# What is the remainder of $16!$ is divided by 19?

Can anyone share me the trick to solve this problem using congruence?

Thanks,
Chan

If you've seen Wilson's theorem then you know the remainder when $18!$ is divided by $19$. To get $16!$ mod $19$, then, you can multiply by the multiplicative inverses of $17$ and $18$ mod $19$. Do you know how to find these? (Edit: Referring to both inverses might be a bit misleading, because you really only need to invert their product. See also Bill Dubuque's answer.)

• I knew Wilson's Theorem, but really don't know how to find the inverse of 17, 18 mod 19. What's the inverse? Can you give me one example? Thanks.
– Chan
Feb 26 '11 at 6:17
• @Chan: $\rm 17 \equiv -2,\ 18\equiv -1$ Feb 26 '11 at 6:21
• @Chan: A general technique to find the inverse of $k$ mod $n$ when $k$ and $n$ are relatively prime is to use integer division and the Euclidean algorithm to find $a$ and $b$ such that $ak + bn = 1$. Since $bn$ is a multiple of $n$, $ak$ is congruent to $1$ mod $n$, and thus the congruence class of $a$ is the inverse of the congruence class of $k$. In this particular case, a further hint to make things easier is that $18\equiv -1$ and $17\equiv -2$ mod $19$. (This makes computations easier for inverting their product.) Feb 26 '11 at 6:21
• Many thanks, I got it now ;)
– Chan
Feb 26 '11 at 6:24

Hint  By Wilson's theorem $$\bmod 19\!:\ \overbrace{{-}1}^{\large \color{#0a0}{18}} \equiv 18! \equiv\!\! \overbrace{18\cdot17}^{\large \color{#c00}{(-1)(-2)}}\!\!\cdot 16!\,$$ $$\Rightarrow\,16!\equiv \dfrac{\color{#0a0}{18}}{\color{#c00}2}\equiv 9$$

Generally (Wilson reflection formula) $$\rm\displaystyle\ (p\!-\!1\!-\!k)!\equiv\frac{(-1)^{k+1}}{k!}\!\!\!\pmod{\!p},\,$$ $$\rm\:p\:$$ prime

• @Bill Dubuque: Thank you!
– Chan
Feb 26 '11 at 6:24
• @Bill Dubuque: If I have $2.16! \equiv 18 \pmod{19}$. Can I cancel out the $2$ from both sides?
– Chan
Feb 26 '11 at 6:42
• @Chan: $\rm\ 2\:x\equiv 2\:y\ (mod\ 19)\ \iff\ 19\ |\ 2\ (x-y)\ \iff\ 19\ |\ x-y\ \iff\ x\equiv y\ (mod\ 19)$ Feb 26 '11 at 6:50
• @Bill Dubuque: Thank you. So if $gcd(a, p) = 1$, then $ax \equiv ay \pmod{p} \implies x \equiv y \pmod{p}$, right?
– Chan
Feb 26 '11 at 7:07
• @Chan: $\rm\ n,m\$ coprime $\rm\iff a\ n + b\ m = 1\$ for some $\rm\:a,b\:\ \iff\ a\ n\equiv 1\ (mod\ m)\:,\:$ for some $\rm\:a\:$ $\rm\iff\ n$ is invertible/cancellable $\rm\:(mod\ m)$ Feb 26 '11 at 7:12

I almost think in this case it is just faster to break it down than calculate the inverses: First the factors between 1 and 10:

$$10!= (2\cdot 10)\cdot (4\cdot5)\cdot(3\cdot6)\cdot(7\cdot8)\cdot 9\equiv 1\cdot 1\cdot (-1)\cdot(-1)\cdot 9\equiv9$$

Now we have

$$16!\equiv 9\cdot 11\cdot 12\cdot 13\cdot 14\cdot 15\cdot 16\equiv9\cdot(-8)\cdot(-7)\cdot(-6)\cdot(-5)\cdot(-4)\cdot(-3)$$

$$\equiv 9\cdot(4\cdot5)\cdot(6\cdot3)\cdot(7\cdot8)\equiv 9\cdot(1)\cdot(-1)\cdot(-1)\equiv 9$$

Of course this doesn't generalize, but the computation is faster than finding 3 inverses and multiplying them. (Again of course that is not true for larger $n$...)

A more explicit answer would go like this:

By Wilson's Theorem, $18!$ is congruent to $-1 \pmod {19}$

$$18! = (18\cdot 17)(16!)$$

Then $(18\cdot 17)(16!)$ is congruent to $-1 \pmod{19}$

But note that $18$ is congruent to $-1 \pmod{19}$ and $17$ is congruent to $-2 \pmod{19}$

Then it follows that $(18\cdot 17)(16!)$ is congruent to $((-1)\cdot (-2))(16!)$ is congruent to $-1 \pmod{19}$

Simplifying, we get, $2\cdot(16!)$ is congruent to $-1 \pmod{19}$. But we need the remainder of $16!$, not $2\cdot 16!$. and $-\frac12$ isn't a valid answer.

Then also notice that $18$ is congruent to $-1 \pmod{19}$. Then $2\cdot 16!$ is congruent to $18 \pmod{19}$ by the fact that if $a$ is congruent to $b \pmod{p}$, then $b$ is congruent to $a \pmod{p}$

So we have that $2\cdot 16!$ is congruent to $18 \pmod{19}$ and from there, its just a matter of dividing both sides by $2$ to get $16!$ is congruent to $9 \pmod{19}$