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I have already posted this question time before on stackexchange, but didn't receive a definitive solution.

So this is the game: consider a positive integer number $n$ and divide it in a finite number of parts $m$, then take the product of all the $m$ numbers you obtained and multiply them. For example for $n=10$ and $m=5$ the product will be $32$. The game consists in finding a number $m$ (positive integer) which maximize this product.

Now it's trivial to prove that the function $f(x)=(n/x)^x$, with $(0,+∞)$ as domain, increases on $(0,n/e]$ and decreases on $[n/e,+∞)$, so it has a max in $x=n/e$, thus you'll find the $m$ you're looking for by evaluating the function in the two integers nearest to $n/e$ and comparing these two values.

What is not easy to prove is that if you take the nearest positive integer to $n/e$ it will always be the best $m$ for the game, or at least for all the cases I've taken into consideration (with computational experiments too).

So my question is: how to prove (or disprove) that this fact holds for any positive integer $n$?

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  • $\begingroup$ What exactly is the game? Are these supposed to be $m$ equal parts? Are they supposed to be integers? Then, you also have a divisibility condition. $\endgroup$
    – Alex Degtyarev
    Commented Jan 8, 2015 at 17:34
  • $\begingroup$ Yes, they are $m$ equal parts (but I think it could be proved this other fact: to maximize the product they have to be all equals). The single parts are not necessarily integers, for example for $n=10$ and $m=3$ you get $(10/3)^3$. $\endgroup$
    – asdfghj
    Commented Jan 8, 2015 at 17:39
  • $\begingroup$ Yes, indeed, even by not mentioning that, it follows from the geometric-arithmetic means inequality that the parts must be equals. But still have no idea of how to prove the theorem I'm asking for. $\endgroup$
    – asdfghj
    Commented Jan 8, 2015 at 19:06
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    $\begingroup$ Suppose just for a moment that instead of taking $n$ to be an integer, you take $n=ke/2$, where $k$ is an odd integer. Then $n/e$ is a half-integer, so the two choices for $m$ are equally close, but the higher of the two choices always gives a (slightly) better payoff --- in fact the ratio of the two payoffs goes rapidly to $1$ from above. This suggests that when your $n/e$ is very close to a half-integer, there should be a slight preference for the larger of the two $m$'s, which is almost always outweighed by the strong preference for the closer one.....continued $\endgroup$
    – Steven Landsburg
    Commented Jan 8, 2015 at 21:09
  • $\begingroup$ continued --- so you might expect to find rare counterexamples when $n/e$ is just slightly less than a half-integer. $\endgroup$
    – Steven Landsburg
    Commented Jan 8, 2015 at 21:10

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Good rational approximations to $e$ of the form $2n/(2m+1)$ provide candidates for counterexamples. These are known from the continued fraction for $e$: $$ 2 + \dfrac{1}{1 + \dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{4+\dfrac{1}{1+ \ldots}}}}}}$$

But to have a counterexample, we would need (approximately) $$ \left(m + \dfrac{1}{2}\right)^2 - \dfrac{n}{e} \left(m + \dfrac{1}{2}\right) < \dfrac{1}{24}$$ and this does not appear to be the case: these will always be convergents where the next element of the continued fraction is $1$, so they are not especially good approximations, and it should be possible to prove that they are not good enough.

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