<s>[INST] 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? 
 [/INST] $\phi(4) = 2$ and since $(4,7), (4,9)$ are relatively prime we have $[7^2]_4 = [1]_4$, $[9^2]_4 = [1]_4$ and so
$[6 \cdot 7^{32}+7 \cdot 9^{45}]_4 = [6]_4+[7 \cdot 9]_4=[69]_4 = [1]_4$.
</s>