What is the remainder when $6\times7^{32} + 7\times9^{45}$ is divided by $4$? 
What is the remainder when $6\times7^{32} + 7\times9^{45}$ is divided by $4$ ?

$7 \equiv 3 \pmod 4$
$7^2 \equiv 9 \pmod 4\equiv 1 \pmod 4$
$(7^2)^{16} \equiv 1^{16} \pmod 4$ 
i.e  $7^{32} \equiv 1 \pmod 4$
Similarly $9 \equiv 1 \pmod 4$ implies $9^{45} \equiv 1 \pmod 4$.
But the problem arise with the coefficients and addition sign.
what to do? 

A different approach: One could rewrite the problem as: 
$ 6*7^{32} + 7*9^{45} = 6* (8-1)^{32} + 7* (8+1)^{45} $ and then apply the binomial theorem. 
$ (8-1)^{32} = \binom{32}{0}*8^{32} *(-1)^0 + \binom{32}{0}*8^{31} *(-1)^1 ... + \binom{32}{0}*8^0 *(-1)^{32}  $
$ (8+1)^{45} = \binom{45}{0}*8^{45} *(+1)^0 + \binom{45}{0}*8^{44} *(+1)^1 ... + \binom{45}{0}*8^0 *(+1)^{45}  $ 
It is easy to see that all terms apart from the last one are divisible by 4. 
Therefore one has just to compute the terms: $ 6* \binom{32}{0} *(-1)^{32} + 7 * \binom{45}{0} *(+1)^{45} $, however: $\binom{n}{0} = 1 \ \forall n $ thus one just gets: $ 6*1 +7*(+1) $ = 13 that is equal to a remainder of +1 mod(4)
Which gives the desired result. 
