NCERT Solutions Class 8 Mathematics 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?

Answer :
Since 21y⁵ 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 + 1 + y + 5 = 8 + y

⇒ 8 + y = 9

⇒ y = 1

Since 21y⁵ is a multiple of 9.

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 :
Since 31z⁵ 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.

∴ 3 + 1 + z + 5 = 9 + z

⇒9 + z = 9

⇒ z = 0

If ∴ 3 + 1 + z + 5 = 9 + z

⇒ 9 + z = 18

⇒ z = 9

Hence 0 and 9 are two possible answers.

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 or 12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of (four different values.)

Answer :
Since 24𝓍 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.

∴ 2 + 4 + 𝓍 = 6 + 𝓍

Since𝓍 is a digit.

⇒ 6 + 𝓍= 6 ⇒ 𝓍= 0

⇒ 6 + 𝓍= 9 ⇒ 𝓍= 3

⇒ 6 + 𝓍= 12 ⇒ 𝓍= 6

⇒ 6 + 𝓍=15 ⇒ 𝓍=9

Thus, 𝓍can have any of four different values.