Place 15 different positive integers on the vertices of this graph so that the ten products of three numbers in a straight line are all equal.


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    $\begingroup$ All ones works, but pretty clearly isn't what you are looking for. Might want to specify that they are distinct. $\endgroup$ – hdsdv Aug 16 '19 at 1:22
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    $\begingroup$ Indeed, distinct. $\endgroup$ – Bernardo Recamán Santos Aug 16 '19 at 1:24

There are a number of ways to do this. An easy strategy:

To turn this into a puzzle about addition instead of multiplication, we can solve the addition version of the puzzle and take (for example) 4 to the power of everything.

So here, I put a magic square in the middle 3x3, and then just pick numbers for the left and right corners so that the remaining four are all different. (This is why I chose 4 instead of 2 as a base: this way, I can use halves, ensuring no conflict between the wings and the center 3x3.)

The additive solution I found:

enter image description here
And to turn this into a multiplicative solution, all you have to do is raise 4 to the power of each of the circles.

enter image description here

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    $\begingroup$ What if the smallest possible product is required? $\endgroup$ – Bernardo Recamán Santos Aug 16 '19 at 2:19

Here is a solution with a constant product of


which I think is the minimum possible:

enter image description here

Some partial progress for a lower bound on the product:

First of all, we need a number with at least $15$ divisors: so, nothing less than $120$ will work. But actually, $15$ is not enough. One of the divisors of $n$ is $n$ itself, and that can only be in a triple with $1$ and $1$, which are not distinct, so we can't use $n$. If $p$ divides $n$, then $n/p$ can only be in a triple with $1$ and $p$; if $p^2$ divides $n$, the $n/p^2$ can only be in a triple with $1$ and $p^2$. Such numbers must be the middles of the sides of a triangle, and since they can only be in a triple with $1$, at most two of them can be used. So we need at least $16$ divisors, plus one for every prime factor past the first two, plus one for every prime square factor past the first two.

This leaves only a few possibilities for improvement:

Of the integers less than $360$, only $240$, $288$, and $336$ pass this test. (Note that if $336$ were possible, then $240$ would be, too: just replace all factors of $7$ with factors of $5$.)

  • $\begingroup$ This is also the minimum solution known to me. In mine, all entries are less than 100. $\endgroup$ – Bernardo Recamán Santos Aug 16 '19 at 17:15
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    $\begingroup$ I have some complicated casework that I'm pretty sure eliminates the smaller products, but I don't think writing it up would be all that exciting to anyone. $\endgroup$ – Misha Lavrov Aug 16 '19 at 17:38

Edit: In an effort to find the minumum, here is a much smaller solution in which the mutual product is



enter image description here

As MKBakker pointed out we could further reduce this by dividing each of the entries 4,8,16,96 and 192 by 2 to get a mutual product of


although they have subsequently improved on this.

  • $\begingroup$ Nice! But this is still not the minimum known to me. $\endgroup$ – Bernardo Recamán Santos Aug 16 '19 at 12:07
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    $\begingroup$ Nice work. If you divide 192, 16, 4, 8 and 96 all by 2, you get a new solution with all products equal 864. I think we may be able to go lower with reducing the number of repeated prime factors, and including prime factors 5 and 7. But I haven't found a solution yet $\endgroup$ – P1storius Aug 16 '19 at 13:50
  • $\begingroup$ @MKBakker Nice spot, I had not even noticed that. $\endgroup$ – hexomino Aug 16 '19 at 15:12

I found a solution trying to minimize prime factors. And finding a balance between the minimum value and minimum number of factors.

The product of each three numbers is 720

I noticed that some fields are connected, in that multiplying one of them results in the multiplication of fixed other fields. There are three such patterns:
-C, D, H, L, and M (and any mirror image of that)
-A, E, G, L, M (and any mirror image of that)
-A special case is A,B,C, where multiplication of A can be compensated by division of both B and C (Again, also works for the opposite side and the inverse)

enter image description here

Using this, I was able to assign the minimum number of factors required for a unique solution. I think my solution is the lowest possible, but I'm not 100% confident. Perhaps clever rearrangement could lead to another prime factor being reduced.

The solution:

enter image description here

Now it's time for weekend!

  • 1
    $\begingroup$ Beautiful, but an even smaller solution is known to me! $\endgroup$ – Bernardo Recamán Santos Aug 16 '19 at 16:38

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