Verify that 4(29!)+5! is divisible by 31. How do I work this out? Step by step explanation please!
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1And how do we know you aren't cheating? – Caleb Stanford May 06 '14 at 10:40
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Hint: $(p-1)! \equiv -1$ (mod $p$). – Geoff Robinson May 06 '14 at 10:41
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so then I have (29!)= -1(mod 29) where do I go from here with simplifying? – Lil May 06 '14 at 10:42
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also should i do the 29! and 5! separately? – Lil May 06 '14 at 10:43
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1Why are you asking the same question twice? – Caleb Stanford May 06 '14 at 11:09
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1http://math.stackexchange.com/questions/778918/verify-that-4295-is-divisible-by-31 – Caleb Stanford May 06 '14 at 11:10
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@Goos, has OP posted the same question twice? If so, then please flag for moderator attention. – Gerry Myerson May 06 '14 at 11:10
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@GerryMyerson Yes. And OP has also copied the nonsense comment on my answer below from here. – Caleb Stanford May 06 '14 at 11:11
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@Goos, OP seems to be quite a piece of work. – Gerry Myerson May 06 '14 at 11:12
2 Answers
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Hint
\begin{align*} 30! &\equiv -1 \pmod{31} \\ 30 &\equiv -1 \pmod{31} \\ \implies 29! &\equiv \quad ? \pmod{31} \\ \end{align*}
Caleb Stanford
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@Lil You know what $30!$ is congruent to. You also know that $30! = 30 \cdot 29!$. – Caleb Stanford May 06 '14 at 10:47
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ok so far I have this... 15!≡(15!)(-1)^2(mod17) ≡ (15!)(16)^2 (mod 17) ≡ 16! (16) (mod17) ≡ (-1)(-1) (mod 17) is this correct so far? – Lil May 06 '14 at 10:57
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Why, Lil, are you working modulo 17? 17 has nothing to do with the problem. Forget 17, and look at what Goos is telling you. – Gerry Myerson May 06 '14 at 11:09
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$30! \equiv -1 \pmod{31}$, so $29! \equiv \frac{30!}{30} \equiv \frac{-1}{30} \pmod{31}$. So,
$4(29!) + 5! \equiv 4(\frac{-1}{30}) + 120 \equiv \frac{-4}{30} - 4 \equiv \frac{120}{30} - 4 \equiv 0 \pmod{31}$.
zscoder
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