<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] Reduction modulo an integer is a homomorphism of rings ie $$a+b \pmod n=a \pmod n+b\pmod n$$ and $$a\times b\pmod n=a\pmod n\times b \pmod n$$
These facts are easy to check directly e.g. $(a+pn)(b+qn)=ab+(aq+bp+pqn)n$ and $(a+pn)+(b+qn)=a+b+(p+q)n$.
This means that it doesn't matter whether you do your reductions first and the arithmetic after, or do the arithmetic first and then the reduction, or a combination of the two.
Note also that the reduction $7\equiv -1$ is allowed. So I would be doing this as $$6\times 7^{32}+7\times 9^{45}$$ and do all the reductions you can to get something like $$2 \times (-1)^{32}+(-1)\times 1^{45}=2-1$$
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