GSEB Class 8 Maths Solutions Chapter 16 Playing with Numbers Exercise 16.2

Question 1. If 21y5 is a multiple of 9, where y a digit, what is the value of y?
Answer: For a number to be a multiple of 9, the sum of its digits must also be divisible by 9. Here, the number is 21y5. The sum of its digits is \( 2 + 1 + y + 5 = 8 + y \). Since 21y5 is a multiple of 9, the sum \( 8 + y \) must be divisible by 9. As \( y \) is a digit, its value can range from 0 to 9. Possible multiples of 9 for \( 8 + y \) are 9, 18, 27, and so on. If \( 8 + y = 0 \), then \( y = -8 \), which is not a digit. If \( 8 + y = 9 \), then \( y = 9 - 8 \), which means \( y = 1 \). If \( 8 + y = 18 \), then \( y = 18 - 8 \), which means \( y = 10 \), not a single digit. Therefore, the only possible value for \( y \) is 1.
In simple words: For a number to be perfectly divided by 9, all its digits added together must also be divisible by 9. In the number 21y5, the digits sum up to 8 plus 'y'. For this sum to be a multiple of 9, and since 'y' is a single digit, 'y' has to be 1.

Exam Tip: Remember the divisibility rule for 9: the sum of the digits must be divisible by 9. Always consider the range of 'y' (0-9) when finding possible values.

Question 2. If 31z5 is a multiple of 9, where z is a digit, what is the value of z? You will find that there are two answers for the last problem. Why is this so?
Answer: For a number to be a multiple of 9, its digits' sum must be divisible by 9. The number is 31z5. The sum of its digits is \( 3 + 1 + z + 5 = 9 + z \). Since 31z5 is a multiple of 9, the sum \( 9 + z \) must be divisible by 9. As \( z \) is a digit, its value can be from 0 to 9. Possible multiples of 9 for \( 9 + z \) are 9, 18, 27, etc. If \( 9 + z = 9 \), then \( z = 9 - 9 \), which means \( z = 0 \). If \( 9 + z = 18 \), then \( z = 18 - 9 \), which means \( z = 9 \). If \( 9 + z = 27 \), then \( z = 18 \), which is not a single digit. So, the possible values for \( z \) are 0 and 9. There are two possible answers for \( z \) because both 0 and 9 are single digits that, when added to 9, produce a sum that is divisible by 9 (9 and 18 respectively).
In simple words: For 31z5 to be divisible by 9, the total of its digits (9 plus 'z') must also be divisible by 9. Since 'z' must be a single digit, it can be either 0 or 9. Both these numbers make the sum a multiple of 9, which is why there are two correct answers.

Exam Tip: When applying divisibility rules, always consider all possible single-digit values (0-9) for the unknown digit that satisfy the condition. Remember to explain *why* multiple solutions exist if the question asks for it.

Question 3. If 24x is a multiple of 3, where x is a digit, what is the value of x?
Answer: For a number to be a multiple of 3, the sum of its digits must also be divisible by 3. The number is 24x. The sum of its digits is \( 2 + 4 + x = 6 + x \). Since 24x is a multiple of 3, the sum \( 6 + x \) must be divisible by 3. As \( x \) is a digit, its value can range from 0 to 9. Possible multiples of 3 for \( 6 + x \) are 0, 3, 6, 9, 12, 15, 18, etc. If \( 6 + x = 0 \), then \( x = -6 \), not a digit. If \( 6 + x = 3 \), then \( x = -3 \), not a digit. If \( 6 + x = 6 \), then \( x = 0 \). If \( 6 + x = 9 \), then \( x = 3 \). If \( 6 + x = 12 \), then \( x = 6 \). If \( 6 + x = 15 \), then \( x = 9 \). If \( 6 + x = 18 \), then \( x = 12 \), not a single digit. Therefore, \( x \) can be 0, 3, 6, or 9. There are four different possible values for \( x \).
In simple words: A number is divisible by 3 if its digits add up to a number that is also divisible by 3. For 24x, the sum of digits is 6 plus 'x'. Since 'x' must be a single digit, the possible values for 'x' that make this sum divisible by 3 are 0, 3, 6, and 9.

Exam Tip: Always list all single-digit possibilities (0-9) that satisfy the divisibility rule. For divisibility by 3, all numbers that result in a sum of digits divisible by 3 are correct.

Question 4. If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Answer: For a number to be a multiple of 3, the sum of its digits must be divisible by 3. The number is 31z5. The sum of its digits is \( 3 + 1 + z + 5 = 9 + z \). Since 31z5 is a multiple of 3, the sum \( 9 + z \) must be divisible by 3. As \( z \) is a digit, its value can range from 0 to 9. Possible multiples of 3 for \( 9 + z \) are 0, 3, 6, 9, 12, 15, 18, 21, etc. If \( 9 + z = 0 \), then \( z = -9 \), which is not a digit. If \( 9 + z = 3 \), then \( z = -6 \), which is not a digit. If \( 9 + z = 6 \), then \( z = -3 \), which is not a digit. If \( 9 + z = 9 \), then \( z = 0 \). If \( 9 + z = 12 \), then \( z = 3 \). If \( 9 + z = 15 \), then \( z = 6 \). If \( 9 + z = 18 \), then \( z = 9 \). If \( 9 + z = 21 \), then \( z = 12 \), which is not a single digit. Therefore, the possible values for \( z \) are 0, 3, 6, or 9.
In simple words: For 31z5 to be divisible by 3, the total of its digits (9 plus 'z') must be divisible by 3. Since 'z' has to be a single digit, the possible values for 'z' are 0, 3, 6, or 9.

Exam Tip: When working with divisibility rules, always ensure your calculated values for the unknown digit fall within the valid range of single digits (0-9).