Susan was about to have her puzzle published as New Scientist's Enigma for the week when she discovered that one of the numbers she had pre-filled in was smudged beyond recognition. It looked like this:
"Fill in each of the empty spaces with a non-zero digit so that the rows, columns and diagonals add up to the totals shown:
| | | | | 20 |
| 1 | | | | 15 |
| | | | 3 | 8 |
| | 7 | | | 20 |
| | | X | | 29 |
| 20 | 22 | 15 | 15 | 9 |
please send the completed grid."
where the numbers at the bottom represent the sums of the columns in the 4x4 grid; those at the right represent the totals of the rows; the 9 at the lower right is the sum of the main diagonal and the 20 at the upper right represents the total of the diagonal going from lower left to upper right.
The x represents the smudged digit.
The puzzle Susan had in mind had only one solution. Can you find that solution?
(In reply to
re: Solution (by hand) by Dej Mar)
Consider the following grid:
1abc
def3
g7hi
jkXl
d+e+f=5. So d,e,f must be in [1,3].
d+g+j must be in [16,19]. So g,j must be in [7,9].
e+h+l=8. So h,l must be in [1,6].
Since l is max 6, j+k+X is min 23. So k,X must be greater than 4.
There exists at least 2 solutions for X = 6,7,8,9 (see below).
Since Susan's puzzle had only one solution, X=5.
1473, 2213, 8714, 9965.
1662, 2123, 8714, 9866.
1752, 2123, 8714, 9776.
1464, 2213, 9713, 8975.
1473, 2123, 9713, 8786.
1842, 2123, 8714, 9686.
1644, 2213, 9713, 8795.
1743, 2213, 8714, 9695.
More solutions exist, but this should be enough to prove X=5.
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Posted by Guest
on 2007-06-05 16:42:33 |