The sum of n positive integers is 19. What is the maximum possible product of these n numbers?

Can you also please explain why this is so?

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- Jul 23rd 2007, 11:33 PMDivideBy0Greatest product from a sum
*The sum of n positive integers is 19. What is the maximum possible product of these n numbers?*

Can you also please explain why this is so? - Jul 24th 2007, 11:30 PMCaptainBlack
Well as nobody else has answered this I will tell you what I know though its not a complete solution.

The arithmetic-geomentric mean inequality tells us that if there are $\displaystyle N$ numbers involved, and that these are $\displaystyle n_i,\ i=1, .. N$, then:

$\displaystyle

\left[ \frac{19}{N} \right]^N \ge \prod_1^N n_i

$

Regarding the lefthand side of this inequality as a function of N this has a maximum $\displaystyle \approx 1085.4$ when $\displaystyle N=7$. So we have that the maximum posible value of the product of an integer partition of $\displaystyle 19$ is less than or equal $\displaystyle 1085$.

The partition $\displaystyle 3,3,3,3,3,2,2$ has product $\displaystyle 972$ and I would not be supprised if this were the maximum, but have not done enough to convince even myself that this is the case.

RonL - Jul 25th 2007, 02:06 AMCaptainBlack
Now as

$\displaystyle

\left[ \frac{19}{9} \right]^9 \approx 832.9

$

and

$\displaystyle

\left[ \frac{19}{5} \right]^5 \approx 792.4

$

The above example of a 7-partition with a product of $\displaystyle 972$ shows that the partition that gives the maximum product must be a 6, 7 or 8-partition.

(Exhaustive search shows that $\displaystyle 972$ is indeed the maximum product and it can be achived with either a 6 or a 7-partition.

RonL - Jul 25th 2007, 03:25 AMCaptainBlack
The reason for guessing that this partition is a good candidate for the maximum product is that the arithmetic-geometric mean inequality is an equality when all the numbers are equal. We cant achive that in this case as 19 is prime, but we can look at partitions where the elements are as near equal as possible, and that is what I did here.

The 6-partition of 19 which achives the maximum product can also be found im this way.

RonL - Jul 27th 2007, 10:38 AMray_sitf
What about [19/(19/e)] to the power of (19/e)?

This gives 1085.405992...

I know the OP wanted integers, but this suggests the following procedure.

If the number leaves a remainder of 2 when divided by three, then the max. answer is 3*3*3*...*2.

If the number leaves a remainder of 1 when divided by three, then the max. answer is 3*3*3*...*2*2.

If there is no remainder on division by three, then it's 3*3*3*...*3.

E.g. for 23 it's 3^7 * 2, and for 25 it's 3^7 * 2 * 2 - Jul 27th 2007, 10:52 AMCaptainBlack
- Jul 27th 2007, 11:09 AMCaptainBlack
- Jul 27th 2007, 11:59 PMDivideBy0
You're correct Captainblack, the answer is 972.

I had a hard time understanding the inequality, but I guess as long as I take it for granted I should be fine.

Also, is it an actual fact that 3 is the best integer and $\displaystyle e$ is the best number to use for something like this? - Jul 28th 2007, 06:14 AMCaptainBlack
The logic says that the nearest integer to N/e is a good choice for the

number of terms in the sum and product, and that lots of 3's and a few 2's

will get you close to the maximum for the product, but its not definitive.

For instance you can get as large a product with 6 terms and with 7 when

the sum is 19.

RonL