modular arithmetic (number theory) Assume that $$7^{64} = 1 \mod 120.$$
I am trying to find $$7^{62} \mod 120.$$
In my maths text, I was told that:
$$\begin{align}
7^{62} & = 7^{64} \cdot 7^{-2} \\
& = 7^{-2} \quad \\
&= 49^{-1} \quad \, \, \mod 120 
\end{align}$$ 
I do not understand why $7^{64} \cdot 7^{-2}$ can be reduced to just $7^{-2}$. Can anyone explain to me?
 A: $7^{64} \equiv 1 \mod 120$. Thus, $7^{64} \times 7^{-2} \equiv 1\times7^{-2} \equiv 7^{-2} \mod 120$.
This is a basic property of modolar arithmetic.
A: If $a\equiv b\bmod m$, then $ca\equiv cb\bmod m$ for any integer $c$. Thus, supposing that 
$$7^{62}\equiv x\bmod 120,$$
we have that
$$7^{64}=7^2\cdot 7^{62}\equiv 7^2\cdot x\bmod 120,$$
and since you're told that
$$7^{64}\equiv 1\bmod 120,$$
you have that
$$1\equiv 7^2\cdot x\bmod 120.$$
This is the sense in which
$$7^{62}\equiv x\equiv 7^{-2}\bmod 120.$$
A: It's just scaling an equality (congruence)
$\rm\quad \begin{array}{rl}\rm a \times (b\equiv c)\!\!\! &\to&\rm  ab\equiv ac\\
\\
7^{-2}\! \times (7^{64}\!\equiv 1)\!\!\! &\to& 7^{62}\equiv 7^{-2}\end{array}$
Remark $\ $ Congruences $\rm\:a\color{#C00}{\equiv} b\:$ behave similar to equalities $\rm\:a\color{#C00}= b\:$ in that they can be added, and multiplied. Scaling is a special case of multiplication, i.e. above is the product of $\rm\:a \equiv a\:$ by $\rm\:b\equiv c.\:$ When working with congruences it is essential to conceive them as generalized equalities so that you can transfer all of your well-honed equational arithmetical skills to congruence arithmetic.
A: Multiply by $7$ at each step and reduce modulo $120$:
$$
1 \mapsto 7\mapsto 49 \mapsto 103\mapsto1 \mapsto 7\mapsto 49 \mapsto 103\mapsto1\mapsto\cdots\cdots
$$
As soon as you see the pattern, you'll know how to do the problem.
