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

Thanks,
Chan

roxrook
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4 Answers4

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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.)

Jonas Meyer
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  • 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. – roxrook Feb 26 '11 at 06:17
  • @Chan: $\rm 17 \equiv -2,\ 18\equiv -1 $ – Bill Dubuque Feb 26 '11 at 06:21
  • @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.) – Jonas Meyer Feb 26 '11 at 06:21
  • Many thanks, I got it now ;) – roxrook Feb 26 '11 at 06:24
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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
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  • @Bill Dubuque: Thank you! – roxrook Feb 26 '11 at 06:24
  • @Bill Dubuque: If I have $2.16! \equiv 18 \pmod{19}$. Can I cancel out the $2$ from both sides? – roxrook Feb 26 '11 at 06:42
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    @Chan: $\rm\ 2:x\equiv 2:y\ (mod\ 19)\ \iff\ 19\ |\ 2\ (x-y)\ \iff\ 19\ |\ x-y\ \iff\ x\equiv y\ (mod\ 19) $ – Bill Dubuque Feb 26 '11 at 06:50
  • @Bill Dubuque: Thank you. So if $gcd(a, p) = 1$, then $ax \equiv ay \pmod{p} \implies x \equiv y \pmod{p}$, right? – roxrook Feb 26 '11 at 07:07
  • @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 $\rm:(mod\ m)$ – Bill Dubuque Feb 26 '11 at 07:12
  • @Bill Dubuque: Thanks, I will try to understand this. Congruence is so confusing :( – roxrook Feb 26 '11 at 07:18
  • @Chan: Above $\rm\ a \equiv n^{-1}\ (mod\ m)\ $ so to cancel factors of $\rm:n:$ simply multiply thru by $\rm\ a\equiv n^{-1}\ $, which can be computed via the extend Euclidean algorithm. It yields the Bezout identity $\rm\ a\ n + b\ m = gcd(n,m):,:$ which, when $= 1$, yields the sought inverse. – Bill Dubuque Feb 26 '11 at 07:28
  • @Chan $\gcd(a,p)=1,, ax\equiv ay\pmod{! p}\implies x\equiv y\pmod{! p}$ is also a direct consequence of Euclid's lemma. – user26486 Jun 25 '15 at 16:52
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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 $n$...)

Eric Naslund
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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 $9 \pmod{19}$

Siong Thye Goh
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user503962
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