Solution to Problem 2: Cross Sums

It's no wonder that Dell puzzle magazines don't write out the solutions to their Cross Sums puzzles: they're very tedious to read. I suspect that I'm making some leaps in logic, but I have no fear that you'll all be able to keep up with me. When I write something like "7 in three digits is {124}", I mean that an entry headed with a 7 that has three squares must be filled with the digits 1, 2, and 4 in some order.

24 in three digits is {789}, AQ must be 7, so BQ is 1. Therefore, AR=8, BR=2, and AP=9. 23 three digits is {689} and 21 in six digits is {123456}, so BP=6, CP=8. BM must be 5 and AM=9.

GR is at most 3, HR is at most 2, and IR is at most 9. These must all be true for their sum to be 14. IQ=8, HQ=1, GQ=2, and GP=2. FQ cannot be 8, so FP=3, FQ=7, and EP=2. 17 in two digits is {89}, so DQ=9, DR=8, ER=9, and EQ=6.

4 in two digits is {1,3}. FJ cannot be 1, so FJ=3, FK=1, EJ=8, DK=3, DL=1, BK=2, BJ=1. AJ is at most 4 and AK is at most 6: both of these must be true to get a sum of 10.

3 in two digits is {12}, so IO=1, IN=2, HO=3. 38 in 6 digits is {356789}, so HL=5

BO cannot be 3, since 3+7+8+9 is only 27. BO=4, BN=3. CO, DO, and EO are 7,8,9 in some order. DN cannot be 2, so DN=1, DO=8. CN is not 6, so CN=4, CO=9, EO=7. EN can only be 5, so EK=4, CK=5, CL=7, EL=3, EM=1.

IK is at least 5, and IL is at least 1, so IJ cannot be 9. IJ=7, HJ=9. HK is not 9, so HK=8, IK=6, IL=1, GL=2. HN must be 7, so HM=6, FN=9, GN=8, GM=4, and FM=7.

The diagonal entries are 8 1 8 8 5 7 2 8 7.

Notes: This was the second problem that I solved, and it took me less than ten minutes. I've been solving Cross Sums puzzles for over three-quarters of my life, so this was very familiar to me, and this was not a very difficult example of the genre.