Originally Posted by

**cassiopeia1289** For n is greater than or equal to 1, use congruence theory to find remainder of:

$\displaystyle 7 | 5^{2n} + (3)(2^{5n-2})$

so here's what I did - I'm new to mod(s) so I wanted to make sure it goes right:

so:

$\displaystyle = (5^2)^n + (3)(2^{5n})(2^{-2})$

$\displaystyle = (25)^n + (3)(2^5)^n (2^{-2})$

$\displaystyle \equiv 4^n + (-4)(4)^n (2^{-2})$ (mod 7)

$\displaystyle \equiv 4^n + (-1)(4^n) $(mod 7)

$\displaystyle \equiv 4^n (1 + -1)$ (mod 7)

$\displaystyle \equiv 0$ (mod 7)

yes??? does it work?