### Puzzle solution process

This is one possible solution path. In several cells the number of digits can be determined immediately. For example, 33ac is a six-digit number in three cells, so it must be entered as nn|nn|nn (using vertical bars to show cell divisions), which means 24dn and 25dn are n|n|nn and 17dn is n|n|n|nn, forcing 25ac to be n|n|nn, 26dn to be nn|n|n and so on. Because the cell contents must eventually be replaced by letters representing 1 to 19, single-digit cells can only contain 1 to 9 (not 0) and two-digit cells must contain 10 to 19.

Rearranging 21ac, it’s (RR + R) + (TA + IE) − S and has four digits. The maximum for (TA + IE) is (16×17 + 18×19) = 614. If R is 4 then (44 + 4) + 614 = 874 is too small; if R is 6 then (66 + 6) = 46662 is too big; therefore R = 5.

For 14ac, the factorial T! has no more than four digits, so T ≤ 7 (7! = 5040, 8! = 40320). For 30ac, W(A − S) − O + K has three digits, so (A − S) ≥ 2, which means A ≥ 3. From 22dn, (R − A) > 0, so A ≤ 4. Then for 29dn = TA to have three digits the possibilities are 63 = 216 and 73 = 343 (not 44 because T and A must be different, and not 53 or 54 because 5 is already taken); but 29dn is a palindrome, so T = 7 and A = 3 and we can enter 29dn = 343 with one digit per cell. To make (A − S) ≥ 2, we must have S = 1.

1ac is now (15M)C + H, with five digits. If C is 4, the minimum possible is (15×2)4 + 6 = 810006, too big, so C ≤ 3; but 1 and 3 are already taken, so C = 2. 27dn is now 2E(4 + M + I) − 1 and has four digits, entered in three cells. The minimum for 27dn is 10|1|1, so the minimum for 2E is (1011 + 1)/(4 + 19 + 18) ≈ 24.7, so E ≥ 6 (24 = 16, 25 = 32, but 5 is already taken). For 31ac, the factorial E! must have no more than three digits, so E ≤ 6, which forces E = 6.

21ac is now 3150 + 6I, in the range 3150 + 6×4 = 3174 to 3150 + 6×19 = 3264. To fit into three cells, each containing 1 to 19, it must be either 3|17|1 to 3|19|9, or 3|2|10 to 3|2|19, which makes 22dn start with 17, 18, 19 or 2 in the first cell. 22dn is 2B3 − 7 and has four digits, so B is between ((1711 + 7)/2)(1/3) ≈ 9.5 and ((2919 + 7)/2)(1/3) ≈ 11.4, ie one of {10, 11}. But 2×113 − 7 = 2655 can’t be partitioned into three cells, so B = 10 and 22dn = 1993, entered as 19|9|3. 21ac is now in the range 3|19|1 to 3|19|9, and 3150 + 6I = 6(525 + I) is a multiple of both 2 and 3, which allows only {3192, 3198}, so I is one of {7, 8}; but 7 is taken, so I = 8 and 21ac = 3198.

In 33ac, the E(I − G) term is now 6(8 − G), so to avoid a fractional value we need G < 8 and the only available value is G = 4. The clue is then 10(O − 10)L + 1314 + 7H + M, with six digits in three cells, so 10(O − 10) is between (101010 − 1314 − 7×18 − 17)/19 ≈ 5239.6 and (191919 − 1314 − 7×11 − 12)/9 ≈ 21168.4; the only power of 10 in that range is 104 = 10000, so O = 14. Now 31ac = 392L − 1, to fit 43_ in the grid, with two digits in the last cell; 392×9 − 1 = 3527 is too low and 392×12 − 1 = 4703 is too high, so L = 11 and 31ac = 4311. That also completes 26dn = 1856, which we can enter as 18|5|6.

34ac = 8N2 + 5 and has three digits, so N < √((1000 − 5)/8) ≈ 11.2; the only available value is N = 9, giving 34ac = 653, 25ac = 7|3|18, 14ac = 5|15|1, 3dn = 7|11|1|18, 1dn = 8|8|14|6, and 7ac = 19|6|5. For 7dn we have 53M2 to fit 19_33 in the grid, so M2 ends in 1 and M ends in 1 or 9, for which the only available value is M = 19, making 7dn = 19|1|3|3, 9dn = 5|13|1, 4ac = 6|9|16, 10ac = 823, 11dn = 2|5|19, 18ac = 18|8|9|4, 27dn = 19|8|3 and 35ac = 4389.

In the grid we have 1|_|13 for 13ac, so 8dn is 6__ with two digits in the last cell. It’s a palindrome, so the last cell must contain 16, ie 8dn = 6|1|16; its clue is 44(123 + U), so U = 16, which makes 13ac = 1|1|13, 5dn = 9569, 6dn = 16|14|2, 16ac = 6|19|3 and 24dn = 1|1|11.

The grid now has 15ac = 2316 and its clue is 8V2 + 4, so V = 17, which gives 17dn = 3|3|3|17, 19dn = 8675 and 23dn = 8|1|11|9.

1ac is 81225 + H, to fit 8__7 in the grid, so H ends in 2 and can only be H = 12, making 1ac = 8|12|3|7 and 33ac = 11|14|17. The grid now has 2dn = 3|3|15, for which the clue is 255W, so W = 13 and the only remaining variable is D = 15. This completes the grid with 12ac = 11|2|5, 4dn = 6|2|10, 30ac = 173, 25dn = 7|7|14, 31ac = 789 and 20dn = 4694.

Finally, changing each cell’s contents to the corresponding letter spells out the message I HATE NUMERICAL CROSSWORDS BECAUSE MAKING A MISTAKE MEANS STARTING ALL OVER AGAIN.