# Maths puzzle

1. May 30, 2004

### jcsd

I've nicked this puzzle (and slightly modified it) from another forum:

If the sum of a set of numbers is 100, what is their highest possible product if:

a) they are non-negative reals

b) they are non-negative integers

2. May 30, 2004

### AKG

Would it be from this thread?

3. May 30, 2004

### jcsd

No, it's not from that forum. The original puzzle only had part b) in it, part a) is my own addition, but I see that someone's already posted a very simlair puzzle.

4. May 30, 2004

### mathman

Since the maximum is always achieved when the numbers are all equal, the problem reduces to finding the maximum of x100/x, and then calculating the products for the integers bracketing 100/x. The answer is x=e. I'll let you work out the final answer.

5. May 31, 2004

### jcsd

I've solved it already, it was meant to be a quiz.

The answer to part a) as the fact that the maximum value of x^1/x is e^1/e, is quite well known, though it's helpful for part b) it's not necessrily needed to be known.

Last edited: May 31, 2004
6. May 31, 2004

### uart

Yes but e^(100/e) is not a valid solution in this case as the numbers must add to 100, and no integer multiple if e is equal to 100. Note that the number of numbers is an integer regardless of whether or not the actual numbers themselves are constrained to be integers.

Anyway I think the solution to "part a" is (100/37)^37 and the solution to "part b" is 4 * 3^32, though I haven't totally confirmed these values.

Last edited: May 31, 2004
7. May 31, 2004

### jcsd

Of course yes.

well your defintely right on part b), and you certainly look right for part a)

the question I really wnated to ask for part a) I suppose was then which value of xy = 100 gives the highest vlaue for x^y.