<s>[INST] $7^{10}$ by $51$ what rest? How to find the remains of Division $7^{10}$ by $51$ using arithmetic debris?
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$$7\equiv51\;\text{mod}11$$
 [/INST] $7^3=343=17\cdot20+3$, 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.
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