# Ncert Solutions For Class 8 Maths Ex 16.2

## Ncert Solutions For Class 8 Maths Chapter 16 Ex 16.2

1. What would be the value of m to make 23m3 to be a multiple of 9?

As, 23m3 is a multiple of 9

Therefore according to the divisibility rule of 9, the sum of all the digits should be a multiple of 9.

∴ 2 +3 +m + 3 = 8 + m

m + 8 = 9

m = 1

1. If 31p5 is a multiple of 9, where p is a digit, find the value of p?

In this problem you will get two results. Explain why??

Since 31p5 is a multiple of 9

So now, if a number is a multiple of 9, then the sum of the digits will be divisible by 9.

3+1+p+5=9+p$3+1+p+5=9+p$

9+p = 9      [Since, 9+p should be multiple of 9]

p = 0

If 3+1+p+5=9+p$3+1+p+5=9+p$

9 +p = 18

p = 9

However, since p is a single digit number, the sum can be either 9 or 18.

But in this case, 0 and 9 are two possible answers.

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

Since 24p is a multiple of 3.

Therefore according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.

Here, the sum of digits of 24p is = 2 + 4+ p

∴  2 + 4+ p=6 + p

Since ‘p’ is a digit.

6 +p = 6

p = 0

6 + p = 9

p = 3

6 + p = 12

p = 6

6 + p = 15

p = 9

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

Therefore, p can have any of the four different values.

1. If 31p5 is a multiple of 3, where p is a digit, find out the values of p.

Since 31p5 is a multiple of 3.

Therefore according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.

Here p is a digit.

∴ 3+1+p+5=9+p

9 + p = 9

p = 0

∴ 3+1+p+5=9+p

9 + p = 12

p = 3

∴ 3+1+p+5=9+p

9 + p = 15

p = 6

∴ 3+1+p+5=9+p

9 + p = 18

p = 9

Since p is a single digit number, the sum of the digits can be 9 or 12 or 15 or 18 and thus, the value of p comes to 0 or 3 0r 6 or 9 respectively.

Therefore, p can have any of the four different values.