### Puzzle explanation

Letters stand for 2 to 27 in the order TSIKQGDAPFNCEJXBRUYLMVWHZO. Ambiguities at 11dn and 25dn are resolved by Edna O’Brien’s apt title, August is a Wicked Month. The puzzle’s title is a misprinted version of Bad Summer or Rad Summer, depending on which meaning of “wicked” is taken.

### Puzzle solution process

This is one possible logical path for solving the puzzle.

For 18ac we have x(yz) giving a 5-digit number, with x, y and z all different. Since 217 = 131072 has 6 digits, yz must be ≤ 16, so it’s one of {8, 9, 16} = {23, 32, 24 = 42}. Since 58 = 390625 has 6 digits, x must be ≤ 4; it can’t be 2, because either y or z is 2 for all possible values of yz; it can’t be 3 because the only 5-digit power is 3(32), which makes x and y the same; so x = 4 and the only 5-digit power possible is 4(23) = 65536. Therefore two of the following are true: I = 4, L = 2, S = 3.

For 14dn, IIIT definitely has a factor of I2 (and possibly I3) and it ends in 6. I can’t be 2 or 3 (because then two of {I = 4, L = 2, S = 3} would be false) or 5 or greater (because 25, 36, 49, 64, ... don’t have any 2-digit multiples ending in 6); thus I = 4.

For 16dn we have 5 distinct numbers with a 3-digit product. Since 2×3×4×6×7 = 1008 has 4 digits, the lowest four factors must be 2×3×4×5; the possibilities are {2×3×4×5×6 = 720, 2×3×4×5×7 = 840, 2×3×4×5×8 = 960}, so we can enter the final 0. Since I doesn’t appear in the clue, one of the letters must be a misprint for I = 4 and the other four must be correct. Since L doesn’t appear in the clue, it can’t have the value 2, so the second true result from 18ac is S = 3.

For 14dn, there’s only one 2-digit multiple of 16 ending in 6, so the entry is 96 = 2×3×4×4 = 2×S×I×I, so the misprint is I for S and the remaining factor is T = 2. Then 18ac is 4(23) = I(TS) so its misprint is L for T.

26ac is the product of 5 factors, at least 4 of which are different, so it’s at least 2×2×3×4×5 = 240. 19dn is then in the range [320, 399], and IST uses the correct values for two of {I = 4, S = 3, T = 2}. T = 2 can’t make a product of 300 or more (2×4×27 = 216), so the true product is ISn = 4×3×n; if n = 26, the product 4×3×26 = 312 is too small, so 19dn must be 4×3×27 = 324 and its misprint is T for whichever letter stands for 27. We now have 26ac = 2_0, and 2×2×3×4×10 = 480 is too big, so one of its factors is 5; 2×2×3×5×6 = 360 is also too big, so the only possibility is 240 = 2×2×3×4×5 = TTSIn and either the K or the X is a misprint for S and the other one is 5.

From the grid, 20dn is now 64 = 4×16 (the products 2×32 and 8×8 aren’t possible), so the T in the clue is a misprint, which makes the X = 16 and the misprint is T for I = 4. Then in 26ac we have K = 5 and the misprint is X for S = 3.

10dn is __5 and is the product of 4 distinct (odd) numbers; it must be 3×5×7×9 = 945 (since 3×5×7×11 = 1155 is too big). Since 2 isn’t a factor, the T must be the misprint, so we have GKSn = G×5×3×n and G is one of {7, 9}. 13ac is now _94, which can’t be a multiple of 4, so the I is a misprint and it’s GGnT, with n odd (or the product would be divisible by 4). For G = 9, we’d have 162n, which can end in 4 only if n ends in 7; 162×7 = 1134 is too big, so G = 7. Then 13ac = 98n, which can end in 4 only if n ends in 3; 98×13 = 1274, too big, so 13ac is 294 = 7×7×3×2 and the misprint is I for S = 3.

21ac is a 2-digit product of 3 different numbers; the smallest it can be (with the T = 2 being correct, F or V being a misprint for S = 3, and the smallest value still available for the other being 6) is 2×3×6 = 36, which rules out 720 for 16dn. The factors of 16dn are then 2×3×4×5 and one of {7, 8}; the first four are TSIK, so either the D or the Z must be a misprint for I = 4. The other factor can’t be 7 = G (or we’d have two misprints), so 16dn is 2×3×4×5×8 = 960 and either D or Z is 8. In 1dn, if the I isn’t a misprint, the maximum value is (4+26)×27 = 810, which doesn’t have enough digits, so the I is a misprint and the Z is correct. If Z = 8, the maximum value is (26+27)×8 = 424, also too small, so in 16dn we have D = 8 and the Z is the misprint for I.

28ac definitely has a factor of X = 16 and matches _4_; being a multiple of 4, its last digit must be one of {4, 8} (not 0, or 29dn would start with 0), so the other factor ends in one of {3, 4, 8, 9}. The minimum is 144 = 16×9; none of the other possibilities, {13, 14, 18, 19, 23, 24}, gives a product with 4 as the middle digit, so 28ac = 144 and the misprint is X for whichever letter stands for 9.

For 15dn, if the S isn’t a misprint, the maximum possible value is 3×27 = 81, too small to match _5_, so the S is the misprint and the factor is X = 16 is true. Being a multiple of 4, its last digit must be one of {2, 6}, and it can’t be more than 16×27 = 432, so the possibilities are {152, 156, 252, 256, 352, 356}. Only two of those are multiples of 16; 256 = 16×16 isn’t allowed because the misprinted factor must be different from the true X = 16, so 15dn = 352 = 22×16 and the misprint is S for whichever letter stands for 22.

15ac is now in the range [300, 399]. If the T is a misprint, the R value must be correct (one of the available values 6, 9-15 or 17-27) and it’s raised to a power of 3 or more. The cube root of 399 is about 7.4, so R could only be 6, but that doesn’t have a power in the 300-399 range. Therefore the T isn’t a misprint and 15ac is n2 for some n. The only squares in the range are 324 = 182 and 361 = 192, but 324 is already used at 19dn, so 15ac = 361 and the R is a misprint for whichever letter stands for 19.

3dn definitely has a factor of K = 5, so it must end in 5 (not 0, or 9ac would start with 0). 9ac is then 59_ and is a multiple of either I = 4 or K = 5 (not both, as multiples of 20 end in an even digit followed by 0). The multiples of 4 are 592 (which has a prime factor of 37, not possible in a product of numbers ≤ 27) and 596 (which has a prime factor of 149); the multiples of 5 are 590 (which has a prime factor of 59) and 595 = 5×7×17 = KGn, so 9ac = 595 and its misprint is I for G = 7, which leaves B = 17.

24ac is a 3-digit power ending in 2; the only numbers that have powers ending in 2 are numbers that end in 2 or 8, which excludes B = 17, so the B is a misprint and the A is correct. A can’t be 10 or more because 210 = 1024 is too big. If A is 6, 26 = 64 is too small and 86 = 262144 is too big, so the only value remaining is A = 9, giving 24ac = 512 = 29, and the B is a misprint for T = 2.

28dn is 1_, which can’t be a product of B = 17 and any number ≥ 2, so the B is a misprint and the Q is correct. Q can’t be 10 or more, because multiplying by 2 or more would make the first digit ≥ 2, so the only value remaining is Q = 6, and 28dn is 12 or 18.

31ac is then a 5-digit power ending in 2 or 8. Since 217 = 131072 is too big, the B must be a misprint and the M is correct. To have a power ending in 2 or 8, M must end in 2 or 8, so it’s one of {12, 18, 22}. The only 5-digit power of 12 is 20736, which doesn’t fit; 18 has no 5-digit powers; so 31ac = 10648 = 223 with M = 22 and the B is a misprint for S = 3 (in both 31ac and 28dn).

24dn is now 5_6 and definitely has a factor of S = 3, so it’s one of {516, 546, 576}. None of those is a multiple of GSS = 7×3×3, so either the G or an S is the misprint and the C is correct. If the G is the misprint, we have 576 (the only multiple of 9) = CnSS, so Cn = 64, but all the factors of 64 (2, 4, 8, 16) are already assigned, so there’s no possible value for C. So the misprint is an S and we have 546 (the only multiple of 7) = CGSn, so Cn = 26; therefore C = 13 and the misprint is an S for T = 2.

27ac is _4; it can’t have a factor of C = 13 (there are no 2-digit multiples of 13 ending in 4), so the C is a misprint and the I is correct. Thus 27ac is a multiple of I = 4, so its first digit must be even; it can’t be 64 because that’s already used at 20dn, so 27ac is one of {24, 44, 84}. 27dn is then one of {20, 40, 80} and can’t have a factor of C = 13, so the C must be a misprint and the P and T are correct, giving n×P×2, where n ≥ 3 (distinct from the factor of T = 2) and P ≥ 10 (the lowest available value), for which the minimum value is 3×10×2 = 60. Therefore 27dn = 80, making nP = 40, which can only be factored as n = 4, P = 10, and the C is a misprint for I = 4.

23dn ends in 1, so it can’t have a factor of I = 4, so the I is a misprint, giving F×n×3×3 = __1, with F and n both odd and F ≥ 11, and Fn ends in 9. If F = 19, n must end in 1, but 19×11×3×3 = 1881 (too big); if F = 17 or 27, n ends in 7, but 17×7×3×3 = 1071 (too big); if F = 13 or 23, n ends in 3 (but can’t duplicate the 3s we already have), but 13×23×3×3 = 2691 (too big). Therefore F ends in 1 and n ends in 9; 21×9×3×3 = 1701 and 11×19×3×3 = 1881 are also too big, so 23dn = 891 = 11×9×3×3 with F = 11 and the I is a misprint for A = 9.

23ac is now 8_, which can’t have a factor of C = 13, so the C is the misprint and the M = 22 is correct, so this is 88 and the C is a misprint for I = 4. 13dn is now 2_8. The multiples of 22 in the range [200, 299] are {220, 242, 264, 286}, none of which fits, so the M is a misprint and we have CnT = 13×n×2. The multiples of 26 in the range [200, 299] are {208, 234, 260, 286}, of which only 208 fits, so the M is a misprint for D = 8.

21ac is 6_. If the F is a misprint, the T = 2 and V must be correct, so we would have n×2×V where n ≥ 3 (distinct from the factor of T = 2) and V ≥ 12 (the lowest available value), for which the minimum value is 3×2×12 = 72 (too big). Therefore the F is correct and, to be a multiple of 11, 21ac = 66. 22dn is then 6___; it’s definitely a multiple of J2, so the minimum for J is sqrt(6000/27) ≈ 14.9, so J ≥ 15. For 16ac = 9_, if the F is correct, the entry must be 99 (a multiple of 11), which means the T = 2 is a misprint and the J is correct, but the minimum for FJn is 11×15×3 = 495 (too big). Therefore the F is a misprint and the J and T are correct, giving nJT where n ≥ 3, for which the possibilities are 3×15×2 = 90 and 3×16×2 = 96; but J can’t be 16 (because that’s assigned to X), so J = 15 and 16ac = 90 and the F is a misprint for S = 3. 22dn can then only be 6075 = 15×15×27.

30ac is then _7, which can’t have a factor of D = 8, so the D is a misprint and the U is correct, being one of the remaining odd values, {19, 21, 23, 25, 27}. Of those, only 19 has a 2-digit multiple ending in 7, so U = 19 and 30ac = 57 = 3×19.

32ac is then ___5, which can’t have a factor of Q = 6, so the Q is a misprint and we have CGHn = 13×7×H×n with H and n both odd, H being one of {21, 23, 25, 27}, and at least one of H and n ending in 5. If H = 21, 13×7×21×15 = 28665 is too big; 13×7×21×5 = 9555 would fit, but it would make 30dn = 55, which is too big for the sum of two numbers (26+27 = 53). If H = 23, 13×7×23×5 = 10465 is too big, so H = 27 would be too big, too. Therefore, H = 25 and 32ac = 6825 = 13×7×25×3 (as 13×7×25×5 = 11375 is too big). That makes 30dn = 52, which can only be 25+27, so the F is a misprint for H = 25 and O = 27.

29dn is 48, which doesn’t have a factor of J = 15, so the J is a misprint and W is one of {12, 24}. In 4ac, if the I = 4 is correct, the maximum value possible is 4×26×27 = 2808 (too small), so the I is a misprint and the V and W are correct. If W = 12, the maximum value possible is 27×26×12 = 8424 (still too small), so W = 24 and in 29dn the J is a misprint for T = 2.

5dn is __1, so all three factors must be odd, so P is a misprint and the G and L are correct. The only remaining unassigned odd values are {21, 23}. If L = 23, GL = 7×23 = 161, so to have a product ending in 1, the third factor would have to end in 1, but 7×23×11 = 1771 is too big. So L = 21, GL = 7×21 = 147 and the third factor ends in 3; 7×21×13 = 1911 is too big, so 5dn = 441 = 7×21×3.

4ac is then __4__ and we have nVW = n×V×24. If V = 14, the maximum value is 27×14×24 = 9072 (too small), so V is one of {18, 23, 26} (not 20, because that would make 7dn start with 0). If n = 16, the maximum value is 16×26×24 = 9984 (too small), so n is one of {17, 18, 19, 21, 22, 23, 26, 27} (not 20 or 25, because that would make 7dn start with 0). The only combination that has a 4 as the middle digit is 4ac = 10488 = 19×23×24, V = 23 and the I is a misprint for U = 19.

12ac is 4_. The W can’t be correct because the minimum for nSW would be 2×3×24 = 144, so the W is a misprint and 12ac is a multiple of SS = 9, ie, 12ac = 45. 6dn is then 85 = 17×5 = BK, so the C is a misprint for K = 5.

7dn is now 8_0; the nearest multiples of U = 19 ending in 0 are 40×19 = 760 and 50×19 = 950, so the U is a misprint and the F and Y are correct. The only multiple of F = 11 matching 8_0 is 880 = 80×11, so Y is a factor of 80; the remaining values are {12, 14, 18, 20, 26}, of which only 20 is a factor of 80, so Y = 20.

17ac is _0, which can’t have a factor of F = 11, so the F is a misprint and the N is correct. With the remaining unassigned values, the possibilities for 17ac are {60 = 5×12, 70 = 5×14} (not 90 = 5×18, because we already have 16ac = 90), so N is one of {12, 14}.

In 1dn, we know the I is a misprint, so the minimum value for Z is 1106/(27+22) ≈ 22.6. The only remaining value ≥ 23 is 26, so Z = 26. That makes 1dn even, so 17ac = 60 and N = 12. 1dn = (n+22)×26 now ends in 6, so n+22 must end in 1 or 6, so n must end in 9 or 4; its minimum value is 1106/26−22 ≈ 20.5, so 1dn = 1196 = (24+22)×26 and the I is a misprint for W = 24.

8ac is 1_. If the H is correct, the minimum value is 25×2−19 = 31, so the H is a misprint and we have nR−19 where R is one of {14, 18}. If R = 14, the two lowest values are 2×14−19 = 9 and 3×14−19 = 23, so there’s no value in the required range. Therefore R = 18 and 8ac = 17 = 2×18−19, which leaves E = 14 for the remaining value.

2dn is _7. If the O and R are both correct, the minimum value is 27×18+2 = 488 (too big), so one of them is a misprint and the K = 5 is correct, so we have mn+5 = _7, which means mn ends in 2. If the O = 27 is correct, the smallest multiple ending in 2 is 27×6 = 162 (too big), so the O is a misprint and we have nR = n×18 ending in 2; 9×18 = 162 is too big, so the only value in range is 2dn = 77 = 4×18+5 and the O is a misprint for I = 4.

1ac is now 17__. If the O is correct, we have nEO = n×14×27 for some n; 4×14×27 = 1512 and 5×14×27 = 1890, so there’s nothing matching 17__, so the O is a misprint and we have EEn = 14×14×n, for which the only value in range is 1ac = 1764 = 14×14×9, so the O is a misprint for A = 9. Then 3dn is 65 = 5×13, so its misprint is K for C = 13.

The grid is now filled except for 11dn and 25dn, which are ambiguous. 11dn = _6 and is clued as H+N, so it’s either n+12 with n being one of {4, 14, 24} or 25+n with n in {11, 21}. 25dn = 1_ and is clued as A, so it’s anything in the range [10, 19] except for {15, 17, 18} which appear elsewhere in the grid. Reviewing the misprint corrections we’ve identified for asterisked clues, we have AUGUSTISAWICK_DMO_TH. The first gap is from 11dn, where the correction is one of {I, E, W, F, L}; the second gap is from 25dn, where the correction is one of {P, F, N, C, E, X, U}. The most plausible choices give AUGUST IS A WICKED MONTH, which is the title of a novel by Edna O’Brien. The correction for 11dn is then E = 14, so it’s 14+12 = 26; the correction for 25dn is N, so 25dn = 12.