All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Holes in a magic square (Posted on 2008-07-26) Difficulty: 2 of 5
We know that using the numbers from 1 to 25 (once each), we can build a magic square of order 5, being 65 the magic constant.

Your task is to build a magic square of order 5, using only the numbers from 1 to 20 (once each), leaving one cell empty in each row, in each column, in each main diagonal.

Obviously, the magic constant will be [(1 + 20)/2]*20/5 = 42.

Note: This type of magic square has a name.

See The Solution Submitted by pcbouhid    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Another observation | Comment 5 of 8 |
With rotations/reflections aside, there are four basic possible arrangements for the holes:

x _ x x x
_ x x x x
x x _ x x
x x x x _
x x x _ x

x _ x x x
x x x x _
x x _ x x
_ x x x x
x x x _ x

_ x x x x
x x _ x x
x x x x _
x _ x x x
x x x _ x

_ x x x x
x x x x _
x x x _ x
x _ x x x
x x _ x x


brianjn has posted a solution for the third grid. I am curious as to how many of the four arrangements have solutions.
  Posted by Dej Mar on 2008-07-28 23:00:28
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (12)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information