MCQ
Let $abc$ be a three digit number. Then $abc + bca + cab$ is not divisible by
  • A
    $a + b + c$
  • B
    $3$
  • C
    $37$
  • $9$

Answer

Correct option: D.
$9$
We know that, the sum of three-digit numbers taken in cyclic order can be written as $111 (a + b + c)$
i.e. $abc + pea + cab = 3 × 37 × (a + b + c)$
Hence, the sum is divisible by $3, 37$ and $(a + b + c)$ but not divisible by $9.$

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