If
31*z*5
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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