Sad Hummer by Hedgehog
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 5digit number, with x, y and z all different.
Since 2^{17} = 131072 has 6 digits, y^{z} must be ≤ 16,
so it’s one of {8, 9, 16} = {2^{3}, 3^{2}, 2^{4} = 4^{2}}.
Since 5^{8} = 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 y^{z};
it can’t be 3 because the only 5digit power is 3^{(32)}, which makes x and y the same;
so x = 4 and the only 5digit 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 I^{2} (and possibly I^{3}) 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 2digit multiples ending in 6);
thus I = 4.
For 16dn we have 5 distinct numbers with a 3digit 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 2digit 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 2digit 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, 915 or 1727)
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 300399 range.
Therefore the T isn’t a misprint and 15ac is n^{2} for some n.
The only squares in the range are 324 = 18^{2} and 361 = 19^{2},
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 3digit 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 2^{10} = 1024 is too big.
If A is 6, 2^{6} = 64 is too small and 8^{6} = 262144 is too big,
so the only value remaining is A = 9, giving 24ac = 512 = 2^{9},
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 5digit power ending in 2 or 8.
Since 2^{17} = 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 5digit power of 12 is 20736, which doesn’t fit;
18 has no 5digit powers;
so 31ac = 10648 = 22^{3} 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 2digit 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 J^{2},
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 2digit 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.
