Question

1. If3125 is a multiple of 3, then what cannot be the values of z?​

Answer 1

Answer:

Now, we know that if the sum of all digits of a number is divisible by 3, then the number is also divisible by 3. So, we check the divisibility of the given number. ... Now, from obtained sum of digits only 9,12,15,18 are divisible by 3. So, the possible values of z are 0,3,6,

Answer 2

Answer:

0,3or9

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 to 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