Question

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

• A. 22
• B. 23
• C. 1
• D. 21

Answer 1

Answer: (B)

We have, $7^n=(1+6{)}^n$
$={}^n{C}_0+{}^n{C}_1{6}^1+{}^n{C}_2{6}^2+{}^n{C}_3{6}^3+\dots +{}^n{C}_n{6}^n$
$=1+6n+6^2\left[{}^n{C}_2+{}^n{C}_{3}6+\dots +{}^n{C}_n{6}^{n-2}\right]$
$=1+6n+36\lambda [$ where, ${}^n{C}_2+\dots +{}^n{C}_n{6}^{n-2}=\lambda ]$
$\Rightarrow 7^n-6n=36\lambda +1$
$\Rightarrow 7^n-6n-50=36\lambda -49$
$\Rightarrow 7^n-6n-50=36\lambda -72+23$
$\Rightarrow 7^n-6n-50=36(\lambda -2)+23$
$\Rightarrow 7^n-6n-50=36\mu +23[$ where $\lambda -2=\mu ]$
$\therefore$ When $7^n-6n-50$ is divided by $36$, then remainder will be equal to $23$ .