Question

If $7^9+9^7$ is divided by $64$ then the remainder is

• A. 0
• B. 1
• C. 2
• D. 63

Answer 1

Answer: (A)

We have
$7^9+9^7=(8-1{)}^9+(8+1{)}^7=(1+8{)}^7-(1-8{)}^9$
$=\left[1+{}^7{C}_{1}8+{}^7{C}_2{8}^2+\dots ..+{}^7{C}_7{8}^7\right]$
$-\left[1-{}^9{C}_{1}8+{}^9{C}_2{8}^2-\dots ..-{}^9{C}_9{8}^9\right]$
$={}^7{C}_{1}8+{}^9{C}_{1}8+\left[{}^7{C}_2+{}^7{C}_3\cdot 8+\dots .-{}^9{C}_2+{}^9{C}_3\cdot 8-\dots \right]8^2$
$=8(7+9)+64k=\mathrm{8..16}+64k=64q$
where $q=k+2$. Thus, $7^9+9^7$ is divisible by $64$ .