We have to find n numbers such that $$ x_1 + x_2 + \cdots + x_n = k $$ $$ x_1 * x_2 * .....* x_n = maximum $$

What are the values of $x_1, x_2...x_n$ ?

Note that $x_1, x_2...x_n$ are all positive integers .

P.S : I had posted a similar question before. In that question, $x_1, x_2...x_n$ could be any positive numbers and did not necessarily have to be integers. This is the previous question : maximum product of n numbers whose sum is k

  • $\begingroup$ Can someone explain the downvotes please? $\endgroup$ Feb 5, 2021 at 15:45
  • $\begingroup$ @riskymysteries I made a similar post before and people probably thought that this post and my post before are asking the same question ..here is the previous post : puzzling.stackexchange.com/questions/107082/… $\endgroup$ Feb 5, 2021 at 16:23

2 Answers 2


First suppose that the $x_i$ are restricted to non-negative integers. In this case we need the $x_i$ to be

all equal, or at least as close to equal as possible.

The reason is that

For any $d>0$ we have $$(t-d)(t+d) = t^2-d^2 < t\cdot t$$ and $$(t-d)(t+d+1) = t^2+t-d^2-d < t(t+1)$$ So if any two of the $x_i$ differ by 2 or more, increasing the smaller and decreasing the larger will increase their product.

So the values of $x_i$ are:

$\lfloor \frac{k}{n}\rfloor$ or $\lceil \frac{k}{n}\rceil$. There will be $k \bmod n$ of the latter.

Suppose now that the $x_i$ are allowed to be negative.

For $n=1$ or $n=2$ allowing negative values makes no difference, so this case is as above.

For $n\ge 3$ we can choose the following values for $x_i$:

The first three variables have values $x_1=T+k+4-n$, $x_2=-T$, $x_3=-1$, and the remaining $n-3$ variables all have value $1$. Here $T$ is an arbitrary value large enough to make $x_1$ positive. Their sum is $k$ as required, but their product can be made arbitrarily large by choosing $T$ larger.

  • $\begingroup$ Funnily enough, your choices in the last paragraph are exactly the same as those I made on the other question, but in a different order. $\endgroup$
    – hexomino
    Feb 3, 2021 at 17:06
  • $\begingroup$ @hexomino , I just realised that whether k is a positive number or k is a positive integer, in both the cases, xi= k/n . The same logic as those presented in the answers to the previous question : puzzling.stackexchange.com/questions/107082/… are applicable . Am I right ? $\endgroup$ Feb 3, 2021 at 18:35
  • $\begingroup$ @HemantAgarwal integers are numbers so yes. $\endgroup$
    – hexomino
    Feb 3, 2021 at 18:36
  • $\begingroup$ @hexomino , so there is nothing new to be gained from keeping this question up , right ? $\endgroup$ Feb 3, 2021 at 18:37
  • $\begingroup$ @HemantAgarwal Correct. However, if I were you I would delete the third question you posted and edit this one to make x_i integers because Jaap has given a correct answer to that question and it seems unnecessary to make him write up the same thing somewhere else. $\endgroup$
    – hexomino
    Feb 3, 2021 at 18:41

The following answer was given by @GarethMcCaughan .

There is always an optimal answer with no two factors differing by more than 1, because if we have factors a,b with a>b+1 then replacing them with a−1,b+1 leaves the sum unaltered and yields a product that has increased by a−b−1≥0.


consider an optimal configuration with no two xj differing by more than 1. Letting x be the smallest of our factors, some of our n factors are x and the rest are x+1; let's say m of them are x+1 where 0≤m<n. (We can't have m=n because then the smallest number would actually be x+1.) Then the sum (which must equal k) is nx+m, so m=k mod n, and there is exactly one m with this property and 0≤m<n, namely the remainder on dividing k by n. So this configuration is unique (up to permutation of the xs).


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