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

Explanation

Here base `1690` and `2608` are large numbers, so first let us reduce them.
`1690 = 7 xx 241 + 3` and `2608 = 7 xx 372 +4`
Let `S = 1690^(2608) + 2608^(1690)`
`= (7xx241+3)^(2608)+(7xx372+4)^(1690)`
`= 7k + 3^(2608) + 4^(1690)` (where k is some positive integer)
Let `S' = S^(2608) + 4^(1690)`
Clearly, the remainder in S and S' will be the same when divided by 7.
`S' = 3 xx 3^(3 xx 869) + 4 xx 4^(3xx 563)`
` = 3 xx 27^(869) + 4 xx 64^(563)`
`= 3(28-1)^(869) + 4(63+1)^(563`
`= 3[7n - 1] + 4[7m + 1] (m, n in I)`
`= 7p + 1` ( where p is some positive integer)
Hence , the required remainder is 1.