# Finding remainder of a product divided by $8$ with modular arithmetic

I was requested to find the remainder of $$9^{52}*7^{33}$$ when dividing it by $$8$$, using modular arithmetic. Let that remainder be $$r$$. I understand the following will apply:

$$9^{52}*7^{33} \equiv r \mod{(8)}$$

so that all I must find is the congruence of the product in the module $$8$$. But how can one go about solving this? I don't have any tries worth showing, since I simply ignore what procedure could be followed.

• $7\equiv -1$ and $9\equiv 1$ mod $8$, so the expression is equal to $$1^{52}\cdot(-1)^{33}\bmod8=-1$$ – Don Thousand May 29 at 21:27
• But doesn't this mean the remainder of the division is negative? That does not make sense. What am I missing? – lafinur May 30 at 1:53
• $-1\equiv 7\bmod8$. – Don Thousand May 30 at 2:05
• Of course, how silly of me. Thank you very much for answering my question. – lafinur May 30 at 2:26

Hint. Since $$9$$ is $$1$$ modulo $$8$$, so is any power. What can you say about odd powers of $$7$$ using the same kind of reasoning? To find a modulus after multiplying you multiply the two moduli.
$$9\equiv 1,\qquad 7 \equiv -1\mod 8.$$
• Doesn't all this imply that $r = -1$? Since I'm left with $-1 \equiv r \mod (8)$. How can a remainder be negative? – lafinur May 30 at 2:02
• There exists versions of Euclidean division for which the remainder can be negative… This being said, it is only a hint. You still have to find the least positive integer which is congruent to $-1\bmod 8$. This shouldn't be too hard. – Bernard May 30 at 8:36