Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the largest number that can be obtained as the product of two or more positive integers that add up to 20?

share|cite|improve this question
do they have to be distinct? – Jean-Sébastien Sep 30 '12 at 19:10
There are some IMO problems regarding things like this. – Beni Bogosel Sep 30 '12 at 19:24
The original problem doesn't state if they have to be distint, so I'm assuming it doesn't matter. – Ctrl Sep 30 '12 at 19:46

If a number $m\gt 4$ occurs in a decomposition (of $20$ in this case), the product can be increased by splitting $m$ suitably. It makes no difference whether $4$ is split as $2\cdot 2$ or not, so it might as well be. It is clear that it is no good to use any $1$'s.

Thus we can assume that all the numbers in our splitting of $20$ are $2$'s and/or $3$'s. Then note that $6=2+2+2=3+3$ but $3^2\gt 2^3$. So if there are three or more $2$'s in a splitting, we can do better.

Remark: If you want to prove that any number $m \gt 4$ should be split further, there are two cases to consider. (i) If $m=2k$, show that $k^2\gt 2k$ if $k\gt 2$. (ii) If $m=2k+1$, show that $k(k+1)\gt 2k+1$ if $k\ge 2$.

share|cite|improve this answer
So 1458 seems to be optimal- 6 3s and a 2. – Geoff Robinson Sep 30 '12 at 19:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.