Subject: General Aptitude · Chapter: Number System · Exam: 2010 · Marks: · Difficulty:
The six-digit number $5 A B B 7 A$ is a multiple of 33 for non-zero digits $A$ and $B$. Which of the following could be possible value of $A+B$ ?
(a)8
(b)9
(c)10
(d)14
Answer
Answer (as printed): B
Explanation
We know that 33 = 11 × 3, where 11 and 3 are co-primes. So, the given number must be divisible by both 11 and 3. Since 5ABB7A is divisible by 11, we have (A + B + A) – (7 + B + 5) = (2A – 12) is either 0 or 11. ⇒ 2A – 12 = 0 or 2A – 12 = 11 ⇒ A = 6 A ≠ 23 2 So, the number becomes 56BB76, which is divisible by 3. \ (5 + 6 + B + B + 7 + 6) = (24 + 2B) must be divisible by 3. 3 \ 2B = 6 ⇒ B = 3 B ≠ 0 and B ≠ 2 Hence, (A + B) = (6 + 3) = 9.
Explanation as extracted from the printed page; notation may be imperfect.