A six-digit number 4ABB8A is a multiple of 33 for non-zero digits A and B. Which of the following could be the possible values of A and B?
Show answer & explanation
Correct answer: B
Explanation
The correct answer is Option 2 i.e. A = 6 B = 3.
For 33: Must be divisible by both 3 and 11.
(4 - A + B - B + 8 - A) = 12 - 2A
If A = 6 then 12 - 2A = 0
Sum of digits = 4 + 6 + B + B + 8 + 6 = 24 + 2B (Divisible by 3)
If B = 3 then 24 + 2B = 30
Hence, A = 6 and B = 3