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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.

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  • $\begingroup$ This easily admits infinitely many solutions. Or are the five numbers supposed to be integer? $\endgroup$ – Lynn Oct 7 '16 at 10:41
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    $\begingroup$ @Lynn : Yes integer. $\endgroup$ – Jamal Senjaya Oct 7 '16 at 10:45
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    $\begingroup$ I hope this doesn't requires brute force and or the use of a program and that it can be solved by logic like I tried in my attempt at an answer. $\endgroup$ – stack reader Oct 7 '16 at 11:07
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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.

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  • $\begingroup$ On reflection, I suspect this does happen to be solvable by hand without outrageous good luck (apart from the good luck that there turn out to be nice solutions). E.g., you could look for a solution using no primes other than {3,5} or none other than {3,5,7} and in either case you have rather few options to play with. But discovering by hand the absence of other solutions would be very difficult. $\endgroup$ – Gareth McCaughan Oct 7 '16 at 11:20
  • $\begingroup$ I do not think you need extremely good luck. The "product is odd" mean no even numbers and "product is cube" mean some distribution of prime factors (but not 2) so each prime factor is present three times (and the product for each is still less than 100). Without checking this should be a pretty small number of sets, but you still need to validate the third condition by hand. $\endgroup$ – Thorbjørn Ravn Andersen Oct 7 '16 at 12:08
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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.

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This is an attempt to determine all solutions by hand.


The product of us is an odd number.

Hence all five the numbers are odd.

The product of us is a cube number

This means that each prime factor must occur a multiple of three times in the factorizations of the five numbers. Since all five numbers are odd, this means all primes occuring are smaller than $20$. The arguments below illustrate this more clearly.

The sum of us is a cube number

The sum of the five numbers is odd and less than $500$, so it is either $27$, $125$ or $343$.

Now for some repetitive arguments to reduce the number of options:

Argument 1

If $19$ occurs as a prime factor, then $19$, $3\times19=57$ and $5\times19=95$ must be among the five numbers, which sum to $171$. Then the sum of the other two numbers must be $172$, so one must be at least $86$, hence it is either $99$ or $91$ and the other is $73$ or $81$, but then the product is not a cube. So $19$ does not occur as a prime factor.

Argument 2

Similarly, if $17$ occurs as a prime factor, then $17$, $3\times 17=51$ and $5\times17=85$ must be among the five numbers, which sum to $153$. So the sum of the other two numbers must be $190$; they must be $99$ and $91$. Again the product is not a cube, so $17$ does not occur as a prime factor.

Argument 3

Similarly, if $13$ occurs as a prime factor, then precisely three out of $13$, $39$, $65$ and $91$ must be among the five numbers. Their sum is then one of $117$, $143$, $169$, $195$, so the other two numbers must sum to $8$, $200$, $174$ or $148$, respectively. A sum of $200$ is impossible, and a sum of $8$ implies that the other two numbers are $1$ and $7$ or $3$ and $5$, either way the product is not a cube. The five largest remaining numbers are $99$, $81$, $77$, $75$ and $63$, so the two other numbers must be $99$ and $75$, but then the product is not a cube. So $13$ does not occur as a prime factor.

Argument 4

Similarly, if $11$ occurs as a prime factor, then precisely three out of $11$, $33$, $55$, $77$ and $99$ must be among the five numbers. Then their sum is either $99$, $121$, $143$, $165$, $187$ or $231$, so the sum of the other two numbers must be either $26$, $4$, $200$, $178$, $156$ or $112$, respectively, where $200$ is again clearly impossible. The largest remaining numbers are $81$, $75$, $63$ and $49$, so the other two numbers must be either $81$ and $75$ or $63$ and $49$. In the first case our original three numbers were either $11$, $77$ and $99$ or $33$, $55$ and $99$, in both cases the product is not a cube. In the second case our original three numbers were $55$, $77$ and $99$, and the product is again not a cube. So then the sum of the other two numbers must be $26$ or $4$, so they are $1$ and $25$ or $5$ and $21$ or $1$ and $3$. In none of these cases the product is a cube. So $11$ does not occur as a prime factor.

Finally:

The only remaining possible prime factors are $3$, $5$ and $7$. Some playing around shows that there are only two solutions.

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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.

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

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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;
}
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  • $\begingroup$ Will that link last for posterity? If not, it would be better to edit the essential parts of your solution (e.g. the details of the program you used) into this answer. $\endgroup$ – Rand al'Thor Oct 14 '16 at 14:04
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    $\begingroup$ $\sqrt[3]{49*65*73*75*81} = 1122.00000715...$ In fact, there's no way 73 could be one of the numbers, since it's prime and it has no other multiples less than 100. $\endgroup$ – Michael Seifert Oct 14 '16 at 15:03
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    $\begingroup$ @kurella : Be careful when calculating with floating point numbers. You have to double check the result. $\endgroup$ – Jamal Senjaya Oct 14 '16 at 15:19
  • $\begingroup$ @randal'thor That link lasts for life time. But I edited and added the code here itself $\endgroup$ – Kurella Hemanth Oct 15 '16 at 7:26
  • $\begingroup$ @MichaelSeifert Because of precision, the code is wrong. $\endgroup$ – Kurella Hemanth Oct 15 '16 at 7:29

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