Elisabeth, yes, it is possible not to use any 16, but it took me a while to figure it out. 1) -> the 9-boxes in the corners can not have 1 or 2. -> in the left and right column 1 and 2 must be green/pink, so you can erase 1 and 2 in the second columns (maybe not needed) 2) Now look at the bottom row and/or the top row (I prefer the bottom row - do the same symmetrically in the top row). The three squares next to 11 have to sum up to 10. That only leaves 3 or 4 for the corner squares in the 9 boxes, because otherwise 12 can't be reached 3) Neither the 11- nor the 12-box (with gray squares in bottom row) can have both 1 and 2. That's why the 2x2 square or the 5x5 square can not have 1 or 2 because the 12 would eat up too many high numbers (6 and (4 or 5)) and you can't add up to 11 using the 1. These 2x2 and 5x5 squares can not be 3 or 4, because they have the same color as the corner sqares and it wouldn't sum up to 10 again. 4) with 5/6 taken in blue (and symmetrical) orange areas you can now can proceed.
If you still get stuck, it might help to eliminate high numbers in the 10 box in the pink area (now you might need the deleted 1/2 in the pink area, second column from the right).
Posted 14th Jul 2019 at 15:47
Elisabeth Daily subscriber Rated puzzle: Moderate Completion time: 12:40 Used 'check puzzle' when incorrect
Thanks, JoergWausW, I thought you might respond! We had a conversation about another 6x6 puzzle some little time ago. Earlier I had obtained the 34 in the corners and the two sets of 56 and your suggestions set me off acually on a different path which I discovered eliminated the 5 from each 16 automatically. Getting rid of 5 in both 16s is probably the quickest way of solving though. Happy solving:)
Posted 14th Jul 2019 at 17:08
JoergWausW Daily subscriber Completion time: 2:25
Getting rid of 5 in both 16s is probably the quickest way of solving though. <-- agreed. I used it, too.
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yes, it is possible not to use any 16, but it took me a while to figure it out.
1)
-> the 9-boxes in the corners can not have 1 or 2.
-> in the left and right column 1 and 2 must be green/pink, so you can erase 1 and 2 in the second columns (maybe not needed)
2)
Now look at the bottom row and/or the top row (I prefer the bottom row - do the same symmetrically in the top row).
The three squares next to 11 have to sum up to 10. That only leaves 3 or 4 for the corner squares in the 9 boxes, because otherwise 12 can't be reached
3)
Neither the 11- nor the 12-box (with gray squares in bottom row) can have both 1 and 2.
That's why the 2x2 square or the 5x5 square can not have 1 or 2 because the 12 would eat up too many high numbers (6 and (4 or 5)) and you can't add up to 11 using the 1.
These 2x2 and 5x5 squares can not be 3 or 4, because they have the same color as the corner sqares and it wouldn't sum up to 10 again.
4) with 5/6 taken in blue (and symmetrical) orange areas you can now can proceed.
If you still get stuck, it might help to eliminate high numbers in the 10 box in the pink area (now you might need the deleted 1/2 in the pink area, second column from the right).
Earlier I had obtained the 34 in the corners and the two sets of 56 and your suggestions set me off acually on a different path which I discovered eliminated the 5 from each 16 automatically. Getting rid of 5 in both 16s is probably the quickest way of solving though.
Happy solving:)
You can however view other players' statistics and comments in the tables above.