When 335 is added to 5A7, the result is 8B2. If 8B2 is divisible by 3

3 3 5


+ 5 A 7


------------------------------------------------------------------------

8 B 2


Check the unit digit. Since 7+5 =12,the carry 1 is added in the 2nd column and te net sum of second column is A+3 + 1(carry) =A+4

A+4 =B

In the hundreds place,there is no carry and thatswhy the sum is 5+3=8.

This implies A cannot be 6 or 8 or else the second column sum would be 6 + 3 + 1(carry) =10(creating 1 as carry) or a 8 + 3 +1=12(Creating 2 as carry).

Eliminate A and B.

8 B 2 is divisible by 3. The digit sum has to be divisible by 3 and hence the value of B can be 5 or 8.It cannot be 2 or else A will be negative.

Both 5 and 8 are admissible values of B but at B=8, A= 8-4 =4 and at B=5,A=5-4=1. (Since A+4=B)

The largest value thus is A=4.

(option c)

Devmitra Sen
GMAT SME