# Wilson's theorem related problem

prove $$18! \equiv -1 \pmod{437}$$

I do not want full solution to the above problem but if anybody can tell me how we can approach to it, I will really appreciate that.

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Note $437 = 19\cdot 23$. – Sean Eberhard Aug 26 '12 at 9:26

hints :

• $437=19\cdot 23$ (as proposed by Sean)
• Wilson's theorem :-)
• $19\cdot 20\cdot 21 \cdot 22=(-4)(-3)(-2)(-1)\pmod{23}$
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As noted by Sean, $\,427=19\cdot 23\,$, thus using Wilson' theorem twice: $$(1)\,\,\,18!\cdot 19\cdot 20\cdot 21\cdot 22=22!=-1\pmod {23}\Longrightarrow$$ $$\Longrightarrow 18!=\frac{-1}{(-4)(-3)(-2)(-1)}=-\frac{1}{24}=-1\pmod {23}$$ $$(2)\,\,\,\,18!=-1\pmod {19}$$

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Is this not a full solution? – awllower Aug 26 '12 at 10:43
Not yet...but almost. The gist of it all is number (1), as (2) is directly Wilson's Theorem – DonAntonio Aug 26 '12 at 10:48
Indeed, this cqn be interpreted this way!! – awllower Aug 26 '12 at 10:49

Wilson's Theorem: (p-1)! = -1 (mod p), p: is a prime number

Note: When I write the "equal" (=) sign, I mean the congruence sign.....

437 is not a prime number: 437 = 19*23

For: 19 By Wilson's Theorem => 18! = -1 (mod 19)

For 23: By W.T => 22! = -1 (mod 23) => 22*21*20*19*18! = -1 (mod 23)

For 22 = -1 (mod 23) For 21 = -2 (mod 23) For 20 = -3 (mod 23) For 19 = -4 (mod 23)

We multiply our reminders => (-1)(-2)(-3)*(-4) = 24

Back to our equation: 22*21*20*19*18! = -1 (mod 23) transformed to => (-1)(-2)(-3)*(-4) = 24 * 18! = -1 (mod 23)

For 24 = 1 (mod 23) (replace 24 by 1)

1 * 18! = -1 (mod 23)

So, 18! = -1 (mod 23)

Hope it helps you understand it more.

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