Find the remainder when $2(26!)$ is divided by $29$.
So I know I'm going to use Wilson's theorem and then I would have $28!=-1(\mod29\:)$ but what is the next step? Step by Step explanation please!
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4If $28! \equiv -1 \pmod{29}$, what is $27!$ congruent to. And what $26!$? – Daniel Fischer May 12 '14 at 22:20
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27!≡-1 (mod 28), 26!≡-1(mod27) correct? – Lil May 12 '14 at 22:23
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See also http://math.stackexchange.com/questions/99876/closed-form-for-p-n-pmodp-where-p-is-prime. – lhf May 12 '14 at 22:26
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1@lhf See also this answer. – Bill Dubuque May 12 '14 at 22:36
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$\begin{eqnarray} {\bf Hint}\ \ \ {\rm mod}\ 29\!:\,\ {-}1\! \overset{\rm Wilson}\equiv 28!\, \equiv &&\ (\color{#c00}{28})\ (\color{#0a0}{27})26!\\ \equiv&& (\color{#c00}{-1})(\color{#0a0}{-2}) 26!\\ \equiv &&\qquad\ \ \,2\cdot 26!\end{eqnarray}$
Bill Dubuque
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@Lil mod $,29!:\ 28\equiv -1,$ by $,20\mid 28-(-1)\ $ and $\ 27\equiv -2,$ by $,29\mid 27-(-2)\ \ $ – Bill Dubuque May 12 '14 at 22:27
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$$28!\equiv -1\bmod 29$$ or $$28\cdot27 \cdot 26!\equiv -1\bmod 29$$ which is $$2\cdot 14\cdot27 \cdot 26!\equiv -1\bmod 29$$ Now let $a$ and $b$ be modular inverses of $14,27$ respectively.(such $a,b$ exist because $gcd(14,29)=1$ and $gcd(27,29)=1$).
Multiply both sides by $a\cdot b$. The result is what you want.
Fermat
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