Question

1E. The number 2A5A0A2 is divisible by 99. What is the digit A?

Answer 1

We know, a number is divisible by 9 if the sum of digits is also divisible by 9.

Also, since A is an digit, so A is in between 0 to 9.

2+A+5+A+0+A+2  = 3A + 9

3A + 9 should be divisible by 9.

3A + 9 = 9n

A = (9n -9)/3

For n= 1 :

A = 0

For n=2 :

A = 3

For n= 3 :

A = 9

After this A > 9 which is not accepted.

Therefore, the digit A can be 0, 3 and 9 .

Hence, this is the required solution.