7x7 KenKen*
2 comments | Leave a comment
The Art of Puzzles - 3/13/09 - Bracelet
![[info]](http://l-stat.livejournal.com/img/userinfo.gif){kind=small}
ericberlin on March 14th, 2009 02:15 am (UTC)What, no comments? Okay -- I really liked this one.
OK, since nobody else did it...
We have a free 7 in the center. Yay.
The 1- can only be a 7 at the bend with 1,2 in some order on each arm. The 1/ needs a 1 on each arm, with either 4 or 6 at the end and 2,2 or 2,3 in the remaining cells.
In the first row, 7 cannot be part of either 60x so it is in the 1- (7-6 in some order). Likewise in the first column 7 cannot be part of 60x or 72x, so one of the 2- regions in the lower left must be 7-5.
18+ needs at least one 7, and with one placed in one of the 2- regions, it can only have one, so it is 5,6,7 in some order. But the 5 and 7 must go in the same row, since one row has both 5 and 7 in the 2-, and this allows both these regions to be filled.
Now 18x cannot have a 4, so the 4 in row 6 must go in the 1/, at the bend, and it has 1,2 on each arm, leaving a 3 in the 18x. That 18x can only be 1x1x2x3x3, so 2 goes in the corner, with 1,3 on each arm; the forced 1s in the 1/ mean the 18x can be filled, as can the rest of the 1/, and also the 1-.
The last two columns must sum to 56. The numbers already placed add to 16; the 1- adds to 13, and the 17+ adds to 17, making 46; the 2- thus adds to 10 and must be 6-4. Then the 17+ must be 5+7+5 and everything in these columns can be placed.
Now the 3-cell 60x cannot contain a 6, so it must be 3x4x5 in some order. Now we can take the sum of the entire middle three columns: 32 already placed, 12 in the 60x, and 38 in the two central sums add up to 82. In add they add to 84, so the right-most digit of the 5-cell 60x is a 2.
In column 2, the unknown cells (first, fifth, last) contain 3,4,6 in some order. If the 3 went into the 72x, the other two cells of the 72x would have to be 4 and 6. But the 2- at the bottom has 4 and 6 in some order so 72x cannot have 4 and 6 in the same column. The 3 cannot go into the 2-, so it must go at the top. Now both 60xs can be filled.
Sums on the 3rd and 5th columns determine which numbers must go into the middle column, and then the rest can be filled.
1324576
5712364
2143657
6237415
3456721
7561243
4675132
We have a free 7 in the center. Yay.
The 1- can only be a 7 at the bend with 1,2 in some order on each arm. The 1/ needs a 1 on each arm, with either 4 or 6 at the end and 2,2 or 2,3 in the remaining cells.
In the first row, 7 cannot be part of either 60x so it is in the 1- (7-6 in some order). Likewise in the first column 7 cannot be part of 60x or 72x, so one of the 2- regions in the lower left must be 7-5.
18+ needs at least one 7, and with one placed in one of the 2- regions, it can only have one, so it is 5,6,7 in some order. But the 5 and 7 must go in the same row, since one row has both 5 and 7 in the 2-, and this allows both these regions to be filled.
Now 18x cannot have a 4, so the 4 in row 6 must go in the 1/, at the bend, and it has 1,2 on each arm, leaving a 3 in the 18x. That 18x can only be 1x1x2x3x3, so 2 goes in the corner, with 1,3 on each arm; the forced 1s in the 1/ mean the 18x can be filled, as can the rest of the 1/, and also the 1-.
The last two columns must sum to 56. The numbers already placed add to 16; the 1- adds to 13, and the 17+ adds to 17, making 46; the 2- thus adds to 10 and must be 6-4. Then the 17+ must be 5+7+5 and everything in these columns can be placed.
Now the 3-cell 60x cannot contain a 6, so it must be 3x4x5 in some order. Now we can take the sum of the entire middle three columns: 32 already placed, 12 in the 60x, and 38 in the two central sums add up to 82. In add they add to 84, so the right-most digit of the 5-cell 60x is a 2.
In column 2, the unknown cells (first, fifth, last) contain 3,4,6 in some order. If the 3 went into the 72x, the other two cells of the 72x would have to be 4 and 6. But the 2- at the bottom has 4 and 6 in some order so 72x cannot have 4 and 6 in the same column. The 3 cannot go into the 2-, so it must go at the top. Now both 60xs can be filled.
Sums on the 3rd and 5th columns determine which numbers must go into the middle column, and then the rest can be filled.
1324576
5712364
2143657
6237415
3456721
7561243
4675132