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

Dear Student

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.
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.

Thanks