If n is natural number , then find the remainder when `(37)^(n+2)+(16)^(n+!)+(30)^n` is divided by 7.

Explanation

`(37^(n+2)+16^(n+1)+30^n)/7`

=`((35+2)^(n+2)+(14+2)^(n+1)+(28+2)^n)/7`

`=((C(n+2,0)35^(n+2)2^0+...C(n+2,n+2)35^0 2^(n+2))+(C(n+1,0)14^(n+1)2^0+...C(n+1,n+1)14^0 2^(n+1))+(C(n,0)28^(n)2^0+...C(n,n)28^0 2^(n)))/7`

So, only terms that are not divided by `7` are,

`(C(n+2,n+2)35^0 2^(n+2))+C(n+1,n+1)14^0 2^(n+1))+C(n,n)28^0 2^(n)))`

`=2^(n+2)+2^(n+1)+2^n = 2^n(2^2+2^2+1) = 7*2^n,`

Which is also divisible by `7`.

So, required remainder will be `0`.