Find $55! \bmod 61$

Question

I am asked to find the smallest positive $x$ such that $x \equiv 55! \pmod{61}$.

This invokes Wilson's theorem where $(p-1)! \equiv -1 \pmod p$.

This means $60! \equiv -1 \pmod{61}$.

But where to go from here?

Answer 1

Doing arithmetic modulo $\;61\;$ all along:

$$55!=\frac{60!}{(-5)(-4)(-3)(-2)(-1)}=\frac{-1}{-120}=-\frac1{2}=-31=30$$

Answer 2

We have $60! = 55! \times 56 \times 57 \times 58 \times 59 \times 60$. We have $60! \equiv -1 \pmod{61}$ from Wilson's theorem. Hence, $$55! \equiv - 56^{-1} \times 57^{-1} \times 58^{-1} \times 59^{-1} \times 60^{-1} \pmod{61}$$

Answer 3

You know that: $$60!\equiv(55)! *(-1)(-2)(-3)(-4)(-5)\mod 61$$ hence $2x\equiv -120x\equiv -1 \mod 61$ you need only to find $x$