Question

If n is a positive integer, prove that $3^{3n}-26n-1$ is divisible by 676.

Answer 1

$3^{3n}-26n-1={27}^n-26n-1...\left(1\right)$
$$\begin{gathered} \mathrm{Now},\mathrm{we}\mathrm{have}: \\ {27}^n=(1+26{)}^n \\ \mathrm{On}\mathrm{expanding},\mathrm{we}\mathrm{get} \\ (1+26{)}^n={}^{n}C_0\times {26}^0+{}^{n}C_1\times {26}^1+{}^{n}C_2\times {26}^2+{}^{n}C_3\times {26}^3+{}^{n}C_4\times {26}^4+...{}^{n}C_n\times {26}^n \\ \Rightarrow {27}^n=1+26n+{26}^2[{}^{n}C_2+{}^{n}C_3\times {26}^1+{}^{n}C_4\times {26}^2+...{}^{n}C_n\times {26}^{n-2}] \\ \Rightarrow {27}^n-26n-1=676\times \mathrm{an}\mathrm{integer} \\ {27}^n-26n-1\mathrm{is}\mathrm{divisible}\mathrm{by}676 \\ \mathrm{Or}, \\ {3}^{3n}-26n-1\mathrm{is}\mathrm{divisible}\mathrm{by}676\left(\mathrm{From}\left(1\right)\right) \end{gathered}$$