2. If 31z5 is a multiple of 9, where z is a digit, what is the value of z?

We know that a number is a multiple of 9 if the sum of its digits is also divisible by 9.

Here, Sum of the digits of 31z5 = 3 + 1 + z + 5

So we have , 3 + 1 + z + 5 = 9k where k is an integer.

Checking for different values of k

For k = 1:

$$\Rightarrow$$ 3 + 1 + z + 5 = 9

$$\Rightarrow$$ z = 9 – 9 = 0

For k = 2:

We have, 3 + 1 + z + 5 = 18

$$\Rightarrow$$ z = 18 – 9 = 9

k = 3 is not possible because 3 + 1 + z + 5 = 27

$$\Rightarrow$$ z = 27 – 9 = 18 which is not a digit.

Hence the required value of z is 0 or 9