## Bearskin by Elap

### Puzzle explanation

In increasing numerical order, the letters used in clues spelt out “pUZzLiNG; dEPIlaTORS; A WorD CuBe”. Replacing the digits in the grid with the letters of DEPILATORS produced a “word cube” of 48 four-letter words, 16 in each of the three grid directions. The title was a homophone of “bare skin”, and “shaving off” in the preamble and the 13ac clue “r − A − Z − O + r” were also hints to the word DEPILATORS.

### Puzzle solution process

This is one possible solving path. There are only 44 cubes of two to five digits, so it’s not too onerous to calculate all the possible values that letters can stand for, namely, {4, 7, 12, 16, 25, 29, 43, 96, 197, 261, 331, 375, 648, 653, 728, 744, 832, 859, 913, 1125, 1952, 2167, 2768, 2875, 3824, 4000, 4088, 4389, 4872, 5184, 5625, 5937, 6656, 7000, 7336, 7576, 8921, 9304, 9319, 9507, 9683, 9791}. The cubes 1000 and 8000 are discounted because they would give a value of 0.

For 4-through, U3 is a three-digit answer, which is only satisfied by U = 7, giving 4-through = 343. The 12-across clue is now 7C − 49 and in the grid we have 4___3, so C = (4___3 + 49)/7, which must end in 6, and C is between (40003 + 49)/7 ≈ 5721.7 and (49993 + 49)/7 ≈ 7148.9; the only value that fits is C = 6656, giving 12-across = 46543.

9-across is now 6656Z2 − i − Z and has six digits. Z can’t be 16 or greater, because the minimum possible value would be 6656×16×16 − 9791 − 16 = 1694129, too big; so Z is one of {4, 12} (as 7 is already taken by U). If Z is 4, the maximum possible for 9-across is 6656×4×4 − 12 − 4 = 106480, but that’s too small because the last four digits of 9-across coincide with 10-across = 6656 + L2 + T, which has to be > 6656. Therefore Z = 12.

14-down is now 7N + 144 and from the grid we have ___3, so N ends in 7 and is between (1003 − 144)/7 ≈ 122.7 and (9993 − 144)/7 = 1407. The only value that fits is N = 197, giving 14-down = 1523.

From the grid, 2-through is __6 and the clue is Lp2 + 12, so the maximum for p is √((996 − 12)/4) ≈ 15.7, and the only value available is p = 4. Now L = (__6 − 12)/16, which must end in 4 or 9, and the maximum is (996 − 12)/16 = 61.5; the only available value is L = 29, making 2-through = 476 and 19-across = 5815 (6656 − 292).

11-down is 5684 + T2, matching 7___6_ in the grid, so T is between √(700060 − 5684) ≈ 833.3 and √(799969 − 5684) ≈ 891.2; the only available value is T = 859, giving 11-down = 743565 and 10-across = 8356. That means 9-across = 958452 − i is one of {948356, 958356}, but 958452 − 948356 = 10096 is too big for i, so 9-across = 958356 and i = 96 (958452 − 958356).

17-down = E − 16 must match 3__ in the grid, so E is in the range [316, 415], for which the available values are {331, 375}, so 17-down is one of {315, 359}. 8-through is 7z − 16, matching _55 or _59, so z = (_55 + 16)/7, ending in 3, or (_59 + 16)/7, ending in 5. 14-through = 5z + 29 must match 1__ from the grid, so z ≤ (199 − 29)/5 = 34; the only value ending in 3 or 5 is z = 25, giving 8-through = 159 and 14-through = 154. The grid now has 17-down = 359, so E = 375.

10-down = 197o − 6459, to match 8____4 in the grid, so o is between (800004 + 6459)/197 ≈ 4093.7 and (899994 + 6459)/197 ≈ 4601.3; the only value in that range is o = 4389, giving 10-down = 858174. The grid now gives 8174___ for 13-down = 860e − 1718, so e is between (8174100 + 1718)/860 ≈ 9506.8 and (8174999 + 1718)/860 ≈ 9507.8, ie e = 9507 and 13-down = 8174302.

14-across = 2425I matches 15_35__ in the grid, so I is between 1503500/2425 = 620 and 1593599/2425 ≈ 657.2, ie one of {648, 653}. But 648×2425 = 1571400 doesn’t fit, so I = 653 and 14-across = 1583525. Now 2-across = S − 1494, to match 4__, so S is in the range 400 + 1494 = 1894 to 499 + 1494 = 1993; the only possible value is S = 1952, making 2-across = 458.

For 1-across = 1927G − 1952 we have ____95 in the grid, so G = (____95 + 1952)/1927 ends in 1 and is between (100095 + 1952)/1927 ≈ 52.96 and (999995 + 1952)/1927 ≈ 519.95, so it’s one of {261, 331}. 1927×331 − 1952 = 635885 doesn’t fit, so G = 261 and 1-across = 500995.

For 2-down = 7D + 4438 we have 4__97 in the grid, so D ends in 7 and is between (40097 − 4438)/7 ≈ 5094.1 and (49997 − 4438)/7 ≈ 6508.4; the only matching value is D = 5937, giving 2-down = 45997.

For 6-down = 8453 − W we have __85 in the grid, so W ends in 68; the only matching value is W = 2768, making 6-down = 5685.

For 15-across = 375l + 28521 we have 3_15__ in the grid, so l is between (301500 − 28521)/375 ≈ 727.9 and (391599 − 28521)/375 ≈ 968.2, ie one of {728, 744, 832, 913} (859 is already taken). The corresponding values for 375l + 28521 are {301521, 307521, 340521, 370896}, of which the only match is 15-across = 301521, from l = 728.

For 18-down we have 3021__ in the grid, and the clue is O(d − 1) + 859, so O(d − 1) is in the range 302100 − 859 = 301241 to 302199 − 859 = 301340. Either O or (d − 1) is less than √301241 ≈ 548.9, so either O or d is one of {16, 43, 331}. If one of them is 16 then the other is around 301290/16 ≈ 18830.6, far too big. If O is 43 then d is between 301241/43 + 1 ≈ 7006.6 and 301340/43 + 1 ≈ 7008.9, but there are no suitable values in that range; if d is 43 then O is between 301241/(43 − 1) + 1 ≈ 7172.4 and 301340/(43 − 1) + 1 ≈ 7174.8, again having no suitable values. If O is 331 then d is between 911.1 and 911.4, but there’s no such integer; therefore d = 331 and O is between 912.9 and 913.2, ie O = 913, making 18-down = 302149.

For 7-across = 943130 − u we have 9_6130 in the grid, so the missing digit is 0 to 3 and u is at least 943130 − 936130 = 7000; in fact it can’t be anything else, as the next possibility would be 17000, far too big; thus u = 7000 and 7-across = 936130.

For 3-down = B + 95r − 96 we have 5_35__ in the grid, so r is in the range (503500 + 96 − 9791)/95 ≈ 5197.9 to (593599 + 96 − 16)/95 ≈ 6249.3; 5937 is taken by D, leaving only r = 5625. Then 13-across is 10325 − A, and it’s complete as 8158 in the grid, so A = 2167 (10325 − 8158). 3-down is now B + 534279, so the entry must be in the range [543500, 543599], with B in the range [9221, 9320], ie one of {9304, 9319}, the corresponding 3-down entries being {543583, 543598}. Respectively, that gives 545_43 or 545_48 for 5-across = a + 545099, so a ends in 4 or 9 and is in the range 545143 − 545099 = 44 to 545948 − 545099 = 849; the only matching available value is a = 744, giving 5-across = 545843.

3-down is now 5435_3 in the grid, narrowing the range for B to [9224, 9314] and the only match is B = 9304, giving 3-down = 543583.

For 20-across = R + 866 we have __9_ in the grid, so R is in the range 1090 − 866 = 224 to 9999 − 866 = 9133. For 16-across = P(R + 2) − 2167 we have 72__2_ in the grid, so P is between (720020 + 2167)/(9133 + 2) ≈ 79.1 and (729929 + 2167)/(224 + 2) ≈ 3239.4, ie one of {648, 832, 1125, 2875}. If we approximate 16-across as, say, 725000, then the corresponding values for R = (16-across + 2167)/P − 2 are {1120.2, 871.999, 644.4, 250.9}; 872 and 251 aren’t close to possible letter values, but the first and third values suggest P and R are 648 and 1125 in some order. If P is 1125 and R is 648 then 16-across is 729083, but that has the wrong penultimate digit; therefore P = 648, R = 1125 and 16-across = 728129. That also gives u 20-across = 1991 and the grids are filled, as shown here.

Now, the values in numeric order are p = 4, U = 7, Z = 12, z = 25, L = 29, i = 96, N = 197, G = 261, d = 331, E = 375, P = 648, I = 653, l = 728, a = 744, T = 859, O = 913, R = 1125, S = 1952, A = 2167, W = 2768, o = 4389, r = 5625, D = 5937, C = 6656, u = 7000, B = 9304, e = 9507, and the letters spell out pUZzLiNG (what we have been doing), dEPIlaTORS and A WorD CuBe. As instructed by the preamble, we must replace the digits 0 to 9 with the letters of DEPILATORS respectively. Doing this produces 16 four-letter words reading across, 16 reading down, and another 16 in the “through” direction, thus “a word cube”.