$32^5 + 2^{27}$ is divisible by

$32^5 + 2^{27}$ is divisible by
  1. 3
  2. 7
  3. 10
  4. 11

Solution

$32^5 = (2^5)^5 = 2^{25}$, so $32^5 + 2^{27} = 2^{25} + 2^{27} = 2^{25}(1 + 2^2) = 2^{25} \times 5 = 2^{24} \times 2 \times 5 = 2^{24} \times 10$. Hence the expression is divisible by 10.

Asked in: CSAT 2024

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