Comments on: Numbers in a Square

By: Chris

Chris — Sun, 21 Oct 2007 14:11:17 +0000

Thanks

By: Laurentiu Craciunas

Laurentiu Craciunas — Sun, 21 Oct 2007 11:09:32 +0000

Chris, some of the comments awaite moderation and when they are approved they are listed cronologicaly even if other comments appeared public before them.

By: Chris

Chris — Sat, 20 Oct 2007 22:00:11 +0000

Thanks for the logic/mathmetical proof Jed. I have one question for GeniusBeauty. How can people post responses which appear to be prior to posts that have already been made? In future I will annotate my posts with Zulu time.

By: Jed Hawk

Jed Hawk — Fri, 19 Oct 2007 21:26:29 +0000

It’s Magic! (But let’s see if there’s a way to prove them rather than rely on past experience)
The total of all numbers is 45, therefore every row column and diagonal must total 15.

We can conclude that the centre number must be 5.

Why?

If the middle number was 1, we need 4 combinations of 14 from two different numbers(not 1).
But there’s only 9+5, 8+6.
If the middle number was 2, we need 4 combinations of 13 from two different numbers(not 2).
But there’s only 9+4, 8+5, 7+6.
If the middle number was 3, we need 4 combinations of 12 from two different numbers(not 3).
But there’s only 8+4, 7+5.
If the middle number was 4, we need 4 combinations of 11 from two different numbers(not 4).
But there’s only 9+2, 8+3, 6+5.
If the middle number was 6, we need 4 combinations of 9 from two different numbers(not 6).
But there’s only 8+1, 7+2, 5+4.
If the middle number was 7, we need 4 combinations of 8 from two different numbers(not 7).
But there’s only 6+2, 5+3.
If the middle number was 8, we need 4 combinations of 7 from two different numbers(not 8).
But there’s only 6+1, 5+2, 4+3.
If the middle number was 9, we need 4 combinations of 6 from two different numbers(not 9).
But there’s only 5+1, 4+2.

Further, if we stick a number in a corner, we require at least 3 combinations making a total of 15.
From the above, 5 is central, 2,4,6,8 go in the corners SOMEHOW but we must control the placement. Then 1, 3, 7 and 9 go at the sides.
Further, 1 is opposite 9. 2 is opposite 8. 3 is opposite 7. 4 is opposite 6. (using the centre as the pivot)
If we place 9 at any one side, then 8 cannot go in an adjacent corner. It must go in one of the other two. This brings rise to two unique solutions (any others are simply rotations of existing solutions) as Chris has found.

By: Laurentiu Craciunas

Laurentiu Craciunas — Fri, 19 Oct 2007 20:50:52 +0000

8 1 6
3 5 7
4 9 2

This one is easy

sum: 9 x 10 / 2 = 45 has to be dividen to 3 so on each line sum = 15
9 can’t be on the same line with 3, 6, 7, 8… from there you can do it too

By: hj

hj — Fri, 19 Oct 2007 19:28:18 +0000

took 5 minutes at work
9,1,5
3,4,8
6,7,2

By: Dwight

Dwight — Fri, 19 Oct 2007 17:53:39 +0000

Chris, ignoring rotation and reflection there’s only one
3×3 magic square.

By: steve

steve — Fri, 19 Oct 2007 17:50:43 +0000

9 1 5
6 2 7
8 3 4

By: Chris

Chris — Fri, 19 Oct 2007 11:45:35 +0000

However, I have found two alternate grids which will work – can you?

By: Chris

Chris — Fri, 19 Oct 2007 11:31:49 +0000

Here’s the grid:
2 7 6
9 5 1
4 3 8

Best regards to all,
Chris

By: stargoat

stargoat — Fri, 19 Oct 2007 04:11:39 +0000

4 9 2
3 5 7
8 1 6

every row = 15
every column = 15
both diagonals = 15

By: Rowell

Rowell — Thu, 18 Oct 2007 21:21:33 +0000

hmmm….. I think i’ll wait for your next blog for the answer to this brain game…. only because Im at work at the moment and I left my brains at home )