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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.

ari

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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 at 1:22
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    $\begingroup$ Indeed, distinct. $\endgroup$ – Bernardo Recamán Santos Aug 16 at 1:24
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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 at 2:19
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Here is a solution with a constant product of

360

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 $5$.)

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  • $\begingroup$ This is also the minimum solution known to me. In mine, all entries are less than 100. $\endgroup$ – Bernardo Recamán Santos Aug 16 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 at 17:38
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Edit: In an effort to find the minumum, here is a much smaller solution in which the mutual product is

$1728$

Solution

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

$864$

although they have subsequently improved on this.

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  • $\begingroup$ Nice! But this is still not the minimum known to me. $\endgroup$ – Bernardo Recamán Santos Aug 16 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 at 13:50
  • $\begingroup$ @MKBakker Nice spot, I had not even noticed that. $\endgroup$ – hexomino Aug 16 at 15:12
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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!

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

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