# Thread: Largest Product of Integers

1. ## Largest Product of Integers

Hello, here is a problem that I came across a while ago, I have what I think is the solution, but perhaps there is an easier way to solve it, or perhaps a more precise solution.

The problem is:

What is the largest product of integers whos sum is 2008.

2. Originally Posted by sean92t0
Hello, here is a problem that I came across a while ago, I have what I think is the solution, but perhaps there is an easier way to solve it, or perhaps a more precise solution.

The problem is:

What is the largest product of integers whos sum is 2008.

I have approximated a lot. If it were product of reals, I believe my solution would have been fully justifiable. Let me know if my answer matches yours.

I will assume positive integers because or else we can maximize the product without bounds.

Lets say that there are y integers whose sum is 2008. Denote them by $x_1,x_2,\cdots,x_y$.

So $x_1+x_2+\cdots+x_y = 2008$

We want to maximize the product $P = x_1 x_2 \cdots x_y$

Since by assumption the numbers are positive, we can apply AM-GM inequality.

$\frac{\sum_1^y x_i}{y} \geq \sqrt[y]{P} \Rightarrow P \leq \left(\frac{2008}{y}\right)^y$

Thus the maximum value of the product is $\left(\frac{2008}{y}\right)^y$ and is purely determined by y.

So we want to maximize $f(y) = \left(\frac{2008}{y}\right)^y$

A little bit of calculus establishes that f(y) approaches maxima at $y = \frac{2008}{e}$

Thus the discrete maximum is attained at $y = \lfloor \frac{2008}{e} \rfloor + 1 = 739$(Actually I have done trial error to choose between 738 and 739)

In AM-GM inequality, equality holds when all the numbers are equal.

Thus $739 x_i \approx 2008 \Rightarrow x_i = \lfloor\frac{2008}{739}\rfloor = 3$

Thus I believe that $3^{669}$ is the largest product.

3. this is very similiar to what I did, however I am not familiar with the inequality you used so I did that part slightly differently. You have 3^669, my problem with that is that the sum of those integers only adds up to 2007.

4. Originally Posted by sean92t0
this is very similiar to what I did, however I am not familiar with the inequality you used so I did that part slightly differently. You have 3^669, my problem with that is that the sum of those integers only adds up to 2007.
No i have added 669 3s and one 1.

5. you're right, i read it wrong.

6. Originally Posted by sean92t0
this is very similiar to what I did, however I am not familiar with the inequality you used so I did that part slightly differently.
The inequality is called AM-GM and you can search it on the web. Its pretty popular...

Originally Posted by sean92t0
you're right, i read it wrong.

7. well before that, consider this:

3^668 * 3 * 1 < 3^668 * 2 * 2

8. Originally Posted by sean92t0
well before that, consider this:

3^668 * 3 * 1 < 3^668 * 2 * 2
Hmm ya... I think I should try a new line of approach... I am trying it

The maximum product is indeed $2^2 \cdot 3^{668}$.