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:

(click to download or right-click to save the image!)

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 :).