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

Since 31z5 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.
Since z is a digit.

∴ 3+1+ z +5 = 9+ z ⇒ 9+z=9 ⇒ z=0
if 3+1+ z +5 = 9+ z ⇒ 9 + z = 12 ⇒ z=3
if 3+1+ z +5 = 9+ z ⇒ 9 + z = 15 ⇒ z=6
if 3+1+ z +5 = 9+ z ⇒ 9 + z = 18 ⇒ z=9

Hence, 0, 3, 6 and 9 are four possible answers.

Class 8 Maths Chapter 11 Exercise 16.2 Solution in Video

for more answers vist to:
https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-16/