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If $a_1+a_2+a_3+...+a_n=100$ and $a_i \in \mathbb{Z}$ for all $i$, find the maximum possible product $(a_1)(a_2)(a_3)...(a_n)$

So we could have:

$9+10+11+12+13+14+15+16=100$

or we could have:

$2+6+8+9+10+11+12+13+14+15=100$

There is unlimited choices, so how would i be able to find the maximum possible product.

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    $\begingroup$ The $a_i$ can also be negative? $\endgroup$
    – mathcounterexamples.net
    Oct 10, 2015 at 6:49
  • $\begingroup$ i would guess so since its in the integer group. $\endgroup$
    – Caddy Heron
    Oct 10, 2015 at 6:49
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    $\begingroup$ Then your product is unbounded. $\endgroup$
    – mathcounterexamples.net
    Oct 10, 2015 at 6:50
  • $\begingroup$ We should not allow negatives, for then for most $n$ there will not be a max. $\endgroup$
    – André Nicolas
    Oct 10, 2015 at 6:51
  • $\begingroup$ If $n$ is allowed to vary, nothing bigger than $4$ should be used, and $4$ is kind of useless too. $\endgroup$
    – André Nicolas
    Oct 10, 2015 at 6:53

2 Answers 2

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HINT: For positive integers, where $n$ is not fixed, consider what you can do if there is a large $a_r$ to increase the product (think about what might make a large component). Then investigate how to maximise the product of small terms - e.g. what do you do to increase the product if one of the factors is $1$? Do a bit of experimenting and come back when you can show your work.

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As you allow negative integers and do not limit the number of terms one can see that given a summation $a_1+a_2+\ldots +a_n=100$ with product of $a_i$'s $N$, a positive number, we can insert 4 more terms $10+40 -20-30$ which does not change the total but product gets multiplied by a factor of $24000$. So there is no upper bound, as we can continue this process indefinitely.

(I suppose you have no condition that the terms be distinct).

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