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

Since 31*z*5 is a multiple of 3, the sum of its digits will be a multiple of 3

That is, 3 + 1 + *z* + 5 = 9 + *z* is a multiple of 3

This is possible when 9 + *z* is any one of 0, 3, 6, 9, 12, 15, 18, and so on …

Since *z* is a single digit number, the value of 9 + *z* can only be 9 or 12 or 15 or 18 and thus, the value of *x* comes to 0 or 3 or 6 or 9 respectively

Thus, *z* can have its value as any one of the four different values 0, 3, 6, or 9

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