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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 = 22. A = 6, B = 33. A = 6, B = 24. A = 3, B = 6 |
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Answer» 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. |
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