Complex innies/outies – part two

I saw two comments (by Graham and David) for the puzzle posted on October 25, and I think there’s a little clarification on “complext innies/outies” that should be made here. You should look for innies/outies in more than just one row/column/nonet. This is a good example of such puzzle. See this image:

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To see the solution to this puzzle click here
The cell outlined in red is the only cell sticking out of all cages that belong to the first 5 columns. So it’s value must be: (Sum of All Cages in First 5 Columns) – 5*45. In this case it equates to: (4+9+15+28+10+17+21+12+14+7+24+11+12+21+14+13)-5*45 = 232 – 225 = 7. Of course you could’ve done it the other way – as an innie of the last 4 columns. And remember – whenever you find an innie/outie in a symetrical puzzle, there’s always one more – exactly opposite to the one you just found. I hope you see which one I’m talking about. It’s a coincidence that in this puzzle the value of the opposite innie/outie is identical (it’s also 7). Once you solve those two and solve the two cells in the upper left corner, there’s nothing fancy about yesterday’s puzzle :).