Question

Find the remainder when 1690${}^{2608}$ + 2608${}^{1690}$ is divided by 7.

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is D $1$

Let $S=(1690{)}^{2608}+(2608{)}^{1690}$
$=\left[\right(7\left(241\right))+3{]}^{2608}+[\left(7\right(372\left)\right)+4{]}^{1690}$
$=7k+3^{2608}+4^{1690}$ where k is a positive integer.
Now the unit's digit is given by
$3^{2608}+4^{1690}$
$=3(27{)}^{869}+4(64{)}^{563}$
$=3(28-1{)}^{869}+4(1+63{)}^{563}$
$=3(28m-1)+4(63n+1)$
Hence units digit is
$-3+4$
$=1$