<s>[INST] Finding the greatest common divisor of $(n+1)(n+2)^2(n+3)^3(n+4)^4$ Determine the largest positive integer that divides $(n+1)(n+2)^2(n+3)^3(n+4)^4$ for all positive integers n.

First, I noticed that out of $n+1, n+2, n+3,$ and $n+4,$ there must be one multiple of $4$, at least one multiple of $3$, and a multiple of $2$ that isn't the multiple of $4$. Since the problem is asking for a minimal case that satisfies these conditions, $n+1$ should be the multiple of $4$, $n+2$ should be the multiple of $3$, and $n+3$ should be the multiple of $2$, thus giving $4 \cdot 3^2 \cdot 2^3 = 288.$
However, I am unsure as to whether or not this answer is correct, as testing cases has consistently yielded higher GCDs than this.
 [/INST] Following the comment of lulu, look at numbers of the form $n=19+24k$.
For $k=0$, $$(n+1)(n+2)^2(n+3)^3(n+4)^4=20\cdot 21^2\cdot 22^3\cdot 23^4$$ which has the factorization $2^5\cdot 3^2\cdot 5\cdot 7^2\cdot 11^3\cdot 23^4$ and is not divisible by any number containing more than five factors of $2$ and two factors of $3$
For $k=1$, the product $44\cdot 45^2 \cdot 46^3\cdot 47^4$ has no factors of $7$, so $7$ cannot be a factor of the divisor you seek.
For $k=2$, the product $68\cdot 69^2 \cdot 70^3\cdot 71^4$ has no factors of $11$, so $11$ cannot be a factor of the divisor you seek.
For $k=4$, the product $116\cdot 117^2 \cdot 118^3\cdot 119^4$ has no factors of $5$ or $23$, so neither $5$ nor $23$ can be a factor of the divisor you seek.
So the only factor of $20\cdot 21^2\cdot 22^3\cdot 23^4$ that can and must appear in the divisor you seek is $2^53^2$.
It's hard to see that in comparing two examples at a time, as many of them will have a $5$ or a $7$, as well as the occasional larger prime factor in common.
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