Find X, then put numbers [2,5,6,8,10,12,18,20,30, and X] to each circle in such a way that the numbers in each line with 4 circles, has the same product Z.

To reduce the results, 2 numbers has been put there as guide.

If we rule out reflection, there are 2 solutions.

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  • $\begingroup$ Two solutions? Then how do you expect us to find them logically? If guessing is involved, then the puzzle is not [grid-deduction]. $\endgroup$
    – Deusovi
    Nov 7 '17 at 9:21
  • $\begingroup$ @Deusovi : Ok I have removed the grid-deduction tag. $\endgroup$ Nov 7 '17 at 9:30
  • $\begingroup$ @Deusovi The two solutions are just due to symmetry. You can still find them using only logic. $\endgroup$
    – w l
    Nov 7 '17 at 9:31
  • $\begingroup$ @wl "If we rule out reflection, there are 2 solutions." $\endgroup$ Nov 7 '17 at 13:17

There are indeed two solutions, assuming $X$ is an integer:


Solve path:

Let's first work out $X$:

The product of all the numbers, squared, is equal to a perfect fifth power, since it is the magic constant to the power of five (each number is counted twice).

So we get:

$n^5=622080000X$, where $n$ is the magic product

But then:

The prime factorisation of $622080000$ is $2^{12}\cdot3^5\cdot5^4$, and so to make $622080000X$ a perfect fifth power $X=2^3\cdot5\cdot m^5=40m^5$, for some integer $m$.
First, $m$ has to be positive, otherwise two of the diagonals would be negative or 0 and the others positive.
Secondly, if $m$ was not of the form $2^a3^b5^c$, two of the diagonals would be divisible by some other prime and the others not.
Thirdly, if $m^5$ was divisible by $5^5$ (or $3^5$), all the diagonals would have to be divisible by at least $5^2$ (or $3^2$), but with $X$ in two diagonals the remaining $5^5$ (or $3^5$) wouldn't be enough to cover the three other diagonals.
Fourthly, if $m^5$ was divisible by $2^{10}$, all the diagonals would have to be divisible by at least $2^4$, but with $X$ out in two diagonals the remaining $2^{10}$ wouldn't be enough to cover the three other diagonals.
Fifthly, if $m=2$, all the diagonals would have to be divisible by $2^3$ and not some higher power, but then $X$ would 'overload' two diagonals.


$m=1$, $X=40$ and $n=14400=2^6\cdot3^2\cdot$.

Now let's consider where the multiples of 3 have to go:

The two diagonals emanating from the 18 already have a factor of 9.
So the two multiples of 3 left, 12 and 30, have to go in the bottom-right and top-left corner.
Let's assume (because we don't care about reflections) that 30 is in the top-left corner. Then 12 is in the bottom-right corner.

Now let's take a brief look at the multiples of 5:

Both the empty cells on the top to bottom-right diagonal are multiples of 5, since neither 6 nor 12 are multiples of 5.

This diagonal is therefore only missing:

A factor of 16. We can split this as (1) 5-40, (2) 10-20, (3) 20-10 or (4)40-5.

The cases are all very similar - cases (1) and (4) lead to solutions, cases (2) and (3) lead to contradictions. I don't have the mental energy to write them all out because they are rather repetitive, but here is case (1) as an example:

Consider the horizontal diagonal (oxymoron not intended). It is also missing a factor of 16.
Our remaining numbers have 2, 2, 4 and 8 as their maximal factors which are a power of two. So this diagonal must have a '2' and an '8'.
Now consider the diagonal running from the top-right corner to the bottom-left corner. This is missing only a factor of 4. So this diagonal must have two '2's.

We can tentatively place:


But then:

The remaining multiples of 5, 20 and 10, must go on separate diagonals. So the 20 must be in the top-right corner and the 10 in the bottom-left interior vertex of the pentagon. Then 8 goes at the top-left interior vertex of the pentagon, and 2 in the last remaining cell.

  • $\begingroup$ When I solved this I made a table of all the numbers and their prime factorization factors which made finding the possible decomposition really easy and might help others better understand your explanation instead of the wall of text. $\endgroup$
    – w l
    Nov 7 '17 at 9:30
  • $\begingroup$ @wl feel free to edit, but I am a bit tired now and just wanted to get my ideas out. $\endgroup$
    – boboquack
    Nov 7 '17 at 9:31

Partial to get the ball rolling

I think that X has this form $2^a \times 3^b \times 5^c$ where $a, b, c$ can be anything starting from 0.
Proof: X can go on 2 or 3 lines.
All the other numbers have the same form. If X had any other prime factor (P) other than 2,3 or 5 then it means that P divides Z. But there are at least 3 lines where Z can be formed only out of numbers that have the prime factors 2, 3 and/or 5.

Working on the rest.


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