7. Find the remainder when 53 ! is divided by 59 .
Answer: 30
Solution: Let $x \equiv 53!\bmod 59$. By Wilson's theorem, $58!\equiv-1 \bmod 59$. Now, $58!=58 \cdot 57 \cdot 56 \cdot 55 \cdot 54$. $53!\equiv(-1)(-2)(-3)(-4)(-5) x \bmod 59$. This means that $-120 x \equiv-1 \bmod 59$ or $2 x \equiv 1 \bmod 59$. Thus, $x \equiv 60 x \equiv 30 \bmod 59$.