# Exercise 16.2

Question 1

If 21*y*5 is a multiple of 9, where *y* is a digit, what is the value of *y*?

Sol :

If a number is a multiple of 9, then the sum of its digits will be divisible by 9.

Sum of digits of 21*y*5 = 2 + 1 + *y* + 5 = 8 + *y*

Hence, 8 + *y* should be a multiple of 9.

This is possible when 8 + *y* is any one of these numbers 0, 9, 18, 27, and so on …

However, since *y* is a single digit number, this sum can be 9 only. Therefore, *y* should be 1 only.

Question 2

If 31*z*5 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?

Sol :

If a number is a multiple of 9, then the sum of its digits will be divisible by 9.

Sum of digits of 31*z*5 = 3 + 1 + *z* + 5 = 9 + *z*

Hence, 9 + *z* should be a multiple of 9.

This is possible when 9 + *z* is any one of these numbers 0, 9, 18, 27, and so on …

However, since *z* is a single digit number, this sum can be either 9 or 18. Therefore, *z* should be either 0 or 9.

Question 3

If 24*x* is a multiple of 3, where *x* is a digit, what is the value of *x*?

(Since 24*x* is a multiple of 3, its sum of digits 6 + *x* is a multiple of 3; so 6 + *x* is one of these numbers: 0, 3, 6, 9, 12, 15, 18…. But since *x* is a digit, it can only be that 6 + *x* = 6 or 9 or 12 or 15. Therefore, *x* = 0 or 3 or 6 or 9. Thus, *x* can have any of four different values)

Sol :

Since 24*x* is a multiple of 3, the sum of its digits is a multiple of 3.

Sum of digits of 24*x* = 2 + 4 + *x* = 6 + *x*

Hence, 6 + *x* is a multiple of 3.

This is possible when 6 + *x* is any one of these numbers 0, 3, 6, 9, and so on …

Since *x* is a single digit number, the sum of the digits can be 6 or 9 or 12 or 15 and thus, the value of *x* comes to 0 or 3 or 6 or 9 respectively.

Thus, *x* can have its value as any of the four different values 0, 3, 6, or 9.

Question 4

If 31*z*5 is a multiple of 3, where *z* is a digit, what might be the values of *z*?

Sol :

Since 31*z*5 is a multiple of 3, the sum of its digits will be a multiple of 3.

That is, 3 + 1 + *z* + 5 = 9 + *z* is a multiple of 3.

This is possible when 9 + *z* is any one of 0, 3, 6, 9, 12, 15, 18, and so on …

Since *z* is a single digit number, the value of 9 + *z* can only be 9 or 12 or 15 or 18 and thus, the value of *x* comes to 0 or 3 or 6 or 9 respectively.

Thus, *z* can have its value as any one of the four different values 0, 3, 6, or 9.