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Show that a set of nine consecutive integers cannot be partitioned into two subsets with equal products

Let $\{n,n+1,n+2,n+3,n+4,n+5,n+6,n+7,n+8\}$ be the set of nine consecutive integers. Now this set contains at most a one multiple of $9$ and so when we partition this set into two subsets one of these will contain a multiple of $9$ which contradicts the fact that the products would be equal.

But how can I express this set without having any multiples of nine? What I have is that $$\{9n+1, 9n+2, 9n+3,9n+4,9n+5,9n+6,9n+7,9n+8, \color{red}{9n+9} \}$$ but the last term is a multiple of $9$ as $9n+9=9(n+1)$.

Jiming Le
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  • The set ${2,3,12,18}$ contains only one multiple of $9$, so I guess that proves that it can't be partitioned into two sets with equal sums. – bof Nov 01 '21 at 23:35
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    Your proof doesn't work because the second subset can contain two multiples of $3$. Any set of $9$ consecutive integers must contain a multiple of $9$. – Robert Shore Nov 01 '21 at 23:43
  • Well, this question is closed, but a simple way to bound the number of possibilities a priori is that the product of the integers in such a subset with $4$ or fewer integers cannot be any larger than $2n^4$ [for $n>100$] and in such a subset of $5$ or more integers must be at least $n^5$ [for $n>100$]. So for these products to have any chance of being equal, $n$ must be less than $100$. But then for each set of $9$ consecutive integers must have at least one prime, or exactly $1$ multiple of a prime $\ge 11$. Can you finish from here. – Mike Nov 05 '21 at 19:02

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The trouble with your approach is that $9$ is not a prime number. The set $\{2,3,6,9\}$ contains exactly one multiple of $9$; nevertheless it can be partitioned into two subsets with equal products.

With a little bit of case work, you can show that one of your $9$ consecutive numbers must be relatively prime to all the others. (In fact, if $1\le k\le16$, any set of $k$ consecutive integers contains one which is relatively prime to all the others.) If that number is greater than $1$, then there is a prime number which divides only one of your two products, so they can't be equal. If that number is $1$, then you're looking at the first $9$ positive integers, and instead of $1$ you could choose $5$ as your number which is relatively prime to all the others.

Much more generally (and harder to prove), if $n\gt1$, then the product of $n$ consecutive positive integers can't be a perfect square, or even a perfect $k^\text{th}$ power for any integer $k\gt1$; this is the Erd6s—Selfridge theorem.

bof
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  • Thanks for the answer. May I ask what’s the reason the approach doesn’t work for $9$ since it isn’t a prime? It would seem that this would work if I considered say for example $7$ consecutive integers instead of $9$. – Jiming Le Nov 02 '21 at 09:18
  • The $7$ consecutive integers $3,4,5,6,7,8,9$ can be partitioned into two sets $3,4,5,6$ and $7,8,9$ with respective products $3\cdot4\cdot5\cdot6=360=9\cdot40$ and $7\cdot8\cdot9=504=9\cdot56$. The products are unequal but I don't see how considering $9$ helps to see that. – bof Nov 02 '21 at 17:11