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