Ah, but the size of those cages was fixed (as in constant) before, and you unfixed it. :-) Anyway, as for the solution:

In the last column there are two 2-cell regions with 3- clues. Each of these can only contain 4-1, 5-2, or 6-3. So the last two cells must contain the other pair from this set, with one number that must be the large number in the 2-, so the other cell in the 2- is a 1.

In column 5, the 2- regions can be 4-2, 5-3, or 6-4 (since there is already a 1 there). So one of them must be 5-3, and the other contains a 4.

The number in the 1- at the top must be 2 or 6. If it is 6, it must be with a 5. Then the 3- in row 1 must be 4-1, and the 4- must be 6-2, and the 3- in the bottom row must be 6,3 (since the second digit can only be 3 or 4). But then the two 3- regions in the middle of this column must have the 1 and 4, and also the 1 and 4 of column 3, making it impossible to fill in the 3- in the top row. So the top of column 5 is a 2.

Now there are three possibilities for the three entries in column 4 which are not in a 3-, which forces three possibilities for filling the 3-s.
If the 1- has a 3, the 4- can be 6-2 and the 3-s have 1, 4, 5 in column 4 and 1, 2, 4 in column 3. Then the 1- in column 3 is 6-5 and a 3 is left for the 3- at the top, conflicting with the 3 already in the row. Or the 1- can have a 3 with the 4- being 5-1, so that the 3-s in have 2, 4, 6 in column 4 and 1, 3, 5 in column 3, which doesn't leave any way to fill the 1- in column 3. So the 1- has a 1, and then the 4- is 6-2 and the 3-s have 3, 4, 5 in column 4 and 1, 2, 6 in column 3. The 3- in the top row is 6,3 in this case, and the 1- in column 3 is 5-4.

This tells us that the three 3- regions in columns 3 and 4 collectively use all six digits. But the vertical 3- in column 2 cannot contain either of the pairs from the horizontal 3-s next to it, so it has the same pair as the 3- on the bottom. Then that pair contains neither 6 nor 1, so it is 5-2 (which can be filled into the bottom row).

With 2, 5, and 6 taken in column 2, the 2- can only be 3-1, so we can place a 4 in the bottom cell. The three-cell 1- cannot be 4-2-1 since neither 2 nor 1 can be added to the bottom row, so it is 6-4-1 with 6 on the bottom. Then the other three-cell region can be filled. Elimination fills the rest.

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Well played. There is a shorter solution path which can go like this:

The lone cells (R6C2 and R6C5) in the multi-cell regions can be identified first. In column six, two 3- clues (from 14, 25, or 36)force the remaining two cells in that column in the multi-cell 2- to also be 3-. In each case, the large number is in column 6 and R6C5 is 1. In column one, a 3- and a 1- can leave behind either a 1- (12, 23, 45, 56) in the remaining two cells, which cannot work for the multi-cell cage, or a 5- (16) with (25) and (34) being the 3- and 1-. No other breakdowns are possible. This means the 1- multicell cage is 1 and 6 in column 1 with a 4 in R6C2. Once you have the 4 and the 1 in row 6, the rest of the puzzle will trickle out by friendlier means. Indeed, I used the two multi-cell regions - the new thing I added to the puzzle type - as the two break-in points you are intended to find to get everything else easily.

The puzzle is solvable, but not quite that way. You say:

In column one, a 3- and a 1- can leave behind either a 1- (12, 23, 45, 56) in the remaining two cells, which cannot work for the multi-cell cage, or a 5- (16) with (25) and (34) being the 3- and 1-. No other breakdowns are possible.

However, if both of R56C1 are subtracted from R6C2, there are two other combos: 124 and 236.

By the way, I posted a KenKen on my blog last week, and I'd like to get some feedback on it. Any takers?

Unfortunately (for me) you are absolutely correct and I can actually only limit R6C2 down to two possibilities. It seems I should spend a little more time solving KenKen puzzles that aren't mine to learn the tricks. I just did yours, for example, and enjoyed how that one fell apart.

Another terrific one. I ended up using some Killer techniques due to the homogeneity of the signs.

R6C5 was my first digit, and then it appeared R1C5 was 26. A couple of uses of the "Rule of 21/odd" and "Rule of 42/even" got R1C2 and R6C2 to both be even. The 4 in C4 has to be in one of the 3- cages, meaning the 4 in C2 couldn't be in R1C2 (else two 1's in C3), R23C2 (at least one of these would interfere with the 3- that has a 4), or R45C2 (not enough even digits to go around in C2). So it's in R2C6, and whether the 4 is a minuend or subtrahend (*cough*), a 1 is in R5C1. Lots of pairs start to creep in, and off we go.

Anyhow, I don't know if you intended any sort of parity summations to be used, but they worked reasonably nicely. I don't recall using column sums in a Kenken (but then, usually not all the signs are the same).

Hadn't intended parity as a work-in, but this post obviously tells me I can try to work off that in the future.

I've definitely put some column/row sum constraints into some KenKen I wrote, but not ones I posted here, generally as I do that in a + only puzzle as a teaching method and I haven't posted + only puzzles here.