This is my first question, so sorry if I'll make any mistake in using the site formatting.
I found this game on a book by Yacov Perelman and I thought it could be nice to introduce Nepero number to secondary school students, but I wanted to be sure about a couple of things before showing it to them. Not that I'll ever give them proofs of these facts, they have not enough basic analysis knowledge to understand them, but it's just a personal curiosity.
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 $2^5=32$. The game consists in finding a number $m$ (positive integer) which maximize this product.
Now it's easy to prove that the function $f(x)=(n/x)^x$, with $(0,+\infty)$ as domain, increases on $(0, n/e)$ and decreases on $(n/e,+\infty)$, so it has a max in $x=n/e$, and you'll find the $m$ you're looking for by evaluating the function in the two integers nearest to $n/e$.
For example for $n=10$ as above you obtain $n/e \approx 3.7$ and so $m$ could be $3$ or $4$, by evaluating $f$ in these two points you find out that the product is maximized for $m=4$.
What strikes me is that if you take the nearest integer to $n/e$ it will always be the best $m$ for the game too (except for $n=1$, but then $m=0$ is not even a valid input for the game), or at least for all the cases I've taken into consideration. Well it could seem quite intuitive, but it's not true for all the convex/concav functions and I was not able to show it for this particular one. So my questions are:
1) Is it true for all the $n$? If so can you prove it? If not could you show me a counterexample?
2) Let's suppose it's true and you can prove it... Is it possible that, by errors due to floating point approximations, someone finds a number which seems to be a counterexample but it's really not? (Well I admit this second question is not so well defined since I'm not even referring to what kind of approximations calculators do, but if you have any clue or could put this question in a more formal way just say let me know, I'll appreciate it)