15 - Solution

5 Responses to “ 15 - Solution ”

1. wbeard
   January 2nd, 2008 | 9:28 pm

   This is one solution to the problem. How many other solutions are there?

   There are eight solutions all together, how?

2. Ganesh Patil
   April 2nd, 2008 | 4:03 am

   You can transpose the matrix and get the other solutions.
   
Alternate Soln - 1:
6 1 8
7 5 3
2 9 4

   Alternate Soln - 2:
8 1 6
3 5 7
4 9 2

   etc.

3. josh
   May 8th, 2009 | 10:00 pm

   It’s easy once you find the center number, which can be found by process of elimination.
   4 and 7 can’t be center because 4 + 7 + 4 = 15 and each number can only be used once. The same applies to 6 and 3, as well as 3 and 9. 8 is to large to be center, and 1 and 2 are to small. 5 is your only choice, then pair up the numbers so their sums equal 10 and place them across from each other. It won’t take long to figure out the placement because we already know 6 and 3 can’t be parallel, neither can 9 and 3 or 7 and 4.

4. franknstein
   July 12th, 2009 | 3:16 am

   I have a method to find the center number.Let center number be y. Now let the sum parallel or diagonal , which is 15, be x .Now adding all the numbers we get 45.We get the equation
   4x-3y=45. Substitute x and get y =5.

5. Derek
   October 24th, 2009 | 7:19 am

   I figured this one out by starting with the largest usable number(9) and figuring out how many ways it adds up to 15. It has two ways, so there’s no way it can be in the center(needs to be able to add up to 15 eight different ways) or a corner(three different ways). It has to be in one of those spots either directly above/below or to the left/right of the center.

   After placing the 9, I put 1, 5, 2, 4 in the column and row of 9 and guessed from there to the solution given in the post.

   Fun.

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