Can’t You Do Division? by Oyler

Puzzle solution process

This is one possible solution path. A previous Listener Crossword editor has prepared a document containing a list of prime factorisations for numbers up to 1000, but such factoring isn’t necessary for a logical solution. The starting point is to consider how the number of factors given in each clue may be expressed as a product of terms of the form (1 + n), where n is the exponent of one of the answer’s prime factors. Note that no answer in this puzzle can have both 2 and 5 as prime factors, or it would contain a 0.

If a clue X is a prime then it’s not the product of more than one number, so the answer must be a prime number raised to the power of (X - 1). This applies to G = x2, K = x6, S = x4, for some values of x. K has two digits, so it can only be 26 and we can enter K = 64. C, F, Q, f, s and t are all clued by 2 and must therefore be primes themselves, ending in one of {1, 3, 7, 9}. S has two digits so it could be 24 = 16 or 34 = 81, but because t is prime S can’t start with 8, so S = 16.

m is clued by 21, which can only be broken down as 7×3, so m is either x20 or x6y2; we can rule out the former because 220 = 1048576 has far more than m’s three digits. Trying primes for the latter, 26×32 = 576 but all other combinations are too big (26×52 = 1600, 36×22 = 2916 etc), so m = 576.

Similarly, p has four digits and is clued by 35 = 7×5; no value of x34 fits, so it must be x6y4. Trying primes, 26×34 = 5184 but the rest are too big (26×54 = 40000, 36×24 = 11664), so p = 5184.

G is the square of a prime ending in one of {1, 3, 7, 9}, so its last digit must be 1 or 9; it can’t be 1 because we already have a 1 in column 8 (the first digit of S), so G is 5__9. Its square root therefore ends in 3 or 9 and is between √5129 ≈ 71.6 and √5879 ≈ 76.7, ie, it is 73, so G = 5329.

F is a prime, ending in one of {1, 3, 7, 9}; we already have 3 and 9 in the same row, and 1 in the same column as its last digit, so F must end in 7. Similarly, f is a prime and we now have {3, 7, 9} in the same row as its last digit, so f ends in 1. The only digits now missing from row 4 are {4, 6, 8}; in the last column we already have 4 and 6, so the 8 must go there and F is one of {147, 167}. But 147 is a multiple of 3 (the sum of its digits is 12, a multiple of 3), not a prime, so F = 167 and row 4 can be completed with a 4 in the first column.

C is a prime and in the same column as its last digit we already have 1 and 9, so C ends in 3 or 7. Then j is either 3_96 or 7_96. It’s clued by 9, which can only be broken down as 3×3, so j is either x8 or x2y2 and one of its prime factors must be 2 (because j is even); 28 = 256 doesn’t fit, so j = 4y2 for some (odd) prime y; for j to end in 6, y2 must end in 9, so y ends in 3 or 7. If j starts with 3 then y is between √(3296/4) ≈ 28.7 and √(3896/4) ≈ 31.2 but there are no numbers ending in {3, 7} in that range. So j starts with 7 and y is between √(7296/4) ≈ 42.7 and √(7896/4) ≈ 44.4, ie y = 43 and j = 7396.

u has three digits, the first of which is one of {1, 2, 3, 4, 9} and the second {2, 3, 8, 9}, to avoid repeats in the rows or columns. It’s clued by 14, which can only be broken down as 7×2; 213 = 8192 is too big, so u = x6y. If x is 3, 36×2 = 1458 is too big, so x must be 2 and u = 64y. Trying primes, we get {64×3 = 192, 64×5 = 320, 64×7 = 448, 64×11 = 704, 64×13 = 832, 64×17 = 1088, ...} which all contain 0 or repeated digits or are too big, apart from {192, 832}; but u can’t start with 8 so u = 192. The last digit of the prime s now has {1, 3, 9} in the same column or row, so s must end in 7.

c is now some permutation of {3, 4, 8} and we already have a 3 in in row 3, so it’s one of {348, 384, 438, 834}. It’s clued by 16, which can be broken down as {8×2, 4×4, 4×2×2, 2×2×2×2}, so it’s one of {x15, x7y, x3y3, x3yz, xyzw}. As the last digit is 4 or 8, one of its prime factors must be 2, and as the digit sum is 15 another prime factor must be 3, so we can eliminate x15 (only one prime factor) and x3y3 (23×33 = 216); and 5 is not a factor. Repeatedly dividing the four candidates by 2 and then by 3, we get {348 = 22×3×29, 384 = 27×3, 438 = 2×3×73, 834 = 2×3×139}, of which the first and third don’t match any of the prime factorisations. For 139, we only need to try dividing it by 7 and 11 to establish that it’s a prime, so 834 doesn’t match a prime factorisation and we’re left with c = 384.

B is clued as 12, which can be broken down as {6×2, 4×3, 3×2×2}, so its prime factorisation is one of {x11, x5y, x3y2, x2yz}. For the first, 211 = 2048 is too big; for the third, 23×32 = 72 clashes with the 7 already in the row and 33×22 = 108 is too big; for the fourth (avoiding having both 2 and 5 as factors), 22×3×7 = 84 clashes with the 8 in the row, 32×2×7 = 126 is too big. So we’re left with x5y, and 35 = 243 is too big, so the only value that fits is B = 96 (= 25×3).

From the grid, the prime f is now 6_1 and the middle digit is one of {2, 5, 7, 8, 9}. We can rule out 621, 651 and 681 because they’re multiples of 3 (their digit sums are 9, 12 and 15 respectively); 671 is easily recognisable as 11×61 (the middle digit is the sum of the outer ones); so f = 691.

From the grid, J = 57_ and the last digit is one of {1, 2, 8, 9}. It’s clued as 12, which can be broken down as {6×2, 4×3, 3×2×2} so the prime factorisation of J is one of {x11, x5y, x3y2, x2yz}. We can eliminate x11, as 211 = 2048 is too big. For x5y, if x is 2 then y is between 571/32 ≈ 17.8 and 579/32 ≈ 18.1 but there are no primes in that range; if x is 3 then y is between 571/243 ≈ 2.35 and 579/243 ≈ 2.38, again, no primes (or integers); if x is 5 then 55 = 3125 is too big; so J isn’t x5y. For x3y2, if x is 2 then y is between √(571/8) ≈ 8.4 and √(579/8) ≈ 8.5; if x is 3 then y is between √(571/27) ≈ 4.599 and √(579/27) ≈ 4.631; if x is 5 then y is between √(571/125) ≈ 2.137 and √(579/125) ≈ 2.152; if x is 7 then y is between √(571/343) ≈ 1.2902 and √(579/343) ≈ 1.2992; none of those ranges contains integers, let alone primes, and if x is 11 then 113 = 1331 is too big, so J isn’t x3y2 and must be x2yz. Of the four possibilities for J, 571 is not a multiple of 2, 3 (its digit sum is 13) or 5 and the minimum possible with other primes is 72×11×13 = 7007, too big; 579 is not a multiple of 2 or 5 but 579 = 3×193, so we’d need 193 = x2y with primes greater than 5, but 72×11 = 539 is too big; 578 is not a multiple of 3 or 5 but 578 = 2×289 and 289 is recognisable as 172, so 578 = 172×2, which doesn’t match x2yz. Therefore J = 572 (which we can verify as 22×11×13).

H is clued by 4, which can only be broken down as 2×2, so its prime factorisation is either x3 or xy. The only two-digit cube of a prime is 27 but we already have 2 and 7 in the row, so H is xy. It contains only digits from {1, 3, 8} (we already have 9 in both of the first two columns) and the second digit can’t be 1 (which we have in column 2). Of the candidates {13, 18, 38, 83}, the only product of two primes is H = 38 (= 2×19).

N is clued by 18, which breaks down as {9×2, 6×3, 3×3×2} so its prime factorisation is one of {x17, x8y, x5y2, x2y2z}. 217 = 131072 is too big; 28×3 = 768 clashes with the 6 already in the row, and 28×7 = 1792 and 38 = 6561 are too big. For the fourth option, 32×52×7 = 1575 is too big, so one of the prime factors would have to be 2, ruling out 5 (or N would end with 0); the only possibilities with three digits are {22×32×7 = 252, 22×32×11 = 396, 22×32×13 = 468, 22×32×17 = 612, 22×32×19 = 684, 22×32×23 = 828, 22×72×3 = 588, 32×72×2 = 882}, all of which either have a repeated digit or clash with the 6 in the row. So N has to be x5y2; 35×52 = 6075 is too big, so one of the prime factors has to be 2, ruling out 5; the only possibilities with three digits are 25×32 = 288 with a repeated digit and 35×22 = 972, so N = 972.

M is now __16, thus a multiple of 4, and the first two digits are from {3, 4, 5, 8}. It’s clued by 20, which breaks down as {10×2, 5×4, 5×2×2}, so its factorisation is one of {x19, x9y, x4y3, x4yz} with 2 to a power greater than 1 and no factor of 5. For the first option, 219 = 524288 is too big. For the second, 29 = 512, so to give an answer ending in 6 the prime y would have to end in 3, but 29×3 = 1536 and 29×13 = 6656 don’t fit and 29×23 = 11776 is too big. For the third, neither of {24×33 = 432, 24×73 = 5488} fits and 24×113 = 21296 is too big, while 34×23 = 648 is too small and 74×23 = 19208 is too big. So M is 24xy, ie, a multiple of 16 in the range 3416 to 8516 where M/16 ends in 1; of the candidate multiples {4016, 4816, 5616, 6416, 7216, 8016}, all contain 0 or a repeated 6 apart from 7216, which clashes with the 7 and 2 already in the row, and 4816, so M = 4816. The only digits now missing from the row are {3, 5}; we already have a 3 in the first column, so 5 goes there and 3 in the last column.

From the grid, n is now 84__, with the third digit being one of {2, 3, 5, 7} and the last from {2, 3, 5}. Like M, it has 20 factors, so its factorisation is one of {x9y, x4y3, x4yz} (x19 being too big). For the first option, consecutive values are 29×13 = 6656, 29×17 = 8704, neither of which fits, and 39 = 19683 is too big. For the second option, with x = 2 consecutive values are 24×73 = 5488, 24×113 = 21296; with x = 3 consecutive values are 34×23 = 648, 34×53 = 10125; with x = 5 consecutive values are 54×23 = 5000, 54×33 = 16875; with x = 7 the minimum value is 74×23 = 19208; so there are no values that fit. Thus n must be x4yz. If x = 11 then yz < 8475/14641 ≈ 0.58 but there are no such products of two primes; if x = 7, yz is between 8423/2401 ≈ 3.51 and 8475/2401 ≈ 3.53 (no integers); if x = 5, yz is between 8423/625 ≈ 13.48 and 8475/625 = 13.56 (no integers); if x = 3, yz is between 8423/81 ≈ 103.99 and 8475/81 ≈ 104.63, which leaves only 104, which is an obvious multiple of 4 (actually 2×2×2×13); so x must be 2 and yz is between 8423/16 ≈ 526.44 and 8475/16 ≈ 529.69, ie one of {527, 528, 529}. 529 may be recognised as 232, not the product of two different primes, and 528 is a multiple of 4 (its last two digits are 4×7), so xy must be 527 and n = 8432 (= 24×17×31).

The only digits missing from row 8 are {3, 5, 8}; we already have 3 and 5 in the first column, so the 8 has to go there.

For a, we now have _9_435_8_, so its last digit must be one of {1, 6}. Any square ending in 6 is either (10x + 4)2 = 100x2 + 10(8x) + 16 or (10x + 6)2 = 100x2 + 10(12x) + 36, so its penultimate digit is the last digit of an even multiple of x plus one of {1, 3}, ie an odd digit; thus there are no squares ending in 86, so a ends in 1. Similarly, for e we have ___843_6_ and the last digit is one of {5, 9} (as we already have 1 in the bottom row); but any square ending in 5 ends in 25 (because (10x + 5)2 = 100x2 + 100x + 25), so e has to end in 9, which forces the prime Q to end in 7. The last digit of the square A has to be one of {1, 5} but A can’t end in 25 because there’s a 2 in column 8, so it ends in 1. It’s either (10x + 1)2 = 100x2 + 10(2x) + 1 or (10x + 9)2 = 100x2 + 10(18x) + 81, so its penultimate digit is even; similarly, T ending in 9 is either (10x + 3)2 = 100x2 + 10(6x) + 9 or (10x + 7)2 = 100x2 + 10(14x) + 49, so its penultimate digit is also even; that leaves only one place for the 5 in column 7, in t = 251.

For e, we have 1__843769; 125843769 isn’t a square, so e = 152843769. In column 4, the only place for the 9 to go is at the top. The digits missing from that column are then {2, 3, 6} and we already have {2, 3} in row 3, so the 6 must go there; then the 3 can only go at the bottom, which forces g = 267. Then the first digit of D can only be 7. Then a is either 297435681 or 697435281; the former isn’t a square, so a = 697435281. The only place left for the last 2 is at the top of column 3. Then b, with 8 factors, is 2_ and the second digit is one of {1, 3, 4}; 21 = 7×3 has only 4 factors and 23 is prime, so b = 24. Then the third digit of P can only be 9, so P = 2398. That forces the last digit of k to be 1, then its first digit to be 5, ie k = 561. That then forces r = 8937 and R = 234975.

The bottom digit of column 2 can only be 5, leaving the 7 to go at the top. The last 5 has to go at the top of column 6. To complete row 5, the first digit of q has to be 9. Row 2 is missing only {1, 3} and it already has a 3 in column 6, so C = 428137, which forces E = 8132 to complete row 3.

The square A is one of {672935481, 672935841}; 672935841 isn’t a square, so A = 672935481. That forces q = 976 and then v = 58 to complete column 7. Then Q = 14657 is forced, with s = 947, and to complete the grid, T = 157326849.

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