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I have recently got interested in the following problem:

Give a decomposition of a natural number to natural summands whose multiplication is maximal.

I have tried to solve this problem, and have written this.

I would like to hear your opinion - is this (mathematical) solution ture? are all my steps leggit?

Also, I was wandering - is this decomposition unique, in a way? For example, we have $3+3+4=3+3+2+2=10$ which gives the same multiplication. I think that this is the only exception from my solution, am I right?

philosophically, is this problem and solution considered very trivial? If so, what similar (harder) problems of this sort are avalibale?

Aryabhata
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Mike
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    The problem itself has an answer here. Your argument is unnecessarily complicated, but at a very quick glance it appears to be correct; at any rate the conclusion is correct. – Brian M. Scott Jan 23 '15 at 22:36

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The decomposition problem is fairly straightforward.

Any number $n, n>4$ can be decomposed into $(n-2)$ and $2$ and the product $2n-4 = n + (n-4) > n$. So we should never have any integer greater than $4$ or (trivially) less than $2$ in a maximal-product partition of $n>1$. Decomposing $4$ into $(2,2)$ makes no difference. We can observe that the best decomposition of $6$ is $(3,3)$ not $(4,2)$ and that in general taking factors of $3$ is the best option unless it means leaving $1$.

The underlying reason for this is that $\frac{\ln x}{x}$ has a maximum at $e$, and a maximum in integers at $3$ (with $\frac{\ln 2}{2}=\frac{\ln 4}{4}$).

Nobody here is likely to read through lengthy proofs on another site.

Joffan
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