Question

$7^{2n}+16n-1(n\in N)$ is divisible by

• A. $65$
• B. $63$
• C. $61$
• D. $64$

Answer 1

The correct option is D $64$

$7^{2n}+16n-1$
$=(8-1{)}^{2n}+16n-1$
$=[{}^{2n}{C}_0(-1{)}^{2n}+{}^{2n}{C}_1\cdot 8\cdot (-1{)}^{2n-1}+...+{}^{2n}{C}_{2n}\cdot 8^{2n}]+16n-1$
$=[{}^{2n}{C}_0-{}^{2n}{C}_1\cdot 8+...+{}^{2n}{C}_{2n}\cdot 8^{2n}]+16n-1$
$=1-2n\cdot 8+{}^{2n}{C}_2\cdot 8^2...+{}^{2n}{C}_{2n}\cdot 8^{2n}+16n-1$
$={}^{2n}{C}_2\cdot 8^2...+{}^{2n}{C}_{2n}\cdot 8^{2n}$
$=8^2({}^{2n}{C}_2+...+{}^{2n}{C}_{2n}\cdot 8^{2n-2})$
$=64k$