Solution to Problem 21: "Cross Sums Typo"

In Cross Sums puzzles, you're well-off to make (or know) the list of partial sums from 1 to n and from n to 9.   For instance, if you have five digits that equal 35, they must be 56789 in some order.   Also, if you have four digits that equal 11, they must be 1235 in some order.

The overwhelming strategy for this puzzle is to realize, for instance, that a row whose squares cannot exceed 5 intersects a column whose squares must all be at least 5.   Their intersection, of course, should have a 5 filled in.   This schema is assumed throughout the puzzle, almost always without comment.

	The difference between the sum of the blue lines and the sum of the red lines is 4, which must be the value of E9. B11 is 5, so B9 is 6 and B10 is 7.   A9-A10 is 12, so C10 is 1, C11 is 2 and D11-12 is 14.   This means that F12 is 5 and so F13 must be 4. (The remaining entries in these two areas is not important to our solution, but it is fairly easy to work out what they would be.)
	L5-M5 is 13, so K4-M4 is 978 and J3-L3 is 162.   J2 is 4, so L1-L5 is 43271.   If M1 were 9, then K1 would be 6, so M1-M2 is 79 and K1-K4 is 8769.   H2-M2 must be 864729, so H1-I1 is 12. G3-H3 is 89, G4-I4 is 12, and G8-H8 is 97.   J6-J7 is 15, so H7-L7 is 6?5?7.   L10-M10 is 14, so L9-M9 is 28 and K8-M8 is 839.   K6-K8 is 698, I5-I7 is 958, and F5-F6 is 12.
	H13-I13 is 13, so H12-H13 is 21 and L12-L13 is 12. J9-J12 can only be 1273, so I10 is 9, I11-I13 is 879, K13-M13 is 829, and H12-M12 is 2?3516. F10-G10 is 38, so G8-G11 is 9786 and F9-F13 is 23154.   (B is 7!)   E9-J9 is 427351, so I9-I13 is 59843.   (C is 21!!)
	F6-K6 is 234516, so D5-I5 must be 261459.   D4-E4 is 49, E1-F1 is 31, and F1-F2 is 12.   A1-A2 is 96, B1-B2 is 71, so D2-D5 is 3142.   If E2 were 5, then E3 would be 1, so E1-E5 is 34296 and C1-C3 is 854.  (A is 7!!!!)
	The solution is 7 7 21.

Comments: I started on this one in the contest, but didn't get very far.   It's a little surprising to me, because I've been solving Cross Sums since I was about eight years old and this particular example isn't all that difficult.   I think that if I had combined all the efforts from the half-done problems into one or two of them, I'd have another 30 points to show for my efforts.   I suppose the challenge is knowing which ones will bear fruit after ten minutes of work.