For any natural number $n,\left(15 \times 5^{2 n}\right)+\left(2 \times 2^{3 n}\right)$ is divisible by

For any natural number $n,\left(15 \times 5^{2 n}\right)+\left(2 \times 2^{3 n}\right)$ is divisible by
  1. 7
  2. 11
  3. 13
  4. 17

Solution

We have, $\left(15 \times 5^{2 n t}\right)+\left(2 \times 2^{3 n t}\right)$ For $ \begin{aligned} & \text { For } \begin{aligned} n & =1, \\ \text { we get } 15 \times 5^2+2 \times 2^3 & =15 \times 25+2 \times 8 \\ & =375+16=391 \end{aligned} \end{aligned} $ which is divisible by 17 $\therefore\left(15 \times 5^{2 n}\right)+\left(2 \times 2^{3 n t}\right)$ is divisible by $17, \forall n \in N$

Asked in: AP EAMCET 2018 (22 Apr Shift 1)

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