Find the smallest positive integer $x$ such that $2015! \equiv x\pmod{2017}$

Question

Q. The next year that is a prime is $2017$. Find the smallest positive integer $x$ such that $2015! \equiv x\pmod{2017}$.

So, this is what I have;

By Wilson’s theorem, $(2017-1)! \equiv -1 \pmod{2017} ⇒ 2016! \equiv -1 \pmod{2017} ⇒ 2016\times2015! \equiv -1 \pmod{2017} ⇒$

not sure how to do the modular arithmetic to have just $2015!$ on the left side..

Answer 1

As $2017$ is prime, using Wilson's Theorem $$2016!\equiv-1\pmod{2017}$$

$$\iff 2015!\cdot 2016\equiv-1\pmod{2017}$$

$$\iff 2015!\cdot (-1)\equiv-1\pmod{2017}\text{ as }2016\equiv-1\pmod{2017}$$

So, $\cdots$

Answer 2

Hint:

$2016$ is relatively prime to $2017$

If, $c$ is relatively prime to $p$, then,

$$ac\equiv bc\pmod{p}\iff a\equiv b\pmod{p}$$

Also, $$2016\equiv -1\pmod{2017}$$

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Also, if I may lengthen this on purpose, then let's do this without Wilson's Theorem.

Consider $2 \le a\le 2015 $. To each such $a$, we associate it's unique inverse $\overline a$, i.e, $a\overline a\equiv1\pmod{2017}$. Note that $a\ne \overline a$, because then $a^2\equiv 1\pmod{p}$ which would mean $a\equiv \pm 1\pmod{p}$ which is not possible for the $a$ in consideration. Thus in multiplying all $a\in \{2,3,\cdots,2015\}$ we pair them with their inverses, and the net product is $1$.

$$2015!\equiv 1\pmod{2017}$$

QED

Answer 3

Your solution is almost done. Notice that $2016 \equiv -1 \text{ }mod(2017)$ so $$2015!(-1)\equiv (-1) mod (2017)\implies 2015!\equiv 1$$