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Difficult 6x6 July 29 https://www.calcudoku.org/forum/viewtopic.php?f=16&t=814 
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Author:  marblevolcano [ Fri Jul 29, 2016 1:36 am ] 
Post subject:  Difficult 6x6 July 29 
New pattern in the 6x6's... Alright, so this one was pretty difficult, but I somehow managed to get it without guessing. E6 must be 2 because the only possible combination in the 14+ cage is [3,5,6]. This means f6=4. D6 must be 5 because the 16+ cage contains [1,2,3,4,6]. This means that the rest of the 15+ cage contains [1,3,6], and 6 cannot go in c6 because of the 6 in c1. This leaves only 3 and 1. The possible combinations for the 9+ cages are [2,3,4], [1,3,5], and [1,2,6]. Because of the 6 in c1, the ccolumn cage cannot contain [1,2,6], and if it contained [1,3,5] then nothing could go in c6. This means the ccolumn 9+ must contain [2,3,4], leading c6 to have 1 and c5 to hold 5. The combination in the 5th row 9+ cage is [1,3,5] because of the 5 in c5. Because of the 5 in a3 and the fact that there must be a 3 in a5 or a6, the 1 cage at a1a2 must be [1,2], with the 2 in a1 and the 1 in a2. From here, you can find the bottom left foursome, leading a4 to contain 4. From here the rest of the puzzle can be solved without care for cages. The end! 
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