[Guest]

Find a 3x3 magic square that is constituted by distinct positive proper fractions with no denominator greater than 9.

[Wilson]

1/8 3/4 5/8

9/9 1/2 0/9

3/8 1/4 7/8

But using 9/9 and 0/9 breaks the rules. Right?

[superprismatic]

Find a 3x3 magic square that is constituted by distinct positive proper fractions with no denominator greater than 9.

When you say "distinct", do you mean by value or form? So, should I consider 6/9, 4/6, and 2/3 as distinct?

[Guest]

When you say "distinct", do you mean by value or form? So, should I consider 6/9, 4/6, and 2/3 as distinct?

The word "distinct" in terms of the original problem text implies that all the nine positive proper fractions must be distinct in terms of both value and form.

Thus, 6/9, 4/6 and 2/3 are deemed the same fraction in terms of the OP.

[Guest]

I have been working on this for some time now and I come up with

The magic number that they all add to is 3/2


1/6  7/9  5/9

8/9  1/2  1/9

4/9  2/9  5/6

Please let me know how if I did get it right.

Edited January 30, 2010 by stewdeker

[plainglazed]

Yep, stewdeker, nice. Really enjoyed this puzzle K S. Very simple in form and solution yet quite deceptively challenging getting there.

[Guest]

Good job stewdeker.

[superprismatic]

In case anyone is interested, stewdeker's solution is the only solution (up to symmetries of the square). I wrote a little program to check this.

[Guest]

In case anyone is interested, stewdeker's solution is the only solution (up to symmetries of the square). I wrote a little program to check this.

Then that's even more impressive. The pool of numbers is a lot bigger than a magic square with integers. How'd you do it stewdeker?

And good job with the program superprismatic. Did you say you were using Fortran?

[Guest]

I tried many solutions that failed for a bit but then I found the right one.

First I looked at a normal magic square that added to 15. Then I noticed that the middle square was 5 with the number pairs that added to 10 on either side of it. So with the fractions I made a list of the 27 proper fractions and took the middle one which was 1/2 and put it in the middle box. In the normal square the magic number was 3 times that of the middle square. So I used 3/2 as my magic number. I listed the proper fractions with their pair that would add to 1. Then I wrote a short and simple program to try them all until it found one. Once I got the idea and listed out all the proper fractions that fit the OP it took a few minutes to write the code and my pc less than a min to find the first one. The others that it listed were the same, just rotated, so I posted the one it listed first.

As I stated I tried several other ways including pure brute force but the number of permutations was too large to quickly find a solution. So I kept refining the possible permutations until I basically got lucky on a hunch.

[Guest]

... So with the fractions I made a list of the 27 proper fractions and took the middle one which was 1/2 and put it in the middle box....

Just curious -- was that magic number (1/2) the mean or the median of the 27 proper fractions? Hmmm, on 2nd thought, I guess those would be one and the same with that set. Good idea.

Edited February 2, 2010 by xamdam

[superprismatic]

Then that's even more impressive. The pool of numbers is a lot bigger than a magic square with integers. How'd you do it stewdeker?

And good job with the program superprismatic. Did you say you were using Fortran?

Yes, Fortran -- it's fast with no memory leaks. I've tried others like Python, Ruby, Haskell, and C but they all have what I consider fatal problems. For example, there is no attempt to insure that programs written and running well today will still run fine 10 years from now. When word got out that Python would be dropping its lambda calculus capability, I dropped it like a hot potato because most of my Python code used it.

[Guest]

1/5 2/5 3/5

2/5 2/5 1/5

3/5 1/5 1/5

All equal to 5/5 right?

Try....

[Guest]

1/5 2/5 3/5

2/5 2/5 1/5

3/5 1/5 1/5

All equal to 5/5 right?

Try....

Doesn't meet the conditions of the OP -- all must be distinct