Computing large powers $\!\bmod n$ when $n$ has multiple prime factors. How will the congruence modulo works for large exponents? What theorem/s may be used? For example to show that $7^{82}$ is congruent to $9 \pmod {40}$.
 A: Maybe just start calculating.  Note that $7^2=49\equiv 9\pmod{40}$.  So $7^4\equiv 81\equiv 1\pmod{40}$. We got lucky. 
It follows that $7^{80}=(7^{16})^5\equiv 1\pmod{40}$. Thus 
$7^{82}=7^{80}7^2\equiv 49\equiv 9\pmod{40}$.
In general, repeated squaring, and reduction with respect to our modulus (in this case $40$) gets us to high powers fairly quickly. 
A more elaborate way is to use Euler's Theorem. If $a$ is relatively prime to $m$, then $a^{\varphi(m)}\equiv 1\pmod{m}.$
Here $\varphi$ is the Euler $\phi$-function. We have $\varphi(40)=16$, so taking $a=7$ we have $a^{16}\equiv 1 \pmod{40}$. It follows that $a^{80}=(a^{16})^5\equiv 1\pmod{40}$. So $a^{82}\equiv a^2=49\pmod{40}$. But $49\equiv 9\pmod{40}$.
Still another way is to factor $40$ as $2^3\cdot 5$. We then work separately modulo $8$ and modulo $5$.
First modulo $8$: We have $7\equiv -1\pmod{8}$, so $7^{82}\equiv (-1)^{82}\equiv 1\pmod{8}$.  
Next modulo $5$: We have $7\equiv 2\pmod{5}$, so $7^2\equiv 4\equiv -1\pmod{5}$, and therefore $7^4\equiv 1\pmod{5}$. (We could also get this directly by using Fermat's Theorem.) It follows that $7^{80}\equiv 1\pmod{5}$, and therefore $7^{82}\equiv 4\pmod 5$. 
Now we are looking for the numbers that are congruent to $1$ modulo $8$ and to $4$ modulo $5$. For bigger numbers, we can use the Chinese Remainder Theorem. But finding a number congruent to $1$ modulo $8$ and to $4$ modulo $5$ can also be done without much machinery.
In general, the technique to use depends very much on the modulus $m$. If it is huge, and we do not know its prime factorization, then the Binary Method of exponentiation is quite efficient.
A: Besides the methods mentioned by Andre, it's worth mentioning another more powerful technique that often proves crucial. Namely, instead of Euler's theorem, based on his $\phi$ (totient) function, one should use an improvement due to Carmichael, based on his $\lambda$ function (universal exponent).
The big gain in power arises from employing $\rm\: \ell = \color{#C00}{lcm}\:$ vs. product to combine exponents:
$$\begin{eqnarray}\rm a^j&\equiv&\rm 1\pmod m\\ \rm a^k&\equiv&\rm 1\pmod n\end{eqnarray}\Bigg\}\ \Rightarrow\rm\ a^{\color{#C00}{lcm}(j,k)}\equiv 1\pmod{lcm(m,n)}$$
Proof $\,\rm\bmod m\!:\ a^j\equiv 1,\,\ j\mid \ell\:\Rightarrow\: a^{\ell}\equiv 1\,$ by mod order reduction.   Similarly $\rm\bmod n\!:\ a^{\ell}\equiv 1,\,$ so combining: $\rm a^{\ell}\equiv 1\bmod m\ \&\ n\Rightarrow $ $\rm\,a^{\ell}\equiv 1\pmod{\!{\rm lcm}(m,n)}\,$ by CCRT.
For example, if $\rm\:n\:$ is coprime to $2,3,5$ then Euler's theorem implies that $\rm\:n^{64}\equiv 1\pmod{240},\:$ versus Carmichael's theorem, which yields the much stronger result that $\rm\:n^4\equiv 1\pmod{240}.\,$ This allows one to immediately solve the problem there, viz. $\rm\,240\:|\:p^4\!-\!1\,$ for all primes $\rm\,p > 5.$ Compare below the computation of the Euler vs. Carmichael function:
$\qquad\qquad\rm\phi(240) = \phi(2^4\cdot 3\cdot 5) = \ \phi(2^4)\ \phi(3)\ \phi(5) = 8\cdot2\cdot 4 = 64$
$\qquad\qquad\rm\lambda(240) = \lambda(2^4\cdot 3\cdot 5) = \color{#C00}{lcm}(2^{2},\phi(3),\phi(5)) = \color{#C00}{lcm}(4,2,4) = 4$
Thus Carmichael generally yields a much lower bound on the order of elements, which greatly simplifies problems like yours and that above. For more examples see my posts here.
