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The Art of Puzzles - 3/17/09 - Bowtie

4 comments —

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.

14235

42513

31452

25341

53124

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.

14235

42513

31452

25341

53124

Spoiler:

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.

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.

This is the path I took.