Challenging Christmas Puzzle

[Platypus]

In my Christmas stocking, Santa left me this wooden puzzle:



The instructions are:
Can you make every row add up to 15?
Every horizontal and vertical row, and the
two corner to corner diagonal rows must
equal 15.

After some noodling and re-arranging of the 25 tiles, I found a couple of solutions, neither the same as the one solution given in the box.

So I wrote a noddy program to try and find additional solutions, and more were revealed.
Upon reflection, I feel there may be even more solutions

The challenge for the programmers and mathematicians among you then is this:
Can you solve it?
More importantly, how many solutions can you find?

I look forward to seeing your answers and comparing them with my own.

[scooterscot]

Instantly i see 8 solutions.

Take the 2nd horizontal row and move it one cube to the right, the 3rd row two cubes to the right and so on until you reach the bottom.

Do the same on the vertical axis.

oh I just realised my solution does not solve for the diagonal.

[VectraMan]

Instantly i see 8 solutions.

Take the 2nd horizontal row and move it one cube to the right, the 3rd row two cubes to the right and so on until you reach the bottom.

Do the same on the vertical axis.

Just what I was going to say: 12345, 23451, 34521 etc. Do that moving left and right, then vertically, then you can turn each upside down, and you have eight.

And it does work diagonally. TL to BR, you'd have 13524 = 15.

[OwlHoot]

The instructions are:
Can you make every row add up to 15?
Every horizontal and vertical row, and the
two corner to corner diagonal rows must
equal 15.

Well since 1 + 2 + 3 + 4 + 5 = 15 each row, column, and diagonal must contain each of 1, 2, 3, 4, 5.

So the number of solutions must be a multiple of 5, because for each solution you can replace every occurrence of 1, 2, 3, 4, 5 by 2, 3, 4, 5, 1 respectively and preserve all the sums.

As you can also flip solutions about either diagonal, and solutions cannot be symmetric about either diagonal, I'd guess there are 15 solutions in total.

[Platypus]

Well since 1 + 2 + 3 + 4 + 5 = 15 each row, column, and diagonal must contain each of 1, 2, 3, 4, 5.

You'd think so, but is this not a solution, according to the rules?

5 4 3 2 1
4 2 1 3 5
2 1 5 4 3
3 5 2 1 4
1 3 4 5 2

However, the diagonals don't contain one each of 1,2,3,4,5 but they do add up 15 !

(correct me if I'm being stupid)

[TimberWolf]

Here's a hard one:

12
12

All horizontal, vertical and main diagonals must sum to 3, as per rules before*








* Don't spend too much time on this

[OwlHoot]

You'd think so, but is this not a solution, according to the rules?

5 4 3 2 1
4 2 1 3 5
2 1 5 4 3
3 5 2 1 4
1 3 4 5 2

However, the diagonals don't contain one each of 1,2,3,4,5 but they do add up 15 !

(correct me if I'm being stupid)

Flip, yes, you're right - It's me being stupid.

So the diagonals needn't contain 1 - 5, and quite possibly neither do the rows and columns (although obviously if one uses too many large numbers in some rows there's a tendency to run out of large enough numbers to make up the required sum in the others).

[Freamon]

The real challenge would be to write a mathematical proof that demonstrates there are exactly n solutions.

[Platypus]

The real challenge would be to write a mathematical proof that demonstrates there are exactly n solutions.

Yep, but I'm not smart enough to do that. Where's SAS ?

[Platypus]

Flip, yes, you're right - It's me being stupid.

Not at all, that was my intuition also, and I've found 16 solutions which obey that rule. But I've also discovered some which don't obey that rule, which is what piqued my interest and prompted me to create this thread!

So the diagonals needn't contain 1 - 5, and quite possibly neither do the rows and columns (although obviously if one uses too many large numbers in some rows there's a tendency to run out of large enough numbers to make up the required sum in the others).

That's exactly what I'm wondering now !