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Find positive numbers $n$ and $a_1, \dots, a_n$ such that $a_1 + \dots + a_n = 1000$ and $a_1a_2 \cdots a_n$ is as large as possible.

This is a problem from the book "problem-solving through problems" and the key heuristic tried to communicate here is that the choice of $1000$ can be replaced by $2,3,4,5,6,7,8,9, \dots$ to help finding some patterns in the problem.

They conclude that

  1. no $a_i$ will be greater than $4$,
  2. no $a_i$ will be equal $1$,
  3. all $a_i$'s can be taken to be $2$ or $3$,
  4. at most two $a_i$'s will be equal $2$.

Then somehow from these conclusions they end up with the maximum being $3^{332} \cdot 2^2$.

I cannot conclude any of these points they make. If $n = 2$ and $a_1+a_2 = 9$, then surely $a_1 = 4$ and $a_2 =5$ is a solution and already the first point fails?

While this seems like a good book the I think that heuristic here is not being communicated very well. Is there some general pattern one can derive from considering the cases with $2,3,4,5,6,7,8,9, \dots$?

Tenial
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  • I think you should look at the geometric-arithmetic mean inequality to answer this question. – ALNS Dec 16 '22 at 11:35
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    Have you tried to prove any of those claims yourself? To see that no $a_i$ can exceed $4$ note that $5=3+2$ and $3\times 2>5$. The others are similar. Working small examples is always a good way to gain insight, but if you prefer you can just prove the given claims directly. – lulu Dec 16 '22 at 11:36
  • I've tried, but without luck. What is the idea with considering $5 = 3+ 2$ and $3 \cdot 2 > 5$? @lulu – Tenial Dec 16 '22 at 11:37
  • I don't think you have understood the underlying question here. My argument is that, if you had a $5$ in your supposedly maximal list, I could break it into $3+2$ and thereby increase the product...proving that your list could not have been maximal. – lulu Dec 16 '22 at 11:38
  • Your example about $9$ shows, I think, that you don't understand the question. Nobody says some list of $a_i$ might not include a $5$. Obviously it's possible to include a $5$. But such a list could not yield a maximal product. With $9$ you could have $3+3+3$ for which the product is $27$, which is considerable larger than your $5\times 4=20$. – lulu Dec 16 '22 at 11:40
  • I guess that $a_i \in \mathbb{N}$? Otherwise, it's an easy optimization problem. – openspace Dec 16 '22 at 11:41
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    To be clear, you said "If n=2" as though you were looking for a maximal sum and product of only two numbers and you would not allow yourself to consider what happens if it were three numbers... but that is not what the problem is asking. They are saying that we get to find the value of $n$ ourselves which will maximize the product and we are not restricted to keeping the value of $n$ the same. They are asking us to find a collection of arbitrary size such that the numbers in the collection add up to $100$ with the product being maximized – JMoravitz Dec 16 '22 at 13:02

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