Question

A is a digit and 3A15 is a multiple of 9 which of the following can be the value of A a) 1 or 9 b)0 or 8 c)0 or 7 d) 0 or 9​

Answer 1

Step-by-step explanation:

Since 31z5 is a multiple of 3 ,the sum of its digits must be a multiple of 3 i.e. 3+1+z+5=9+z.

Hence 9+z is a multiple of 3

Since z is a single digit which can value from 0−9,

Possible values of z will be,

9+0=9

9+3=12

9+6=15

9+9=18

Thus, there can be four possible values i.e. 0,3,6 or 9