• # question_answer 3)                 If $31z5$ is a multiple of 9, where z is a digit, what is the value of z? You will find that there are two answers for the last problem. Why is this so?

Since $31z5$ is a multiple of 9, its sum of digits $3+1+z+5=9+z$ is a multiple of 9; so $9+z$ is one of these numbers: 0, 9, 18, 27, 36, 45, ... But since z is a digit, it can only be that $9+z=9$ or 18. Therefore, $z=0$ or 9.