I was solving some problems and I came across this problem. I didn't understand how to approach this problem. Can we solve this with out actually calculating $18!\,\,?$
how to prove $437\,$ divides $18!+1$? (NBHM 2012)
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elementary-number-theory
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4$18!+1$ does not divide $437$; do you mean that $437$ divides $18!+1$? – 2012-11-30
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0thanks for the correction. – 2012-11-30
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0i got upto that point. Now we have to prove that 23(because 437=19*23) also divides 18!+1. i am unable to prove that part. – 2012-11-30
2 Answers
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Note that $437=(19)(23)$. We prove that $19$ and $23$ divide $18!+1$. That is enough, since $19$ and $23$ are relatively prime.
The fact that $19$ divides $18!+1$ is immediate from Wilson's Theorem, which says that if $p$ is prime then $(p-1)!\equiv -1\pmod{p}$.
For $23$ we need to calculate a bit. We have $22!\equiv -1\pmod{23}$ by Wilson's Theorem.
Now $(18!)(19)(20)(21)(22)=22!$.
But $19\equiv -4\pmod{23}$, $20\equiv -3\pmod{23}$, and so on. So $(19)(20)(21)(22)\equiv 24\equiv 1\pmod{23}$. It follows that $18!\equiv 22!\pmod{23}$, and we are finished.
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0Could you explain "relatively prime" in this context? 19 and 23 are prime - what's relative about that? – 2012-11-30
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1Two numbers `a`,`b` are relatively prime (or "coprime") iff `gcd(a,b)=1`, see http://en.wikipedia.org/wiki/Coprime_integers. – 2012-11-30
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0@TimPietzcker: Also, the significance in this context is that if `a` and `b` are relatively prime, and both divide `c`, then their product also divides `c` for hopefully obvious reasons. – 2012-11-30
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1@TimPietzcker The implication is still valid even if you were using, say, 12 and 35, even though those are not prime - it is enough that they not share any prime factors, which is what "relatively prime" means. – 2012-11-30
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That's not so big numbers. This is a demonstration: 18! + 1 = 6402373705728001 = 14650740745373 * 437
Of course that's not as interesting as André's one.
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3The OP was asking to solve it without calculation $18!$. – 2012-11-30