## Even Odder by Ploy

### Puzzle solution process

This is one possible solving path. We’ll use subscripts Xa and Xd 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 Oa = 4 and Od = 5.

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

33ac is a three-digit factorial, one of {5! = 120, 6! = 720}, so Pa − Ia is one of {13, 14}; the two possibilities for Pa mean Ia is one of {17, 18, 19, 20} and Id is in {16, 19, 18, 21} respectively. 8dn is a cube matching _2_, ie one of {53 = 125, 93 = 729}, so Sd!/Pd − Id is in {5, 9}. The lower limit for Sd! is 30(5 + 16) = 630 and the upper limit is 32(9 + 21) = 960; the only factorial in that range is 720, so Sd = 6 and Sa = 7. Now 8dn = (720/Pd − Id)3 and for that to be an integer Pd 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 Pd = 30 and Pa = 31, making Ha = 33 and Hd = 32. For 8dn we have 24 − Id = one of {5, 9} and the only value above that fits is Id = 19, with Ia = 18. This makes 8dn = 125, as well as 9dn = 70, 16ac = 58, 33ac = 120 and 22dn = 22479.

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

28dn is now 150(41 − Td), which clearly ends in 0. That makes 90 = 41ac = Ba + 77, so Ba = 13 and Bd = 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 Td = 27 and Ta = 26, which also gives 25ac = 1430 and 26dn = 3684.

For 37ac = 10829 + Ma we have __8_7 in the grid, so Ma ends in 8 and the only available value is Ma = 28, with Md = 29, and 37ac = 10857. Now for 37dn = 5Ud/3 − 29 we have 1_ in the grid (with the second digit not allowed to be 0), so Ud ≥ (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 Ud = 24 and Ua = 25, which gives 37dn = 11 and 14dn = 720. The maximum value for 18dn = 8(Ed + 117) is 1120 (if Ed is 23), so it starts with 1. Now we have 1_2 for 18ac = Ga2 + 22, so Ga2 and therefore Ga must end in 0 and be ≤ √(192 − 22) ≈ 13.04; the only possible value is Ga = 10, with Gd = 11, giving 18ac = 122 and 39dn = 40. For 13dn = Ld + 2 we now have _2 in the grid, so Ld = 20 and La = 21, making 13dn = 22, 40ac = 1046 and 30dn = 95.

In the grid we have 1_1_ for 18dn = 8(Ed + 117), which must be even. Thus 31ac = Ea + 2 ≥ 21; the only available values that make that work are {22, 23}, so 18dn ends in 2, Ed + 117 ends in 4 or 9, and Ed ends in 7 or 2. Thus Ed = 22, Ea = 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 − Aa, so Aa = 15 and Ad = 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 = Ca + 75, so Ca = 17 and Cd = 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.