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

Explanation

Since `31z5` is a multiple of `3` , the sum of its digits must be a multiple of `3` i.e. `3+1+z+5=9+z`
Hence `9+z` is a multiple of `3`.
Since `z` is a single digit that can value from `0 `to `9`.
Possible values of `z` will be,
`9+0=9`
`9+3=12`
`9+6=15`
`9+9=18`
Thus, there can be four possible values i.e. `0,3,6` or`9`.