Question

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

• A. $22$
• B. $23$
• C. $1$
• D. $19$

Answer 1

Answer: (B)

$23$

Expanding $7^n$ as $(1+6{)}^n$ by binomial theorem we get
${=}^n{C}_0{6}^n+{}^n{C}_1{6}^{n-1}+..................{.}^n{C}_{n-2}{6}^2{+}^n{C}_{n-1}6{+}^n{C}_n{6}^0$
$={[}^n{C}_0{6}^n+{}^n{C}_1{6}^{n-1}+..................{.}^n{C}_{n-2}{6}^2]+6n+1$
$=\left[36\lambda \right]+6n+1$ ($36\lambda$ has all the numbers are multiple of 36 as every term is having $6^2$ and higher powers)
Now,
$\Rightarrow 7^n-6n-50=\left[36\lambda \right]+6n+1-6n-50$ $=\left[36\lambda \right]-49=\left[36\lambda \right]-72+23$
So we can clearly see that $\left[36\lambda \right]-72$ is divisble by $36$ and thus $23$ is the required Remainder
Therefore Correct Answer is $B$