# Beginner puzzle

## This puzzle is intended to be suitable for people who are new to puzzle solving.

### Clarification: Both experienced solvers and new solvers are welcome to post solutions to this puzzle.

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.

• I have been able to entertain several non puzzle minded friends with this puzzle and potential more general incarnations and questions such as: where do these numbers come from in the first place and what about another regular polygon with more sides than 3. In short: this is indeed a suitable introduction to PE. Thanks. Apr 8 at 1:35
• @FirstNameLastName I am happy that this was a suitable puzzle for you. Some of my puzzles are easy and some are tough. Nowadays when I post an easy puzzle, I label it “Beginner puzzle”. Apr 8 at 2:36
• minor nuance : this puzzle is suitable, but, perhaps a bit too easy for me personally, if I may say so, and rather, as you, I modestly believe, intended, certainly suitable for beginners, I use it to convince people they can solve puzzles :-) Apr 9 at 1:55

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.

    4
5   3
6   2   10

with 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

2
10-6
3-4-5

• Your final answer is not valid, having products of 60, 24 and 180. The solution you describe is valid, though the product is not as large as possible. Mar 30 at 11:31
• I’m not sure I understand your reasoning leading to “10 can only be flanked by 2 and 3”. How large is “too large”, and why? Mar 30 at 18:57
• Indeed, I screwed up when I wrote the last element. And I completely missed the “product as large as possible” part. (facepalm) Mar 30 at 20:35