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Let $S$ be the set of numbers of the form $n(n + 1)(n + 2)(n + 3)(n + 4),$ where $n$ is any positive integer. The first few terms of $S$ are \begin{align*} 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 &= 120, \\ 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 &= 720, \\ 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 &= 2520, \end{align*} and so on. What is the GCD of the elements of $S$?

Every other number should be even, so $n(n + 1)(n + 2)(n + 3)(n + 4)$ has at least a factor of $2^2=4$ (or $2^3=8$ when $n, n+2$ and $n+4$ are all even). Every third number is divisible by $3$, so $n(n + 1)(n + 2)(n + 3)(n + 4)$ is at least divisible by $3$. Every 5 number is divisible by $5$, so $n(n + 1)(n + 2)(n + 3)(n + 4)$ is divisible by $5$ as well. So I put $3\times 4\times 5 = 60$ as the answer, which is wrong. But I don't think numbers above $5$ are guaranteed to be a factor of $n(n + 1)(n + 2)(n + 3)(n + 4)$...

Américo Tavares
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    See https://math.stackexchange.com/questions/12065/the-product-of-n-consecutive-integers-is-divisible-by-n-factorial – lab bhattacharjee Mar 28 '18 at 07:57
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    As one of the even numbers will be divisible by $4=2^2$, you can increase the power of $2$ that will divide the product, not that that (in general - in this case it's enough) brings you much closer to the result, that can be seen in the question @labbhattacharjee linked above. – Henrik supports the community Mar 28 '18 at 08:19
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    There's a gap in your reasoning: for the entire product $n(n+1)(n+2)(n+3)(n+4)$ to be divisible by $8$ it is not necessary for any single one of $n,n+1,n+2,n+3,n+4$ to be divisible by $8$. The lack of the latter divisibility does not rule out the former. Also, you already know the GCD is a multiple of $60$. It also must be a divisor of $120$, the smallest element of $S$. Not a lot of room in between those values... – Erick Wong Mar 28 '18 at 08:28

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For $n\ge 1$, the number $$\frac{n(n+1)(n+2)(n+3)(n+4)}{120}$$ is just the binomial coefficient $\binom{n+4}{5}$ which is always an integer. Hence all the numbers are divisible by $120$.

Because $120$ is the smallest , the $\gcd$ of all the numbers is $120$

Peter
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