Question

Q2. If 31z5 is a multiple of 9, where z is a digit, what is​

Answer 1

$\bold\red{0~or~9}$

According to the divisibility rule of 9, if the sum of the digits in a number is divisible by 9, then the entire number is also divisible by 9 .

3 + 1 + z + 5 = 9 ➜ divisible by 9 ✔

3 + 1 + 9 + 5 = 18 ➜ divisible by 9 ✔

$\red{So,~z~can ~be ~0 ~or~ 9}$