Kakuro Puzzles

Tough one for me. Could not find a pure logic solution yet.

Mike,
Actually pretty straight forward with possibilities (unless I made a fortuitous goof).
I got HILM right away, then noticed there had to be a '3' in 17D (either P,T) and 24D (either Q,U). Once you eliminate the 3's elswhere in rows 10A, 15A - it falls into place.

This falls into place OK after you make the assumption that 24D must be 1,2,3,4,5,& 9 rather than any other combination. Trouble is this is not always a safe assumption.

2:57

Bluey,
You don't need to make any assumption with 24D - it must be 123459, 123468 or 123567.
Like Steve, I got HILM right away as follows:
a) 17D must be 12347 or 12356, but 35A and 29A rule out 12347 - H&L solved.
b) No 6 in 24D, so 24D can only be 123459 - I&M solved.
Then it is simply a case of eliminating 3's like Steve explained.

7:58

5:07

You are correct. I came back and solved it without much trouble. Must have been having a bad day.

Another way to solve this one. Agreed, H, I, L, M are straightforward.

Reading down, we then have P or U must be 3; likewise Q or V must be 3. Therefore, neither 27D nor 24D can contain a 3. That restricts the choices for 24D such that we can fill in G, whence the whole upper left corner More...

nice puzzle

easy with all those hints above ;-)

6:07

9:26

547

but, Dave, 10A was 5,1!

4:59

14:58 with a guess in 4D

I did it in 6 minutes (from scratch) using the 3s approach and E=8, U=4 (both to prevent a nonunique solution). Man, am I slow in picking things up!

I particularly liked this logic:

After filling in H,I,L,M, have first three columns adding to 68, and also to 10+24+15+6+(N+O+R+S). So N+O+R+S=13. We know these four squares include exactly one 5 and no 3's, so must be 5/4/2/2.

This forces 2's in C and Y, and shows that 6 across is 5/1 (in some order), so 27 down and 24 down each includes both a 7 and a 5. Not hard from here.

Straightforward with lots of crosses as above, 2018.