Question

For all integers $n\ge 1$, which of the following is divisible by $9$?

• A. $8^n+1$
• B. $4^n-3n-1$
• C. $3^{2n}+3n+1$
• D. ${10}^n+1$

Answer 1

The correct option is B $4^n-3n-1$

A) $a\left(n\right)=8^n+1={(9-1)}^n+1=9^n-n_{C_1}{9}^{n-1}+..+{(-1)}^{n-1}{n}_{C_{n-1}}9+{(-1)}^n{n}_{C_n}+1$
$\Rightarrow a\left(n\right)=9(9^{n-1}-n_{C_1}{9}^{n-2}+..+{(-1)}^{n-1}{n}_{C_{n-1}})+{(-1)}^n{n}_{C_n}+1=9\lambda +{(-1)}^n{n}_{C_n}+1$
if $n$ is even, then $a\left(n\right)=9\lambda +2$ which is not divisible by $9$
if $n$ is odd, then $a\left(n\right)=9\lambda$ which is divisible by $9$
B) $b\left(n\right)=4^n-3n-1={(3+1)}^n-3n-1=3^n+n_{C_1}{3}^{n-1}+...+n_{C_2}{3}^2+n_{C_1}3+n_{C_n}-3n-1$
$\Rightarrow b\left(n\right)=9(3^{n-2}+n_{C_1}{3}^{n-3}+...+n_{C_2})=9\lambda$
Therefore, $b\left(n\right)$ is divisible by $9$.
C) $c\left(n\right)=3^{2n}+3n+1=3(3^{2n-1}+n)+1=3\lambda +1$
Therefore, $c\left(n\right)$ is not divisible by $9$.
D) $d\left(n\right)={10}^n+1={(9+1)}^n+1=9^n+n_{C_1}{9}^{n-1}+...+n_{C_n}+1$
$\Rightarrow d\left(n\right)=9(9^{n-1}+n_{C_1}{9}^{n-2}+..+n_{C_{n-1}})+2=9\lambda +2$
Therefore, $d\left(n\right)$ is not divisible by $9$