$7^{10}$ by $51$ what rest?

Question

How to find the remains of Division $7^{10}$ by $51$ using arithmetic debris?

$$$$ $$7\equiv51\;\text{mod}11$$

Answer 1

Note this:

\begin{align} 7^2 &\equiv - 2 &\pmod{51} \\ 7^{10} &\equiv - 32 &\pmod{51} \\ 7^{10} &\equiv 19 &\pmod{51} \\ \end{align}

So this means that the remainder will be 19.

Answer 2

$7^3=343=17\cdot20+3$, so $7^3\equiv3\pmod{17}$, and $7^{10}\equiv3^3\cdot7\equiv27\cdot7\equiv10\cdot7\equiv2\pmod{17}$, since $4\cdot17=68$. Of course $7^{10}\equiv1^{10}\equiv1\pmod3$. Thus, we want an integer $n\in\{0,1,\dots,50\}$ such that $n\equiv2\pmod{17}$ and $n\equiv1\pmod3$; $2+17=19$ clearly does the trick.