Chapter 16 Playing with Numbers

Solution:

We are given the addition problem:

$\begin{array}{cccc} & 3 & 1 & Q \\ + & 1 & Q & 3 \\ \hline & 5 & 0 & 1 \\ \hline \end{array}$

Let's analyze the addition column by column, starting from the rightmost (ones) column.

1. Ones Column:

$Q + 3$ results in a number with the ones digit 1.

Since Q is a single digit, the only possibility is that $Q + 3 = 11$.

So, we find that Q = 8. This gives 1 in the ones place and a carry-over of 1 to the tens column.

2. Tens Column:

We add the digits in the tens column along with the carry-over from the ones column:

$1$ (carry-over) $+ 1 + Q = 0$.

This means the sum results in a number with the ones digit 0. Substituting $Q = 8$:

$1 + 1 + 8 = 10$.

The sum is 10. The 0 is placed in the tens place of the result, and 1 is carried over to the hundreds column. This is consistent with the given sum.

3. Hundreds Column:

We add the digits in the hundreds column along with the carry-over from the tens column:

$1$ (carry-over) $+ 3 + 1 = 5$.

This matches the digit in the hundreds place of the result.

Thus, the value of Q is 8.

Let's verify the addition with Q = 8:

$\begin{array}{cccc} & 3 & 1 & 8 \\ + & 1 & 8 & 3 \\ \hline & 5 & 0 & 1 \\ \hline \end{array}$

The addition is correct. Hence, Q = 8.

Solution:

We are given the addition problem:

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

1. Ones Column:

The sum of the ones column is $A + A + A = 3 \times A$.

The result of this sum is a number that has A as its ones digit. Let's test the possible values for A (from 0 to 9) to see which ones satisfy this condition.

• If A = 0, then $3 \times 0 = 0$. The ones digit is 0. This is a possible solution.
• If A = 1, then $3 \times 1 = 3$. The ones digit is 3, not 1.
• If A = 2, then $3 \times 2 = 6$. The ones digit is 6, not 2.
• If A = 3, then $3 \times 3 = 9$. The ones digit is 9, not 3.
• If A = 4, then $3 \times 4 = 12$. The ones digit is 2, not 4.
• If A = 5, then $3 \times 5 = 15$. The ones digit is 5. This is a possible solution.
• If A = 6, then $3 \times 6 = 18$. The ones digit is 8, not 6.
• If A = 7, then $3 \times 7 = 21$. The ones digit is 1, not 7.
• If A = 8, then $3 \times 8 = 24$. The ones digit is 4, not 8.
• If A = 9, then $3 \times 9 = 27$. The ones digit is 7, not 9.

So, the possible values for A are 0 and 5.

Case 1: A = 0

If A = 0, the sum becomes $0 + 0 + 0 = 0$. The result is B0, which means B must also be 0. This is a trivial solution ($0+0+0=00$).

Case 2: A = 5

If A = 5, the sum in the ones column is $5 + 5 + 5 = 15$.

We write down 5 in the ones place of the result (which matches A) and carry over 1 to the tens column.

2. Tens Column:

The digit B in the result is simply the carry-over from the ones column. So, B = 1.

This gives us the solution A = 5 and B = 1.

Let's verify the addition with these values:

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

The addition is correct.

Solution:

The problem can be written as the multiplication: $(10B + A) \times (10B + 3) = 570 + A$.

Let's analyze the multiplication step-by-step using the standard long multiplication method.

$\begin{array}{cccc} & & B & A \\ & \times & B & 3 \\ \hline \end{array}$

1. Analyze the Ones Digit:

When we multiply the ones digits of the two numbers, $3 \times A$, the resulting number must have A as its ones digit. This is the same condition we saw in Example 2.

The possibilities for A are therefore 0 and 5.

Case 1: A = 0

If A = 0, the multiplication becomes $B0 \times B3 = 570$.

We can write this as an equation:

$(10B) \times (10B + 3) = 570$

$100B^2 + 30B = 570$

Dividing the entire equation by 10:

$10B^2 + 3B = 57$

Let's test integer values for B:

• If B = 1, $10(1)^2 + 3(1) = 13$ (too small)
• If B = 2, $10(2)^2 + 3(2) = 40 + 6 = 46$ (too small)
• If B = 3, $10(3)^2 + 3(3) = 90 + 9 = 99$ (too large)

There is no single-digit integer value for B that satisfies the equation. So, A cannot be 0.

Case 2: A = 5

If A = 5, the multiplication becomes $B5 \times B3 = 575$.

We can write this as an equation:

$(10B + 5) \times (10B + 3) = 575$

$100B^2 + 30B + 50B + 15 = 575$

$100B^2 + 80B + 15 = 575$

$100B^2 + 80B = 575 - 15$

$100B^2 + 80B = 560$

Dividing the entire equation by 10:

$10B^2 + 8B = 56$

Dividing by 2:

$5B^2 + 4B = 28$

Let's test integer values for B:

• If B = 1, $5(1)^2 + 4(1) = 5 + 4 = 9$ (too small)
• If B = 2, $5(2)^2 + 4(2) = 5(4) + 8 = 20 + 8 = 28$ (This is the solution!)
• If B = 3, $5(3)^2 + 4(3) = 45 + 12 = 57$ (too large)

The value B = 2 satisfies the equation. So, the digits are A = 5 and B = 2.

Let's verify the multiplication with these values: $25 \times 23$.

$\begin{array}{cc}& & 2 & 5 \\ \times & & 2 & 3 \\ \hline & & 7 & 5 \\ & 5 & 0 & \times \\ \hline & 5 & 7 & 5 \\ \hline \end{array}$

The result is 575, which matches the form 57A where A = 5. The solution is correct.

Solution:

We are given the addition problem:

$\begin{array}{cc} & 3 & A \\ + & 2 & 5 \\ \hline & B & 2 \\ \hline \end{array}$

Step 1: Analyze the ones column.

The sum in the ones column is $A + 5$, which results in a number ending with the digit 2. This implies that the sum must be 12.

$A + 5 = 12$

$A = 12 - 5 = 7$

So, we find that A = 7. This gives 2 in the ones place and a carry-over of 1 to the tens column.

Step 2: Analyze the tens column.

We add the digits in the tens column, including the carry-over from the ones column:

$1$ (carry-over) $+ 3 + 2 = B$

$6 = B$

So, we find that B = 6.

The solution is A = 7 and B = 6.

Verification:

$\begin{array}{cc} & 3 & 7 \\ + & 2 & 5 \\ \hline & 6 & 2 \\ \hline \end{array}$

The addition is correct.

Solution:

We are given the addition problem:

$\begin{array}{cccc} & & 4 & A \\ + & & 9 & 8 \\ \hline & C & B & 3 \\ \hline \end{array}$

Step 1: Analyze the ones column.

The sum in the ones column is $A + 8$, which results in a number ending with the digit 3. This implies that the sum must be 13.

$A + 8 = 13$

$A = 13 - 8 = 5$

So, A = 5. This gives 3 in the ones place and a carry-over of 1 to the tens column.

Step 2: Analyze the tens column.

We add the digits in the tens column, including the carry-over:

$1$ (carry-over) $+ 4 + 9 = 14$

The result 14 means that the digit in the tens place of the sum is 4, and there is a carry-over of 1 to the hundreds column. Therefore, B = 4.

Step 3: Analyze the hundreds column.

The digit C is the carry-over from the tens column.

So, C = 1.

The solution is A = 5, B = 4, and C = 1.

Verification:

$\begin{array}{cccc} & & 4 & 5 \\ + & & 9 & 8 \\ \hline & 1 & 4 & 3 \\ \hline \end{array}$

The addition is correct.

Solution:

We are given the multiplication problem:

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

Step 1: Analyze the ones digit multiplication.

The product of $A \times A$ must be a number whose ones digit is A. Let's check the possibilities for A:

• $0 \times 0 = 0$ (Possible)
• $1 \times 1 = 1$ (Possible)
• $2 \times 2 = 4$
• $3 \times 3 = 9$
• $4 \times 4 = 16$
• $5 \times 5 = 25$ (Possible)
• $6 \times 6 = 36$ (Possible)
• $7 \times 7 = 49$
• $8 \times 8 = 64$
• $9 \times 9 = 81$

So, A can be 0, 1, 5, or 6.

Step 2: Test the possible values of A.

• If A = 0: $10 \times 0 = 0$. The result should be 90. This is incorrect.
• If A = 1: $11 \times 1 = 11$. The result should be 91. This is incorrect.
• If A = 5: $15 \times 5 = 75$. The result should be 95. This is incorrect.
• If A = 6: $16 \times 6 = 96$. The result is 96, which matches the pattern '9A'. This is correct.

Therefore, the value of A is 6.

Verification:

$\begin{array}{cc} & 1 & 6 \\ \times & & 6 \\ \hline & 9 & 6 \\ \hline \end{array}$

The multiplication is correct.

Solution:

We are given the addition problem:

$\begin{array}{cc} & A & B \\ + & 3 & 7 \\ \hline & 6 & A \\ \hline \end{array}$

Step 1: Analyze the tens column.

The sum in the tens column is $A+3$, possibly with a carry-over from the ones column. The result is 6. This means A must be less than 6. If there's no carry-over, $A+3=6$, which gives $A=3$. If there is a carry-over of 1, $1+A+3=6$, which gives $A=2$.

Step 2: Analyze the ones column using the possible values of A.

The sum in the ones column is $B+7$, which results in a number ending with the digit A.

• Case 1: A = 3. If A=3, then $B+7$ must end in 3. This means $B+7=13$, which gives $B=6$. This would produce a carry-over of 1 to the tens column. But if we have a carry-over, then $1+A+3=6 \implies 1+3+3=7 \neq 6$. So this case is incorrect.
• Case 2: A = 2. If A=2, this implies there was a carry-over from the ones column (since $1+2+3=6$). For a carry-over to exist, $B+7$ must be greater than 9. The sum $B+7$ must end in A, which is 2. This means $B+7=12$. $B = 12 - 7 = 5$. So, we find B = 5.

This gives us the solution A = 2 and B = 5.

Verification:

$\begin{array}{cc} & 2 & 5 \\ + & 3 & 7 \\ \hline & 6 & 2 \\ \hline \end{array}$

The addition is correct.

Solution:

We are given the multiplication problem:

$\begin{array}{cccc} & A & B \\ \times & & 3 \\ \hline C & A & B \\ \hline \end{array}$

Step 1: Analyze the ones column.

The product $3 \times B$ must result in a number whose ones digit is B. Let's check the possibilities for B:

• $3 \times 0 = 0$ (Possible)
• $3 \times 5 = 15$ (Possible, ends in 5)

So, B can be 0 or 5.

Step 2: Test the possible values of B.

• Case 1: B = 5. If B=5, then $3 \times 5 = 15$. We write down 5 and carry over 1. Now for the tens column: $(3 \times A) + 1$ (carry-over) must end in A. Let's test values for A: If A=0, $(3 \times 0) + 1 = 1$. (Not 0) If A=1, $(3 \times 1) + 1 = 4$. (Not 1) If A=2, $(3 \times 2) + 1 = 7$. (Not 2) No single digit A satisfies this condition.
• Case 2: B = 0. If B=0, then $3 \times 0 = 0$. There is no carry-over. Now for the tens column: $3 \times A$ must end in A. From previous examples, we know this works for A=0 (trivial solution) and A=5. If A=5, then $3 \times 5 = 15$. We write down 5 (which is A) and carry over 1. This carry-over becomes the digit C. So, C=1.

This gives the solution A = 5, B = 0, C = 1.

Verification:

$\begin{array}{cccc} & 5 & 0 \\ \times & & 3 \\ \hline 1 & 5 & 0 \\ \hline \end{array}$

The multiplication is correct.

Solution:

We are given the multiplication problem:

$\begin{array}{cccc} & A & B \\ \times & & 5 \\ \hline C & A & B \\ \end{array}$

Step 1: Analyze the ones column.

The product $5 \times B$ must end in B. This is only possible if B is 0 or 5.

Step 2: Test the possible values of B.

• Case 1: B = 5. If B=5, then $5 \times 5 = 25$. We write down 5 and carry over 2. For the tens column: $(5 \times A) + 2$ must end in A. Let's test values for A: If A=1, $(5 \times 1) + 2 = 7$. (Not 1) If A=2, $(5 \times 2) + 2 = 12$. This ends in 2 (which is A). So, A=2 is a possibility. In this case, the product is 12, so the carry-over is 1. This carry-over becomes C. Thus C=1. This gives a solution: A=2, B=5, C=1.
• Case 2: B = 0. If B=0, then $5 \times 0 = 0$. There is no carry-over. For the tens column: $5 \times A$ must end in A. This is possible if A=0 (trivial solution) or A=5. If A=5, then $5 \times 5 = 25$. We write down 5 (which is A) and carry over 2. This carry-over becomes C. So C=2. This gives another solution: A=5, B=0, C=2.

Both are valid solutions. We can present either one. Let's use the first one we found.

The solution is A = 2, B = 5, C = 1.

Verification:

$\begin{array}{cccc} & 2 & 5 \\ \times & & 5 \\ \hline 1 & 2 & 5 \\ \end{array}$

The multiplication is correct.

Solution:

We are given the multiplication problem:

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

Step 1: Analyze the ones column.

The product $6 \times B$ must end in B. Let's check possibilities for B:

• $6 \times 0 = 0$ (Possible)
• $6 \times 2 = 12$ (Possible)
• $6 \times 4 = 24$ (Possible)
• $6 \times 6 = 36$ (Possible)
• $6 \times 8 = 48$ (Possible)

Step 2: Analyze the entire product.

The result of the multiplication is BBB, which is a number like 111, 222, 333, etc. This means the result is $B \times 111$. So, $(10A + B) \times 6 = 111 \times B$.

Let's test the possible values of B from Step 1.

• If B = 2, the result is 222. We check if 222 is divisible by 6. $222 \div 6 = 37$. The number AB would be 37. So A=3, B=7. This contradicts B=2.
• If B = 4, the result is 444. We check if 444 is divisible by 6. $444 \div 6 = 74$. The number AB would be 74. So A=7, B=4. This is consistent.
• If B = 6, the result is 666. $666 \div 6 = 111$. AB must be a 2-digit number. This is not possible.
• If B = 8, the result is 888. $888 \div 6 = 148$. AB must be a 2-digit number. This is not possible.

The only consistent solution is A = 7 and B = 4.

Verification:

$\begin{array}{cccc} & 7 & 4 \\ \times & & 6 \\ \hline 4 & 4 & 4 \\ \hline \end{array}$

The multiplication is correct.

Solution:

We are given the addition problem:

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

Step 1: Analyze the ones column.

The sum $1 + B$ results in a number ending with 0. This implies that the sum must be 10.

$1 + B = 10$

$B = 10 - 1 = 9$

So, B = 9. This gives 0 in the ones place and a carry-over of 1 to the tens column.

Step 2: Analyze the tens column.

We add the digits in the tens column, including the carry-over:

$1$ (carry-over) $+ A + 1 = B$

Substitute the value of B we found:

$1 + A + 1 = 9$

$A + 2 = 9$

$A = 9 - 2 = 7$

So, A = 7.

The solution is A = 7 and B = 9.

Verification:

$\begin{array}{cc} & 7 & 1 \\ + & 1 & 9 \\ \hline & 9 & 0 \\ \hline \end{array}$

The addition is correct.

Solution:

We are given the addition problem:

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

Step 1: Analyze the ones column.

The sum is $B + 1 = 8$.

$B = 8 - 1 = 7$

So, B = 7.

Step 2: Analyze the tens column.

The sum $A + B$ results in a number ending with 1. Since B=7, this means $A+7$ ends in 1. This implies the sum is 11.

$A + 7 = 11$

$A = 11 - 7 = 4$

So, A = 4. This gives 1 in the tens place and a carry-over of 1 to the hundreds column.

Step 3: Analyze the hundreds column.

We add the digits in the hundreds column, including the carry-over:

$1$ (carry-over) $+ 2 + A = B$

Let's check if this holds true with the values we found for A and B:

$1 + 2 + 4 = 7$

This gives 7, which is the value of B. The values are consistent.

The solution is A = 4 and B = 7.

Verification:

$\begin{array}{cccc} & 2 & 4 & 7 \\ + & 4 & 7 & 1 \\ \hline & 7 & 1 & 8 \\ \hline \end{array}$

The addition is correct.

Solution:

We are given the addition problem:

$\begin{array}{cccc} & 1 & 2 & A \\ + & 6 & A & B \\ \hline & A & 0 & 9 \\ \hline \end{array}$

Step 1: Analyze the tens column.

The sum $2 + A$, possibly with a carry-over, results in a number ending in 0. This implies the sum is 10.

Let's assume there is no carry-over from the ones column for now. Then $2 + A = 10$.

$A = 10 - 2 = 8$

So, let's tentatively say A = 8. This would give 0 in the tens place and a carry-over of 1 to the hundreds column.

Step 2: Analyze the ones column.

The sum is $A + B = 9$.

Using our value A = 8:

$8 + B = 9$

$B = 9 - 8 = 1$

So, B = 1. The sum is exactly 9, so there is no carry-over to the tens column. This is consistent with our assumption in Step 1.

Step 3: Analyze the hundreds column.

We add the digits in the hundreds column, including the carry-over from the tens column:

$1$ (carry-over) $+ 1 + 6 = A$

$8 = A$

This result, A=8, is consistent with the value we found in Step 1.

The solution is A = 8 and B = 1.

Verification:

$\begin{array}{cccc} & 1 & 2 & 8 \\ + & 6 & 8 & 1 \\ \hline & 8 & 0 & 9 \\ \hline \end{array}$

The addition is correct.

To check the divisibility of the number 21436587 by 9, we will use the divisibility rule for 9.

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

---

The given number is 21436587.

Let's find the sum of the digits of this number.

Sum of digits $= 2 + 1 + 4 + 3 + 6 + 5 + 8 + 7$

Sum of digits $= 3 + 4 + 3 + 6 + 5 + 8 + 7$

Sum of digits $= 7 + 3 + 6 + 5 + 8 + 7$

Sum of digits $= 10 + 6 + 5 + 8 + 7$

Sum of digits $= 16 + 5 + 8 + 7$

Sum of digits $= 21 + 8 + 7$

Sum of digits $= 29 + 7$

Sum of digits $= 36$

---

Now, we check if the sum of the digits, which is 36, is divisible by 9.

We know that $36 = 9 \times 4$.

So, 36 is divisible by 9.

---

Since the sum of the digits of 21436587 (which is 36) is divisible by 9, the number 21436587 is also divisible by 9.

$\textbf{Conclusion:}$ The number 21436587 is divisible by 9.

To check the divisibility of the number 152875 by 9, we will use the divisibility rule for 9.

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

---

The given number is 152875.

Let's find the sum of the digits of this number.

Sum of digits $= 1 + 5 + 2 + 8 + 7 + 5$

Sum of digits $= 6 + 2 + 8 + 7 + 5$

Sum of digits $= 8 + 8 + 7 + 5$

Sum of digits $= 16 + 7 + 5$

Sum of digits $= 23 + 5$

Sum of digits $= 28$

---

Now, we check if the sum of the digits, which is 28, is divisible by 9.

We can divide 28 by 9:

$28 \div 9 = 3$ with a remainder of $28 - (9 \times 3) = 28 - 27 = 1$.

Since the remainder is not 0, 28 is not divisible by 9.

---

Since the sum of the digits of 152875 (which is 28) is not divisible by 9, the number 152875 is also not divisible by 9.

$\textbf{Conclusion:}$ The number 152875 is not divisible by 9.

Given:

A three-digit number $24x$.

The number $24x$ is divisible by 9.

---

To Find:

The value of the digit $x$.

---

Solution:

We know the divisibility rule for 9, which states that a number is divisible by 9 if the sum of its digits is a multiple of 9.

For the given number $24x$, the sum of its digits is:

$2 + 4 + x = 6 + x$

According to the divisibility rule, for $24x$ to be divisible by 9, the sum $(6 + x)$ must be a multiple of 9.

Since $x$ is a single digit, its value can be any integer from 0 to 9. That is, $0 \le x \le 9$.

Let's find the possible values for the sum $(6 + x)$:

If $x=0$, the sum is $6+0 = 6$.

If $x=9$, the sum is $6+9 = 15$.

So, the sum $(6+x)$ must be a multiple of 9 that lies between 6 and 15 (inclusive).

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

The only multiple of 9 in the range [6, 15] is 9 itself.

Therefore, we must have:

$6 + x = 9$

Solving for $x$, we get:

$x = 9 - 6$

$x = 3$

The value of $x$ must be 3.

---

Verification:

If $x=3$, the number is 243.

The sum of the digits of 243 is $2 + 4 + 3 = 9$.

Since 9 is divisible by 9, the number 243 is also divisible by 9.

Hence, the value of $x$ is 3.

To check the divisibility of the number 2146587 by 3, we will use the divisibility rule for 3.

Divisibility Rule for 3: A number is divisible by 3 if the sum of its digits is divisible by 3.

---

The given number is 2146587.

Let's find the sum of the digits of this number.

Sum of digits $= 2 + 1 + 4 + 6 + 5 + 8 + 7$

Sum of digits $= 3 + 4 + 6 + 5 + 8 + 7$

Sum of digits $= 7 + 6 + 5 + 8 + 7$

Sum of digits $= 13 + 5 + 8 + 7$

Sum of digits $= 18 + 8 + 7$

Sum of digits $= 26 + 7$

Sum of digits $= 33$

---

Now, we check if the sum of the digits, which is 33, is divisible by 3.

We know that $33 = 3 \times 11$.

So, 33 is divisible by 3.

---

Since the sum of the digits of 2146587 (which is 33) is divisible by 3, the number 2146587 is also divisible by 3.

$\textbf{Conclusion:}$ The number 2146587 is divisible by 3.

To check the divisibility of the number 15287 by 3, we will use the divisibility rule for 3.

Divisibility Rule for 3: A number is divisible by 3 if the sum of its digits is divisible by 3.

---

The given number is 15287.

Let's find the sum of the digits of this number.

Sum of digits $= 1 + 5 + 2 + 8 + 7$

Sum of digits $= 6 + 2 + 8 + 7$

Sum of digits $= 8 + 8 + 7$

Sum of digits $= 16 + 7$

Sum of digits $= 23$

---

Now, we check if the sum of the digits, which is 23, is divisible by 3.

We can divide 23 by 3:

$23 \div 3 = 7$ with a remainder of $23 - (3 \times 7) = 23 - 21 = 2$.

Since the remainder is not 0, 23 is not divisible by 3.

---

Since the sum of the digits of 15287 (which is 23) is not divisible by 3, the number 15287 is also not divisible by 3.

$\textbf{Conclusion:}$ The number 15287 is not divisible by 3.

Solution:

We are given that the number 21y5 is a multiple of 9.

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

The sum of the digits of 21y5 is:

$2 + 1 + y + 5 = 8 + y$

For 21y5 to be a multiple of 9, the sum $(8 + y)$ must be a multiple of 9.

Since y is a single digit, its value must be an integer from 0 to 9. So, the possible values for the sum $(8 + y)$ are from $8+0=8$ to $8+9=17$.

The only multiple of 9 between 8 and 17 is 9.

Therefore, we set the sum equal to 9:

$8 + y = 9$

$y = 9 - 8$

$y = 1$

Hence, the value of y is 1.

Solution:

We are given that the number 31z5 is a multiple of 9.

By the divisibility rule of 9, the sum of the digits must be divisible by 9.

The sum of the digits of 31z5 is:

$3 + 1 + z + 5 = 9 + z$

For 31z5 to be a multiple of 9, the sum $(9 + z)$ must be a multiple of 9.

Since z is a single digit, its value must be an integer from 0 to 9. The possible values for the sum $(9 + z)$ are from $9+0=9$ to $9+9=18$.

The multiples of 9 in the range [9, 18] are 9 and 18. This gives us two possibilities.

Case 1:

$9 + z = 9$

$z = 0$

Case 2:

$9 + z = 18$

$z = 18 - 9 = 9$

Thus, the possible values of z are 0 and 9.

---

Reason for two answers:

The sum of the known digits of the number 31z5 is $3+1+5=9$. This sum is already a multiple of 9. For the total sum $(9+z)$ to remain a multiple of 9, the digit 'z' must also be a multiple of 9. The single-digit multiples of 9 are 0 and 9. Therefore, z can have two possible values (0 or 9), and in both cases, the sum of the digits will be a multiple of 9 (either 9 or 18).

Solution:

We are given that the number 24x is a multiple of 3.

According to the divisibility rule of 3, the sum of the digits of a number must be divisible by 3 for the number to be divisible by 3.

The sum of the digits of 24x is:

$2 + 4 + x = 6 + x$

For 24x to be a multiple of 3, the sum $(6 + x)$ must be a multiple of 3.

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

The sum of the known digits is $2+4=6$, which is already a multiple of 3. For the total sum $(6+x)$ to remain a multiple of 3, the digit 'x' must also be a multiple of 3.

The single-digit multiples of 3 are 0, 3, 6, and 9.

Therefore, the possible values of x are 0, 3, 6, or 9.

We can also check this by setting the sum to possible multiples of 3:

• $6 + x = 6 \implies x = 0$
• $6 + x = 9 \implies x = 3$
• $6 + x = 12 \implies x = 6$
• $6 + x = 15 \implies x = 9$

Any higher multiple of 3 (like 18) would give a two-digit value for x, which is not possible.

Solution:

We are given that the number 31z5 is a multiple of 3.

By the divisibility rule of 3, the sum of the digits must be divisible by 3.

The sum of the digits of 31z5 is:

$3 + 1 + z + 5 = 9 + z$

For 31z5 to be a multiple of 3, the sum $(9 + z)$ must be a multiple of 3.

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

The sum of the known digits is $3+1+5=9$, which is already a multiple of 3. For the total sum $(9+z)$ to remain a multiple of 3, the digit 'z' must also be a multiple of 3.

The single-digit multiples of 3 are 0, 3, 6, and 9.

Therefore, the possible values of z are 0, 3, 6, or 9.

We can verify this by checking the possible sums:

• If z = 0, sum = 9 (Multiple of 3)
• If z = 3, sum = 12 (Multiple of 3)
• If z = 6, sum = 15 (Multiple of 3)
• If z = 9, sum = 18 (Multiple of 3)