carry} = M \Rightarrow 2 + 4 + 1 = 7$. Correct.

Final Answer (ii): L = 4, M = 7

Given:

The number is $148101B095$. It is divisible by $33$.

Solution:

Divisibility by $33$ implies the number is divisible by both 3 and 11.

Divisibility by 11:

$\text{Sum of odd places (from right)} = 5 + 0 + 1 + 1 + 4 = 11$

$\text{Sum of even places (from right)} = 9 + B + 0 + 8 + 1 = 18 + B$

$\text{Difference} = (18 + B) - 11 = 7 + B$

For divisibility by $11$, $7 + B$ must be $11$ (as $B$ is a digit).

$B = 11 - 7 = 4$.

Divisibility by 3:

$\text{Sum of digits} = 1 + 4 + 8 + 1 + 0 + 1 + 4 + 0 + 9 + 5 = 33$.

Since $33$ is divisible by $3$, the value $B = 4$ is correct.

Final Answer:

The value of B is 4.

Given:

The number $123123A4$ is divisible by $11$.

Solution:

$\text{Sum of odd places (from right)} = 4 + 3 + 1 + 2 = 10$

$\text{Sum of even places (from right)} = A + 2 + 3 + 1 = 6 + A$

For divisibility by $11$, the difference must be $0$.

$(6 + A) - 10 = 0$

$A - 4 = 0 \Rightarrow A = 4$.

Final Answer:

The value of A is 4.

Given:

The number $56x32y$ is divisible by $18$.

Solution:

Divisibility by $18$ implies the number is divisible by both 2 and 9.

1. Divisibility by 2: The unit digit $y$ must be even ($0, 2, 4, 6,$ or $8$).

2. Divisibility by 9: The sum of digits must be a multiple of $9$.

$\text{Sum of digits} = 5 + 6 + x + 3 + 2 + y = 16 + x + y$.

We want the least value of y. Let's test even digits starting from $0$.

If $y = 0$, then sum $= 16 + x$. For this to be divisible by $9$, $16 + x = 18 \Rightarrow x = 2$.

Since $x = 2, y = 0$ is a valid solution, the least value of $y$ is $0$.

Final Answer:

The least value of y is 0.