Question

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

Answer 1

Answer:

Given, 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.

That is, 3 + 1 + z + 5 = 9 + z

Therefore, 9 + z is a multiple of 3.

This is possible when the value of 9 + z is any of the values: 0, 3, 6, 9, 12, 15, and so on.

At z = 0, 9 + z = 9 + 0 = 9

At z = 3, 9 + z = 9 + 3 = 12

At z = 6, 9 + z = 9 + 6 = 15

At z = 9, 9 + z = 9 + 9 = 18

The value of 9 + z can be 9 or 12 or 15 or 18.

Hence 0, 3, 6 or 9 are four possible answers for z.