## MP Board Class 8th Maths Solutions Chapter 16 Playing with Numbers Ex 16.2

Question 1.

If 21y5 is a multiple of 9, where y is a digit, what is the value of y?

Solution:

The given number is 21y5

For this number to be multiple of 9, the sum of its digits should be a multiple of 9.

∴ 2 + 1 + y + 5 is a multiple of 9.

or 8 + y is a multiple of 9.

For y to be a single digit number,

8 + y = 9 ⇒ y = 1

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?

Solution:

The given number is 31z5.

For this number to be multiple of 9, the sum of its digits should be a multiple of 9.

∴ 3 + 1 + z + 5 is a multiple of 9.

or z + 9 is a multiple of 9.

∴ For z to be a single digit number,

z + 9 = 9 or z + 9 = 18

⇒ z = 0 or z = 9

Hence, z = 0, 9

We get two answers, because 9 and 18, both are multiples of 9.

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Question 3.

If 24x is a multiple of 3, where x is a digit, what is the value of x? (Since 24x 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 o r12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of four different values).

Solution:

The given number 24x is a multiple of 3.

⇒ Sum of the digits of this number are also a multiple of 3.

∴ 2 + 4 + x = x + 6 is a multiple of 3.

For x to be a single digit number,

x + 6 = 6, 9, 12, 15 or x = 0, 3, 6, 9.

Question 4.

If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?

Solution:

As the given number 31z5 is a multiple of 3, the sum of its digits should also be a multiple of 3.

∴ 3 + 1 + z + 5 is a multiple of 3.

or z + 9 is a multiple of 3.

Now, for z to be a single digit number,

z + 9 = 9, 12, 15, 18 or z = 0, 3, 6, 9