Question

If ${1690}^{2608}+{2608}^{1690}$ is divided by $7$, then the remainder is

• A. $5$
• B. $2$
• C. $0$
• D. $1$

Answer 1

The correct option is D $1$

We know that,
$1690=7\times 241+3,2608=7\times 372+4$

Now, let
$A={1690}^{2608}+{2608}^{1690}=(7\times 241+3{)}^{2608}+(7\times 372+4{)}^{1690}=7k_1+3^{2608}+7k_2+4^{1690},k_1,k_2\in N=7k+3^{2608}+4^{1690},k\in N$

$B=3^{2608}+4^{1690}$
Remainder of $A$ and $B$ are same, so
$B=3\times 3^{3\times 869}+4\times 4^{3\times 563}=3\times {27}^{869}+4\times {64}^{563}=3(28-1{)}^{869}+4(63+1{)}^{563}=3(28k_3-1)+4(63k_4+1),k_3,k_4\in Z=7m+1,m\in Z$

Hence, the remainder is $1$.