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
$7^9+9^7={\left(8-1\right)}^9+{\left(8+1\right)}^7={\left(1+8\right)}^7-{\left(1-8\right)}^9$
$=\left[1{+}^7{C}_{18}{+}^7{C}_2{8}^2+.....{+}^7{C}_7{8}^7\right]$
$-\left[1{-}^9{C}_{1}8{+}^9{C}_2{8}^2-.....{-}^9{C}_9{8}^9\right]$
${=}^7{C}_{18}{+}^9{C}_{18}+\left[{}^7{C}_2{+}^7{C}_3.8+....{-}^9{C}_2{+}^9{C}_3.8-.....\right]8^2$
$=8\left(7+9\right)+64k=\mathrm{8..16}+64k=64q,$ where $q=k+2$
Thus, $7^9+9^7$ is divisible by $64.$