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17 March 2009 @ 08:58 am
3/17/09 - Bowtie  
5x5 KenKen*

( 4 comments — Leave a comment )
devjoedevjoe on March 17th, 2009 04:28 pm (UTC)
Lovely large-scale logic in this one.

The 96 must be 1x2x3x4x4. That means that the first row of the 12 and first column of the 15 both contain a 5 and one of 1, 2, or 3. The 15 could be 1x1x3x5 or a sum, but a 5 in the 12 means it must be a sum.

The 24 could possibly be 1x1x1x2x2x2x3 (which fits in just two ways, both with 3 in the center) or various sums. But fitting the product in the center blocks both 1 and 2 from 4 cells of the 96, making it impossible to fill. So it is a sum.

The 1 can only be 4/2/2/1/1 with 4 at the bend. Now the sum of all the digits not in the 15 is 14 + 12 + 24 + 10 = 60, so the 15 must be a sum. This also puts a 3 and 5 into the last column of the 12 and the last row of the 15. The 12 cannot contain two 5s and a 3, so it has one in the upper right corner. The last two cells of the 12 must add to 4, and 3+1 is the only way to do that without a conflict. This forces the 3 in the 96 to the 3rd row, and forces the 15 to have a 5 in the lower left corner; now each other cell must be at its maximum (2+5+5+3) to reach a sum of 15.

There is now only one way to place the remaining 5s, and the rest fills in by elimination.

thedan on March 17th, 2009 07:01 pm (UTC)

You can show much more easily that the 15 is a sum; since the 1 has to be 4/2/2/1/1, there's a 3/5 in one row of the 15, so if it were a product there would be two 1's in the same row. I don't think I actually used the 24 clue to solve.
ericberlin on March 17th, 2009 08:37 pm (UTC)
This is the path I took.
motrismotris on March 17th, 2009 08:39 pm (UTC)
... and the "Tuesday" one I intended, but I don't mind the detail devjoe gives.
( 4 comments — Leave a comment )