Question

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

Answer 1

We have,

${1690}^{2608}+{2608}^{1690}$

Here base $\left(1690\&2608\right)$ is too big, so first let us reduce it

$1690=7\times 241+3$

$2608=7\times 372+4$

Consider that,

$S={1690}^{2608}+{2608}^{1690}$

$={(7\times 241+3)}^{2608}+{(7\times 372+4)}^{1690}$

$=7k+3^{2608}+4^{1690}$ (where k is some positive integer)

Let,

$S^{\prime }=3^{2608}+4^{1690}$

Clearly, the remainder in S and $S^{\prime }$ will be the same when divide by 7.

$S^{\prime }=3\times 3^{3\times 867}+4\times 4^{3\times 563}$

$=3\times {27}^{867}+4\times {64}^{563}$

$=3{(28-1)}^{867}+4{(63+1)}^{563}$

$=3[7n-1]+4[7m+1](m,n\in I)$

$=7p+1$ (where p is some positive integer )

The remainder is $1$.

Hence, this is the answer.