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Wilson's theorem states that if $n$ is a prime number, it will divide $(n-1)! + 1$, using this find the smallest divisor of $12!+6! +12!×6! + 1$?

I checked yahoo answers and there someone gave the answer as $7$(which is wrong). The answer is $91$. Please explain the method. Thanks

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Kousik
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    $91=7\times 13$ isn't prime. It's not the smallest divisor of anything. – mjqxxxx May 10 '15 at 17:43
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    The method is this: $12! + 6! + 12!\times 6! + 1 = (12! + 1)\times (6! + 1)$, so Wilson's theorem states that $7$ and $13$ divide it. And it is equal to $1$ modulo $6!$, so it's not divisible by $2$, $3$, or $5$. Its smallest divisor is therefore $7$. – mjqxxxx May 10 '15 at 17:45
  • @Arthur it is mod 6!, so mod 720. Then it does say something about 5. – wythagoras May 10 '15 at 18:00
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    Some rewording may be useful. The smallest (positive) divisor of our number is $1$, – André Nicolas May 10 '15 at 18:10
  • To see that $91$ isn't prime, $91=10^2-3^2=(10-3)(10+3)=7\times 13$. – user45220 May 10 '15 at 18:11
  • You really don't need to include a statement of Wilson's theorem in your title. It makes it appear that you put no effort at all into even formulating your question, let alone trying to answer it. – WillO May 10 '15 at 18:20

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If $\,f(x)\,$ is a polynomial with integer coefficients and $\,\color{#c00}{f(0) = 1},\ \color{#0a0}{f(-1) = 0}\,$ then the prime $\,p\,$ is the least factor $> 1$ of $\,f((p\!-\!1)!)\, $ since, $\,{\rm mod}\ p\!:\ f((p\!-\!1)!)\equiv \color{#0a0}{f(-1)\equiv 0}\,$ by Wilson, and if prime $\,q< p\,$ then $\,q\mid (p\!-\!1)!\,$ therefore $\,{\rm mod}\ q\!:\ f((p\!-\!1)!) \equiv \color{#c00}{f(0)\equiv 1},\,$ so $\,q\nmid f((p\!-\!1)!).$

OP is special case $\,p=7,\,\ f(x) = (12!/6!\cdot x +1)(x+1),\, $ so $\,f(6!) = (12!+1)(6!+1)$

Above we used standard Congruence Rules, most notably the Polynomial Congruence Rule $\, a\equiv b\,\Rightarrow\, f(a)\equiv f(b)\ \pmod{\! m},\ $ for any polynomial $\,f(x)\,$ with integer coefficients.

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Bill Dubuque
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