Find the remainder when `1690^(2608)+2608^(1690)` is divided by 7.

Explanation

We have ,`1690^(2608) + 2608^(1690) = (1690^(2608) - 3^(2608))`
` + (2608^(1690)-4^(1690)) + (3^(2608) + 4^(1690))`
The number ` (1690^(2608) - 3^(2608))` is divisible by
`1690 - 3 = 1687 = 7 xx 241` which is divisible by 7 , the
difference ` (2608^(1690) - 4 ^(1690) )` is also divisilble by 7 , since it is
divisible by ` 2608 - 4 = 2604 = 7 xx372`
As to sum `3^(2608) + 4^(1690)` , it can be rewritten as
` 3*(3^(3)) + 4*(4^(3))^(563)`
`= 3(38-1)^(869) + 4 (63 +1)^(563)`
` = 3= (7m-1) + 4 (7n +1)`
where , m and n are some positive integers ]
where p is some positive integer .
Hence , the remainder is . 1