Question

The remainder when $7^{n}-6 n-50(n \in N)$ is divided by $36$, is

• A. 22
• B. 23
• C. 1
• D. 21

Answer 1

Answer: (B)

We have, $7^{n}=(1+6)^{n}$
$={ }^{n} C_{0}+{ }^{n} C_{1}\, 6^{1}+{ }^{n} C_{2}\, 6^{2}+{ }^{n} C_{3} \,6^{3}+\ldots+{ }^{n} C_{n} 6^{n}$
$=1+6 n+6^{2}\left[{ }^{n} C_{2}+{ }^{n} C_{3} \,6+\ldots+{ }^{n} C_{n}\, 6^{n-2}\right]$
$=1+6 n+36 \lambda \left[\right.$ where, $\left.{ }^{n} C_{2}+\ldots+{ }^{n} C_{n} 6^{n-2}=\lambda\right]$
$\Rightarrow 7^{n}-6 n=36 \lambda+1$
$\Rightarrow 7^{n}-6 n-50=36 \lambda-49$
$\Rightarrow 7^{n}-6 n-50=36 \lambda-72+23$
$\Rightarrow 7^{n}-6 n-50=36(\lambda-2)+23$
$\Rightarrow 7^{n}-6 n-50=36 \mu+23[$ where $\lambda-2=\mu]$
$\therefore $ When $7^{n}-6 n-50$ is divided by $36$, then remainder will be equal to $23$ .