# 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 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 12 20 40 and their positions on the corners and sides of a pentagon, and, the common product 240. For a square, however, no distinct positive natural numbers exist that satisfy the common product rule for the sides.

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