Fill a 3 by 3 grid with the digits 1 to 9 using the following five rules:

1. The sum of the top row is twice the sum of the center row.

2. The sum of the left column is twice the sum of the center column.

3. The sum of the right column plus twice the sum of the bottom row is equal to the sum of the whole grid.

4. The sum of the bottom row plus twice the sum of the right column is not equal to the sum of the whole grid.

5. The top row is the only row with both odd and even numbers.

Show that there is only one solution.

Let the box look like this:

a b c
d e f
g h i

1. a+b+c=2(d+e+f)
2. a+d+g=2(b+e+h)
3. g+h+2i=a+b+d+e
4. c+f+2i not equal a+b+d+e
5. 2 odds and 1 even at top row

There are 40 possibilities for 2 odds and 1 even. Out of those 40, there are only 3 that would yield a plausible pair for the second row. Not caring about the order, they are [(abc)(def)(ghi)]=[(589)(137)(246)], [(279)(135)(468)], and [(789)(246)(135)], which I'll call case 1, 2, and 3.

Then I found the maximum and minimum sum of the columns for each possibility. For case 1, it's (8,22). For case 2, it's (12,22). For case 3, it's (10,20). From the second criterion, twice the sum must also be in the range. This eliminates case 2, since the minimum is 12 and twice the minimum 24 is not in the range.

For case 3, that means that the center column must add up to 10 and the right column must add up to 20. That would yield

987
624
513

which does not satisfy criterion number 3.

For case 1, the possibilities for the sum of the first and center columns are (16,8), (18,9), (20,10), and (22,11).  It's impossible to get a sum of 9 for the columns. But the remaining 3 gives:

9 5 8
3 1 7
4 2 6

9 5 8
7 3 1
4 2 6

9 8 5
7 1 3
6 2 4

Of these 3, only the first one satisfies criterion 3. And it also satisfies the criterion 4. Good old fashion elimination shows only 1 solution!

Posted by np_rt on 2004-05-16 03:22:48