04 Feb '03 22:58

Ok, first up, I don't want to see the solution to this, and if you write a program to crack it (not too tough) that's fine, but please don't post it here, as I'm still working on it.

It goes a little something like this. You have the integers 1-15 arranged in a triangle. Imagine a pool triangle with all the numbered balls inverted, so that the row of 5 is at the top and not the bottom. Can you arrange the numbers in such a way that each number between and below two numbers is the subtraction of those numbers. A visual example would be:

15 14 12 9 13

1 2 3 4

As you can see 15-14=1, 14-12=2, 12-9=3 and so on, until you get to the bottom of the pyramid, where it might end:

11 5

6

In the above case, the system would not work, because then 2-1 would have to equal 1, and the number 1 has already been used. Apparently there is at least one solution that allows all 15 integers to be used (only once) that satisfies the above constraints...

If any of this is not clear, and looking back on what I write it surely can't be that clear, let me know and I'll try to explain in more detail...

good luck!

joe

It goes a little something like this. You have the integers 1-15 arranged in a triangle. Imagine a pool triangle with all the numbered balls inverted, so that the row of 5 is at the top and not the bottom. Can you arrange the numbers in such a way that each number between and below two numbers is the subtraction of those numbers. A visual example would be:

15 14 12 9 13

1 2 3 4

As you can see 15-14=1, 14-12=2, 12-9=3 and so on, until you get to the bottom of the pyramid, where it might end:

11 5

6

In the above case, the system would not work, because then 2-1 would have to equal 1, and the number 1 has already been used. Apparently there is at least one solution that allows all 15 integers to be used (only once) that satisfies the above constraints...

If any of this is not clear, and looking back on what I write it surely can't be that clear, let me know and I'll try to explain in more detail...

good luck!

joe