<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] $$
6 \cdot 7^{32} + 7 \cdot 9^{45} \equiv 6 \cdot 1 + 7 \cdot 1 \equiv 6+7 \equiv 13 \equiv 1 \mod 4
$$
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