Remainder on division with $22$ What is the remainder obtained when $14^{16}$ is divided with $22$? 
Is there a general method for this, without using number theory? I wish to solve this question using binomial theorem only - maybe expressing the numerator as a summation in which most terms are divisible by $22$, except the remainder? 
How should I proceed?
 A: Since $$ 14^2 \equiv -2 $$ so $$14^{16} \equiv (-2)^8 \equiv 16^2\equiv (-6)^2 \equiv 14$$

or
$$ 14^2 = 22k -2 $$ so $$14^{16} = (22k-2)^8 = 22l+2^8=22l+22\cdot 11+14$$
A: A method that uses FLT.
Finding the remainder on dividing $14^{16}$ by $22$ is equivalent to twice that on dividing $7\cdot14^{15}$ by $11$.
By FLT, $14^{10}\equiv1\pmod{11}$ so $$7\cdot14^{15}\equiv7\cdot14^5\equiv98\cdot196\cdot196\equiv-1\cdot(-2)\cdot(-2)\equiv-4\equiv7\pmod{11}$$ hence the required remainder is $7\times 2=14$. 
A: You can use binomial expansions and see that 
$$14^{16} = (22 - 8)^{16}$$
implies that the remainder is just the remainder when $(-8)^{16}( = 8^{16})$ is divided by $22.$
Proceeding similarly, 
$8^{16} = 64^8 = (66 - 2)^8 \implies 2^8 = 256 \text{ divided by } 22 \implies \text{remainder = 14}$
A: Using the Euclidean Algorithm
Note that
$$
\begin{align}
14^{16}&\equiv0&\pmod2\tag1
\end{align}
$$
Reducing mod $11$ and using Fermat's Little Theorem, we get
$$
\begin{align}
14^{16}
&\equiv3^6&\pmod{11}\\
&\equiv3&\pmod{11}\tag2
\end{align}
$$
Solving $11a+2b=1$ using the Euclidean Algorithm as implemented in this answer,
$$
\begin{array}{r}
&&5&2\\\hline
1&0&1&-2\\
0&1&-5&11\\
11&2&1&0\\
\end{array}\tag3
$$
we have
$$
11(1)+2(-5)=1\tag4
$$
from which we get
$$
\begin{align}
-10&\equiv0&\pmod2\\
-10&\equiv1&\pmod{11}
\end{align}\tag5
$$
Multiplying $(5)$ by $3$ gives
$$
\begin{align}
-30&\equiv0&\pmod2\\
-30&\equiv3&\pmod{11}
\end{align}\tag6
$$

Using the Chinese Remainder Theorem
The Chinese Remainder Theorem says that the solution to $(6)$ is unique mod ${22}$. Thus, we get the smallest positive solution to be
$$
\begin{align}
14&\equiv0&\pmod2\\
14&\equiv3&\pmod{11}
\end{align}\tag7
$$
Thus, $(1)$, $(2)$, and $(7)$ yields
$$
\begin{align}
14^{16}&\equiv14&\pmod{22}\tag8
\end{align}
$$
