This is one possible solving path. We’ll use subscripts X_a and X_d to refer to the respective across and down values of X. It will be useful to compile a list of the first factorial values, {1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320, 9! = 362880, …}.

From 12ac, which has two digits, we have (using across values throughout) O! = 12ac − A − T − M − C + I, so O! ≤ 99 − 2 − 4 − 6 − 8 + 33 = 112. Since O ≥ 2, O! must be one of {2, 6, 24}, with O in {2, 3, 4}. For 11ac to be positive we need O > N, so N = 2|3 (not yet knowing which is across or down) with O_a = 4 and O_d = 5.

11ac is a factorial and must be 8! = 40320 (the only one with five digits), so (P_a + H_a)(4 − N_a)/4^N_a = 8, which means P_a + H_a = 8×4^N_a/(4 − N_a). If N_a is 3 that gives 512, but the maximum sum for any two distinct variables is 31 + 33 = 64; so N_a = 2, N_d = 3 and P_a + H_a = 64, ie {P_a, H_a} = {31, 33} but we don’t know which is which.

33ac is a three-digit factorial, one of {5! = 120, 6! = 720}, so P_a − I_a is one of {13, 14}; the two possibilities for P_a mean I_a is one of {17, 18, 19, 20} and I_d is in {16, 19, 18, 21} respectively. 8dn is a cube matching _2_, ie one of {5³ = 125, 9³ = 729}, so S_d!/P_d − I_d is in {5, 9}. The lower limit for S_d! is 30(5 + 16) = 630 and the upper limit is 32(9 + 21) = 960; the only factorial in that range is 720, so S_d = 6 and S_a = 7. Now 8dn = (720/P_d − I_d)³ and for that to be an integer P_d must be a factor of 720 = 2⁴×3²×5; it can’t be 32 = 2⁵ because that has too many occurrences of the prime factor 2, so P_d = 30 and P_a = 31, making H_a = 33 and H_d = 32. For 8dn we have 24 − I_d = one of {5, 9} and the only value above that fits is I_d = 19, with I_a = 18. This makes 8dn = 125, as well as 9dn = 70, 16ac = 58, 33ac = 120 and 22dn = 22479.

Now for 6ac we have 7808 + R_a = __17, which can only work if R_a = 9 and R_d = 8, giving 6ac = 7817.

28dn is now 150(41 − T_d), which clearly ends in 0. That makes 90 = 41ac = B_a + 77, so B_a = 13 and B_d = 12, which also gives 27ac = 221 and 34dn = 239. In the grid we have 21_0 for 28dn; the only matching number that’s divisible by both 3 and 50 is 28dn = 2100, making T_d = 27 and T_a = 26, which also gives 25ac = 1430 and 26dn = 3684.

For 37ac = 10829 + M_a we have __8_7 in the grid, so M_a ends in 8 and the only available value is M_a = 28, with M_d = 29, and 37ac = 10857. Now for 37dn = 5U_d/3 − 29 we have 1_ in the grid (with the second digit not allowed to be 0), so U_d ≥ (11 + 29)×3/5 = 24; it can’t be 25 because 5×25/3 isn’t an integer, and all the higher values are already assigned, so U_d = 24 and U_a = 25, which gives 37dn = 11 and 14dn = 720. The maximum value for 18dn = 8(E_d + 117) is 1120 (if E_d is 23), so it starts with 1. Now we have 1_2 for 18ac = G_a² + 22, so G_a² and therefore G_a must end in 0 and be ≤ √(192 − 22) ≈ 13.04; the only possible value is G_a = 10, with G_d = 11, giving 18ac = 122 and 39dn = 40. For 13dn = L_d + 2 we now have _2 in the grid, so L_d = 20 and L_a = 21, making 13dn = 22, 40ac = 1046 and 30dn = 95.

In the grid we have 1_1_ for 18dn = 8(E_d + 117), which must be even. Thus 31ac = E_a + 2 ≥ 21; the only available values that make that work are {22, 23}, so 18dn ends in 2, E_d + 117 ends in 4 or 9, and E_d ends in 7 or 2. Thus E_d = 22, E_a = 23 and 18dn = 1112, with 31ac = 25 as well as 23ac = 9896, 29ac = 694, 38ac = 3417, 3dn = 793, 4dn = 3200, 5dn = 84949 and 15dn = 7581.

Now we have 38 in the grid for 4ac = 53 − A_a, so A_a = 15 and A_d = 14, giving 1ac = 5117, 19ac = 104, 20ac = 5564, 21ac = 5032, 1dn = 569 and 2dn = 17.

The only remaining unassigned variable is C = 16|17. We have 92 in the grid for 12ac = C_a + 75, so C_a = 17 and C_d = 16, giving 10ac = 6579, 35ac = 54, 42ac = 9039, 7dn = 8385, 17dn = 8464, 19dn = 1308 and 36dn = 479 to complete the grid. Converting the sums of vertical triplets to letters, we get THOSE UNSHOCKED DON’T UNDERSTAND IT.