Prove that for all $n \in \mathbb{Z}$, $6! \mid n\cdot(n+1)\cdots(n+5)$ using only criteria of divisibility (without using combinatorial arguments).
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3$$\frac{(n+5)(n+4)(n+3)(n+2)(n+1)n}{6!}=\binom{n+5}6$$ See http://math.stackexchange.com/questions/11601/proof-that-a-combination-is-an-integer – lab bhattacharjee Sep 09 '14 at 16:13
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Hint: $n$, $(n+1)$, $(n+2)$, $(n+3)$, $(n+4)$, $(n+5)$ are all 6 consecutive numbers so one of them is a multiple of 6, and 3 of them are a multiple of 2; and 2 of them are a multiple of 3...
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I was going to give this answer, but I decided not to because it seems to me this is combinatorial. – Git Gud Sep 09 '14 at 15:58
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@GitGud I think that misunderstands "combinatorial." It is combinatorial if you find a set with $n\cdots(n+5)$ elements that partitions into sets of size $6!$. – Thomas Andrews Sep 09 '14 at 16:07
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Technically speaking "at least" is a correct word but as you said "exactly" is much better. Thanks for your suggestion. – Bourbaki is ALIVE Sep 09 '14 at 16:20
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Consider the primes that $6!=2^4\cdot 3^2\cdot 5$ has.
Since $A=n(n+1)(n+2)(n+3)(n+4)(n+5)$ is a product of six consecutive numbers, there are three even numbers, and there is at least one number divisible by $4$.
There are two numbers divisible by $3$. Also, there exists a number divisible by $5$.
Hence, $A$ is divisible by $2^4\cdot 3^2\cdot 5=6!.$
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