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.
