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

Home > Just Math
Zero mod 3 (Posted on 2015-11-04) Difficulty: 2 of 5
After reviewing Charlie’s set of 9 generic solution to my Magic matrix I have found a common feature to all solutions:
The central cell of the matrix is always a multiple of 3.

Show, why it is a must.

No Solution Yet Submitted by Ady TZIDON    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
No Subject Comment 1 of 1
From memory, the earlier problem asked for the number of arrangements of digits 1-9 in 3x3 grid such that each row, column, and diagonal is divisible by 9.

Add up the row, column, and both diagonals passing through the central square.  That total will be divisible by 9 and counts every digit once and the central digit an additional three times, or (sum of all) + (3 * central cell) is divisible by 9.  Since (sum of all)=45 which is divisible by 9, (3 * central cell) will be divisible by 9 as well, requiring central cell=3,6, or 9.

  Posted by xdog on 2015-11-04 12:24:34
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 (8)
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