denis_berthier wrote: .... Which techniques do you use when surface sums are not available or don't allow you to go very far? .... Did you try e.g. the April 14 puzzle? If so, did you find it difficult?

That was a week ago, so I don't remember anymore. On looking at my completed April 14 puzzle, I don't see any cross-outs or any particular evidence that I had any trouble with it.

So I'm going to try a real-time experiment. I just printed out another copy of that puzzle, and as I solve each digit (or small group of digits), I'll list what I did below.

- The lower right corner, r9c9, must be at least 5 because of the 14, and at most 5 because of the 6. So it's 5. That imediately solves the four cells in that corner.
- Where a 2-digit 16 intersects a 2-digit 17, that cell must be 9. That immediately solves the upper left corner.
- Any 3-digit 22 must include a 9. So one of r7c6, r8c6, r9c6 must be 9. It can't be r8c6 because we already have a 9 in that row. It can't be r9c6 because that would put 5 in r9c5, which is impossible because the 4-digit 30 in c5 must consist of 6-7-8-9. So r7c6 is 9.
- r1c4 is at most 5 because of the 8, and r1c5 is at most 4 because of the 10. These maximum values must be the actual values, in order to reach the sum of 9.
- The remaining two cells in that 3-digit 8 must be 1 and 2. Since we already have 2 in r2, we know which is which.
- r2c5 must be one of 1-2-3-4, and 3 is the only one left.
- Likewise, the rest of the 10 in c5 falls into place immediately.
- r9c3 must be either 1 or 3, but 3 would give us a duplicate 4 in r8, so r9c3 must be 1. That also gives us r8c3 and r9c2.
- The 34 in c1 must consist of 9-8-7-6-4. The only one that can still be 4 is r5c1.
- The 33 in c9 must be either 9-8-7-6-3 or 9-8-7-5-4. So r3c9 must be either 3 or 4. That makes r3c8 either 1 or 2. But it can't be 1 because the three missing digits in the 37 in c8 must add up to 8 (45 minus 37), one of which must be 1. So r3c8 and r3c9 must be 2 and 3 respectively.
- r2c7 must be 4 or 5 (we already have 1,2,3 in that row). So r1c7 must be 1 or 2. 1 is impossible since it would force an illegal two-digit number into r1c8. So r1c7 is 2, and that also solves r1c8 and r2c7.
- The two remaining digits in the 19 in r3 must be 7 and 9. But 9 can't go into r3c6, because that would force r2c6 to be 4 or less, which is impossible since we already have all the 4-or-less digits in r2. So r3c3 is 9 and r3c6 is 7.
- r2c6 and r4c6 now fill in easily as 6 and 1 respectively.
- The only place left in r2 for a 9 is r2c9.

Bill Smythe