Remainder of $3^7/8$ I read here that the remainder of $\frac{ab}{c}$ is equal to the remainder of $\frac{a}{c}\frac{b}{c}$ implying that the remainder of $\frac{a^b}{c}$ is equal to the remainder of $[\frac{a}{c}]^b$. However, when I apply this here, I would get remainder of $[\frac{3}{8}]^7=3^7$ (while the correct answer is $3$). What am I doing wrong here? And if this method is not valid, are there other quick and simple methods of finding remainders like this? 

Remainder of $ \frac {ab}{c}$ = remainder of $ \frac {a}{c} $ $\times$ remainder of $ \frac {b}{c} $ is FALSE in most cases.
When remainder of $ \frac {a}{c} $ $\times$ remainder of $ \frac {b}{c} \lt c$, only then it is valid.
However when remainder of $ \frac {a}{c} $ $\times$ remainder of $ \frac {b}{c} \ge c$, then 
Remainder of $ \frac {ab}{c}$ = Remainder of $ \frac {remainder  of  \frac {a}{c} \times remainder of  \frac {b}{c} }{c}$
Since as you wrote: remainder of $ \frac {a}{c} $ $\times$ remainder of $ \frac {b}{c} = 3^7$ and dividing $ 3^7 $ by $8$ is the very problem we started at,
To apply the rule in this case, you could make use of the fact that:
$$ 3^7 = 3^4 \times 3^ 3$$ 
So assume $a$ to be $3^4$ and $b$ to be $3^3$.
Now, it's quite easy to calculate that $3^4 = 81 $ and $3^3 = 27 $
Remainder of $ \frac {3^7}{8} $ = Remainder of $ \frac {3^4 \times 3^3}{8}$ = Remainder of $ \frac {81 \times 27}{8} $
By the rule, Remainder of $ \frac {81 \times 27}{8} $ = Remainder of $ \frac {81}{8} $ $ \times $ Remainder of $\frac {27}{8} $ = $1 \times 3 $ = $3$
Since $ 3 \lt 8$, this is indeed our answer.
