We are 5 different numbers

Question

We are 5 different positive integer numbers smaller than 100.

The product of us is an odd number.
The product of us is a cube number.
The sum of us is a cube number.

Determine what numbers we are.

Answer 1

I shall assume (though the question doesn't currently say explicitly) that the numbers involved are supposed to be integers.

The "sum is a cube" condition seems really difficult to do anything with in terms of actual reasoning, so I suspect the only practical options are computer search and good luck. I opted for computer search and found

two solutions, both with sum 125: (5,7,15,35,63) and (1,3,15,25,81). In both cases it's easy to see that the product is a cube.

Answer 2

Here is one solution (after several subsequent... edits):

a=1, b=2, c=1/2, d=(9-\sqrt{65})/4, and e=(9+\sqrt{65})/4.

Indeed, as

their product is 1, their sum is 8, and they are indeed positive and below 100.

Man, three mistakes in such a short post.

---

Well, OP changed the wording several times... An integer solution seems to be: {1, 3, 15, 25, 81}, summing up to 125.

Answer 3

Here is what I got so far

The product of us is an odd number.

this implies that every numbers are odd.

The sum of us is a cube number.

since the limit is 100 and that every numbers are odd, and adding an odd amount of odd numbers results in an odd number,
we can deduce that only only 1, 27, 125, 343 are possible sums. since every numbers must be different, 1 is impossible.

The product of us is a cube number.

I have no idea how to proceed other than brute force,
but only 1 combination of numbers add up to 27 : 1+3+5+7+11. And their product is not a cube number. Therefor, 27 is also an impossible sum.
which leaves 125 and 343 as possible sums.

Answer 4

1, 3, 15, 25, 81
$1 + 3 + 15 + 25 + 81 = 125$ (a cube number)
$1 × 3 × 15 × 25 × 81 = 91125$ (a cube number)

Answer 5

The only 3 answers for this are

• 1,3,15,25,81
• 5,7,15,35,63
• 49,65,73,75,81

I used a simple C++ program to find these values:

#include <iostream>
#include <cmath>
bool integer(float k)
{
    if( k == (int) k) return true;
    return false;
}
int main()
{
    int a,b,c,d,e;
    for(a=1;a<100;a++)
    {
        for(b=1;b<100;b++)
        {
            for(c=1;c<100;c++)
            {
                for(d=1;d<100;d++)
                {   
                    for(e=1;e<100;e++)
                    {
                        if((a*b*c*d*e)%2 == 1 && integer(cbrt(a*b*c*d*e)) && integer(cbrt(a+b+c+d+e)) && (a!=b && a!=c && a!=d && a!=e && b!=c && b!=d && b!=e && c!=d && c!=e && d!=e) && (a<b && a<c && a<d && a<e && b<c && b<d && b<e && c<d && c<e && d<e))
                        {
                            printf("%d %d %d %d %d\n",a,b,c,d,e);
                        }
                    }
                }     
            }
        }    
    }
   return 0;
}