Find a 3x3 magic square that is composed of 9 prime numbers (not the numbers from 1-9) and show how you found it.
(A magic square, as you may already know, is one in which the respective sums of the numbers in all the rows, columns, and both major diagonals all add up to the same number.)
Since "Magic Square" is a term used outside the scope of this problem, I'm sure you can find an answer on the internet. Please find a solution independently.
2 is out of the question, because it would make some, but not all of the sums even.
I decided that there would have to be at least 4 pairs of primes that add up to the same number so these pairs could be on opposite sides.
I found by brute force what numbers would have many pairs of primes that could sum up to it. After trying it, I eventually came to 90, which has a whopping 10 pairs of primes that will sum to it. That doesn't mean that 90 is the sum for each row, just the sum for of opposite outer primes.
So anyway, expect another comment from me with a solution.
Edited on March 9, 2004, 7:18 pm
Posted by Tristan
on 2004-03-09 19:18:05