Question

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?
1. A = 3, B = 2
2. A = 6, B = 3
3. A = 6, B = 2
4. A = 3, B = 6

Answer 1

Correct Answer - Option 2 : A = 6, B = 3

Given:

Number =  4ABB8A

Concept used:

33 = 3 × 11

So, the divisibility rule of 3 and 11 must be satisfied for the number to be multiple of 3 and 11.

Divisibility rule of 3: The sum of the digits must be a multiple of 3.

Divisibility rule of 11: Difference between the sum of alternate digits of the number = 0 or multiple of 11

Calculation:

Applying divisibility rule of 3:

4 + A + B + B + 8 + A = 12 + 2B + 2A = 12 + 2(A + B)

For the number to be multiple of 3, A + B must be multiple of 3 according to the above sum.

This is satisfied by options 2 and 4 only.

Applying divisibility rule of 11:

Sum of alternate digits = (4 + B + 8) and (A + B + A)

Difference between the sum = 12 + B - (2A + B) = 11N (N = 0, 1, 2, ....)

12 - 2A = 0 (least multiple)

Then, A = 6 

Then, from the options 2 and 4, only option 2 is possible.

∴ The possible option is 2 only.