# A Magic Diamond

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.

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

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

• What if the smallest possible product is required? – Bernardo Recamán Santos Aug 16 at 2:19

Here is a solution with a constant product of

360

which I think is the minimum possible:

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 $$5$$.)

• This is also the minimum solution known to me. In mine, all entries are less than 100. – Bernardo Recamán Santos Aug 16 at 17:15
• 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. – Misha Lavrov Aug 16 at 17:38

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

$$1728$$

Solution

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.

• Nice! But this is still not the minimum known to me. – Bernardo Recamán Santos Aug 16 at 12:07
• 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 – P1storius Aug 16 at 13:50
• @MKBakker Nice spot, I had not even noticed that. – hexomino Aug 16 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)

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:

Now it's time for weekend!

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