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
$7^{9} + 9^{7} = \left(8 - 1\right)^{9 }+ \left(8 + 1\right)^{7} = \left(1 + 8\right)^{7} - \left(1 - 8\right)^{9}$
$= \left[1+ ^{7}C_{18}+ ^{7}C_{2} 8^{2} +.....+ ^{7}C_{7} 8^{7} \right]$
$-\left[1- ^{9}C_{1}8+ ^{9}C_{2} 8^{2} - .....- ^{9}C_{9} 8^{9} \right]$
$= ^{7}C_{18 }+ ^{9}C_{18} +\left[ ^{7}C_{2} + ^{7}C_{3}.8 +....- ^{9}C_{2} + ^{9}C_{3}.8 -.....\right]8^{2}$
$= 8 \left(7 + 9\right) + 64 \,k = 8..16 + 64\, k = 64 \,q,$ where $q = k + 2$
Thus, $7^{9} + 9^{7}$ is divisible by $64.$