Unless this was done on purpose (to be a trick), I think the word "squares" in clues 2 down and 6 down should be changed to "cells." In other clues (2 across and 3 down) the word "cell" is used. So when I saw "squares" I assumed that meant "perfect square numbers like 1,4,9,16…" This completely throws off a person’s ability to solve this puzzle.
Anyways, once you get past that bit of confusion, here is how you can reason it out. Starting with the across clues…
1A: Since no duplicates are mentioned, there are none in this row. Therefore the only group of numbers that can add to 22 are 1,2,3,4,5,7.
4A: Since no duplicates are mentioned, there are none in this row. Since there are exactly four even numbers available, we have 7,9,2,4,6,8.
5A: Since no duplicates are mentioned, there are none in this row. That leaves only 1,4,5,6,7,8.
3A: We know we have two 5s and two 1s and no other duplicates. 275511=15. So we have 5,5,1,1, (9,6 or 8,7) so far.
Check point, we have used all four 1s and all four 5s so far.
2A: Since no duplicates are mentioned, there are none in this row. Since there are only five odd numbers, and 1 and 5 are used up, the three odd numbers must be 3,7,9.
Check point, we have used all four 7s in addition to all the 1s and 5s.
3A: Since the 7s are used up, the remaining sum of 15 that we need can’t come from 8+7. Therefore we have 5,5,1,1,9,6.
6A: We know we have two 3s and no other duplicates. So we have 3,8,3 so far.
Check point, here are all the numbers that are REMAINING. 2,2,4,6,8,9.
6A: Since there are two 2s left, and only two unfilled rows left, 2A and 6A, neither of which can have any duplicates (other than the two 3s mentioned in 6A), there must be one 2 in each. 313832=15. Similar to the reasoning used for 3A, since the 7s are used up, the remaining sum of 15 that we need can’t come from 8+7. Therefore we have 3,8,3,2,6,9.
2A: Which leaves 2,4,8,3,7,9 here.
Moving on to the down clues…
1D: Since no duplicates are mentioned, there are none in this column. That leaves only 1,3,4,5,8,9.
2D: We know we have two 7s and no other duplicates. Since there are exactly four even numbers available, we have 7,7,2,4,6,8.
4D: Since no duplicates are mentioned, there are none in this column. Therefore the only group of numbers that can add to 21 are 1,2,3,4,5,6.
5D: Since no duplicates are mentioned, there are none in this column. That leaves only 1,2,5,6,8,9.
6D: We know we have two 9s and no other duplicates.
Check point, here are all the numbers that are REMAINING. 1,2,3,3,4,5,6,7,7,8.
6D: Since there are two 3s and two 7s left, and only two unfilled columns left, 3D and 6D, neither of which can have any duplicates (other than the two 9s mentioned in 6D), there must be one 3 and one 7 in each. 399937=11. Now that all the 3s and 7s are used up (along with all the 9s still), the only way to get the remaining sum of 11 is with 5 and 6. Therefore we have 9,9,3,7,5,6.
3D: Which leaves 1,2,3,4,7,8 here.
Ok. Now that we know what numbers go in each row and column, we can start placing them.
6D tells us that the two 9s in that column are separated by two cells, so we either have 9s in 1A and 4A, 2A and 5A, or 3A and 6A. Well, there are no 9s in 1A, nor in 5A. Therefore a 9 goes in 3A6D and 6A6D.
1D and 6A only have two numbers in common (now that the 9 in 6A has been placed), and they are 3 and 8. Notice that 6A tells us that the 8 is enclosed by the two 3s. So the 8 cannot be in 1D since it has to have a 3 on either side of it. Therefore a 3 goes in 6A1D.
I am assuming that "Two 3s enclosing an 8" means "directly enclosing" such that 3,2,8,6,3,9 would not count as "Two 3s enclosing an 8." Therefore we know an 8 goes in 6A2D and a 3 goes in 6A3D.
2D tell us that the two 7s in that column are separated by two cells, so we either have 7s in 1A and 4A, 2A and 5A, or 3A and 6A. Well, there are no 7s in 6A (besides, 6A2D is taken by an 8). And 2A tells us that the numbers in the first three cells are even, so a 7 can’t be placed at 2A2D. Therefore a 7 goes in 1A2D and 4A2D.
2D also tells us that, other than the two 7s, the other four numbers are even. There is only one even number in 3A. Therefore6 goes in 3A2D. The two remaining numbers in 2D are 2 and 4, with the two remaining cells in 2A and 5A. 5A has no 2s. Therefore a 2 goes in 2A2D and a 4 goes in 5A2D.
3A tells us that one of the 5s is enclosed by two 1s. Again, going with my "directly enclosing" assumption, there is only one place where 151 could go since 3A2D and 3A6D are taken, and the remaining 5 will go in the leftover cell. Therefore a 1 goes in 3A3D and 3A5D, and a 5 goes in 3A1D and 3A4D.
3D tells us that the lower three cells add to 14. We already know one of the lower three cells is a 3 and the remaining numbers in that column are 1,2,4,7,8. The only way to get the remaining sum of 11 with these numbers is with 7 and 4. 4A only has one 7 in it, which has already been placed in 4A2D (and 5A only has one 4 in it, which has already been placed in 5A2D). Therefore a 7 goes in 5A3D and a 4 goes in 4A3D. The two remaining numbers in 3D are 2 and 8, with the two remaining cells in 1A and 2A. 1A has no 8s. Therefore an 8 goes in 2A3D and a 2 goes in 1A3D.
2A tells us that the numbers in the first three cells are even, and two of the three even numbers have been place. Therefore a 4 goes in 2A1D.
In 1D the three remaining numbers are 1,8,9 with the three remaining cells in 1A, 4A and 5A. 1A and 5A have no 9s. Therefore a 9 goes in 4A1D. 1A has no 8s. Therefore an 8 goes in 5A1D and a 1 goes in 1A1D.
In 2A the three remaining numbers are 3,7,9 with the three remaining cells in 4D, 5D and 6D. 4D has no 9s and 6D has all of its 9 already in place. Therefore a 9 goes in 2A5D. 4D has no 7s. Therefore a 7 goes in 2A6D and a 3 goes in 2A4D.
In 1A the three remaining numbers are 3,4,5 with the three remaining cells in 4D, 5D and 6D. 5D and 6D have no 4s. Therefore a 4 goes in 1A4D. 5D has no 3s. Therefore a 3 goes in 1A6D and a 5 goes in 1A5D.
In 4A the three remaining numbers are 2,6,8 with the three remaining cells in 4D, 5D and 6D. 4D and 6D have no 8s. Therefore an 8 goes in 4A5D. 6D has no 2s. Therefore a 2 goes in 4A4D and a 6 goes in 4A6D.
In 5A the three remaining numbers are 1,5,6 with the three remaining cells in 4D, 5D and 6D. 5D has its 1 already in place and 6D has no 1s. Therefore a 1 goes in 5A4D. 5D has its 5 already in place. Therefore a 5 goes in 5A6D and a 6 goes in 5A5D.
In 6A the two remaining numbers are 2 and 6 with the two remaining cells in 4D and 6D. 4D has its 2 already in place, and 5D already has its 6 in place. Therefore a 6 goes in 6A4D and a 2 goes in 6A5D.
The end!
P.S. The only bit of info I didn’t need to use was in 6A: No 7s. I had already determined that all four 7s were used in other rows before I started to consider row 6A.

Posted by nikki
on 20061016 16:08:28 