Using the numbers 2, 3, 4, 5, 6 and 10 each exactly once, place a number into each circle so that the products of the three numbers along each edge are the same, and as large as possible.
This puzzle is from a UK Junior Mathematical Olympiad.
First observe that
there are two multiples of 3, namely 3 and 6. The two numbers must be placed so that the three sides are multiples of 3 (and none of them multiple of 9 since we have too few factors of 3 for that), and the only way to do that is to place one at a corner and the other at the opposing side. The same applies to 5 and 10.
So we get this far:
2/4 5/10 3/6 3/6 2/4 5/10
In order to maximize the product,
see that each pair is in the form of $(n, 2n)$. The extra factors of 2 can go either to all sides or all corners. Out of the two, putting them at the corners gives greater product.
So the answer is
4 5 3 6 2 10with the side product of 120.
This is intended only as a partial answer:
Bubbler’s answer gave an arrangement of the six numbers where the product of the numbers (for each edge) is:
120
The following is a proof that the common product can’t be larger than what Bubbler achieved:
First of all there must an edge that doesn’t contain the number 10.
The maximum product for this non-10 edge is the product of the three largest numbers other than 10 which is $4 \times 5 \times 6=120$.
Q. E. D.
The above proof idea came from an answer to this question which was later deleted. I don’t know the reason for the deletion.
One derived rule limits the amount of choices for the 6 numbers and their positions
If we call corner values A B C and opposite middle values a b c then one easily proves A/a=B/b=C/c must be equal to some r. But that common ratio r must be bigger than 1 for maximum common product AbC=BcA=CbA. So 10 is a corner value leaving only 5 as opposite middle value and common ratio 2. This further determines 6 as next corner value and all the following values, and, the common product 120=ABC/r. It is not that hard to find similar 6 numbers with such common pairwise rational ratio r. For example 1 2 3 4 6 8 or also n^0...n^5 for some n>0. A similar slightly more difficult reasoning leads to find 1 2 3 4 5 6 8 10 12 15 and their positions on the corners and sides of a pentagon, and, the common product 120. For a square I find in a similar way 1 2 3 4 5 6 10 12 and their positions on the corners, and, the common product 60.
Considering that
10 will give the largest product, it can only be in the middle of a row.
And that
we don’t want the multiple of 10 to be too large,
it means that
10 can only be flanked by 2 and 3, giving 2x3x10 = 60
Then the other numbers place themselves easily:
it’s either 2-4-5, 2-4-6, 2-5-4, 2-6-4, 2-5-6, or 2-6-5 on one row
but
none of the first four give 60 as product,
so we are left with
2-5-6 or 2-6-5
But since this means rejecting
4
in that row, it means the other row must have it, so we have
4 x 3 x 5 = 60
as only possibility, meaning the other row is thus
2 x 5 x 6 = 60
Thus the final answer is
2
10-6
3-4-5
