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**jacquimurphy****Member**- Registered: 2006-10-23
- Posts: 1

my puzzle is this, i have eight houses in a square. nothing in the middle of course. i have 36 people to put into these houses. i can only use each number( 1-8) once and all rows and columns must add up to 15.i can solve the puzzle but iwould like to know if there is a formula that would help if the values were changed. thanks in advance if anyone can help.

regards jacqui

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**pi man****Member**- Registered: 2006-07-06
- Posts: 251

I would like to see your solution. If each column (or each row) must add up to 15 and there are 3 columns, that's 45 people. And you only have 36.

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**pi man****Member**- Registered: 2006-07-06
- Posts: 251

Unless you don't count the row and column which goes through the center square? If that's the case, then there is a solution. Label the houses as follows:

A B C

D E

F G H

The top row (A, B, & C) and the bottom row (F, G & H) add up to 15 each for a total of 30. That means houses D and E must add up to 6. Likewise. houses B plus G have 6 people total. Using each number 1 through 8 once and only once, there are only 2 sets of numbers that add up to 6 ==> 1+5 and 2+4. That leaves 3, 6, 7 and 8 for the corner squares. Let's go ahead and put 5 in spot B. That means A+C = 10 and therefore A and C must be 3 and 7.

3 5 7

D E

F 1 H

Now look at the column with the 3. We know D and E are going to be filled in with 2 and 4. D can't be the 2 since that would make F=9. So D=4, E = 2, F = 8 and H = 6.

3 5 7

4 2

8 1 6.

So there's really only 1 solution. Any other "solution" is just some rotation or mirror image of the above.

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