This is one possible solution path. Clues of the form “n : 0” say that there are n prime factors and they are all the same, which means the answer is pⁿ for some prime p. Thus 18ac is p¹⁰ and has four digits; 3¹⁰ = 59049 is too big, so 18ac = 1024 = 2¹⁰. Similarly, 5dn is p¹¹ in four digits, so 5dn = 2048 = 2¹¹. And 20ac is p⁵ in two digits; 3⁵ = 243 is too big, so 20ac = 32 = 2⁵. For 17dn, from the sequence {…, 5⁴ = 625, 7⁴ = 2401, 11⁴ = 14641, …} the only one that fits is 17dn = 2401. For 21dn, p⁴ in two digits is one of {2⁴ = 16, 3⁴ = 81}.

For clues where the difference between the lowest and highest prime factors is odd, the lowest factor must be 2 (because all other primes are odd). For 22ac, clued as “2 : 1427”, the (only) two prime factors must be 2 and 1429, so 22ac = 2858 = 2×1429. For 13dn “3 : 1281”, the answer is 2×p×1283, to fit _132 in the grid, so the middle prime p must end in 2 or 7; 2×7×1283 = 17962 is too big, so 13dn = 5132 = 2×2×1283.

Clues of the form “n : 2” have an answer made up of only two (odd) primes with a difference of 2. For 2dn “2 : 2”, if the two primes are 11 and 13, the answer is 11×13 = 143, too big for the two-digit answer; so 2dn is one of {3×5 = 15, 5×7 = 35}. For 1ac “6 : 2”, if the two primes are 5 and 7, the minimum value is 5⁵×7 = 21875, too big for the four-digit answer; so the two primes are 3 and 5, making the answer 3^r×5^{(6 − r)}, where r is in the range [1, 5]. The possible values are {9375, 5625, 3375, 2025, 1215} but 2dn needs the second digit to be 1 or 3, so 1ac is one of {9375, 3375}, which makes 2dn = 35. Now 3dn matches 7___ and the clue “10 : 1” means it’s 2^r×3^{(10 − r)}, where r is in the range [1, 9]; 2⁴×3⁶ = 11664 is too big and 2⁶×3⁴ = 5184 is too small, so 3dn = 7776 = 2⁵×3⁵.

For 10ac “3 : 104”, the lowest and highest prime factors are p and p + 104, which must be either {3, 107} or {5, 109} (7 + 104 = 111 is a multiple of 3; 11 + 104 = 115 is a multiple of 5, and the minimum of 11²×115 = 13915 is too big anyway). For {5, 109}, the product 5×p×109 must end in 75 (because we have the 7 in the grid) and therefore be a multiple of 25, requiring p to be 5, but 5×5×109 = 2725 doesn’t match. So we have 3×p×107; for primes from 3 upwards, the possible products are {963, 1605, 2247, 3531, 4173, 5457, 6099, 7383, 9309, 9951, 11877, …}, of which the only match is 10ac = 4173 = 3×13×107.

1dn “4 : 16” has to match 3_4_ or 9_4_ in the grid. Its extreme prime factors p and p + 16 must be either {3, 19} or {7, 23} (5 + 16 = 21 and 11 + 16 = 27 are multiples of 3, and for the latter the minimum of 11³×27 = 35937 is too big anyway). For {7, 23}, the minimum products are {7³×23 = 7889, 7²×11×23 = 12397, …}, none of which fit the grid, so 1dn is 3×pq×19, with the (odd) product pq either between 3040/3/19 ≈ 53.3 and 3949/3/19 ≈ 69.3, or between 9040/3/19 ≈ 158.6 and 9949/3/19 ≈ 174.5. Of the odd numbers from 159 to 173, the only ones that are products of two primes are {159 = 3×53, 161 = 7×23, 169 = 13×13}, but we need both primes to be  <= 19, which rules out the first two; for the other, 3×13×13×19 = 9633 doesn’t match 9_4_, so we need 3_4_, making 1ac = 3375. Of the odd numbers from 55 to 69, the products of two primes are {55 = 5×11, 57 = 3×19, 65 = 5×13, 69 = 3×23}, of which we can rule out the last because 23 > 19; the other three give products of {3135, 3249, 3705}, of which the only match is 1dn = 3249 = 3×3×19×19.

Turning to the northeast corner, 4ac “3 : 69” must be 2×p×71, to fit _2__, so p is between 1202/2/71 ≈ 8.5 and 9298/2/71 ≈ 65.5, ie a prime in the range [11, 61]. 9ac “5 : 1” is made up of 2 and 3 only, giving {2⁴×3 = 48, 2³×3² = 72, 2²×3³ = 108, 2×3⁴ = 162}, ie it’s one of {48, 72}. 11ac “4 : 11” is 2×pq×13 where p and q are primes from 3 to 13. Thus the first three digits of 6dn are all even and its clue “2 : 594” says 6dn = p(p + 594). We can rule out p = 3 because 597 is a multiple of 3 and p = 11 because 605 is a multiple of 5; from the rest, the potential products are {5×599 = 2995, 7×601 = 4207, 13×607 = 7891, 17×611 = 10387, …}, of which only 6dn = 4207 has the required even digits, making 9ac = 72. Now 4ac is 2×p×71 = _2_4, so the prime p ends in 7, limiting it to {17, 37, 47}; of the respective products {2414, 5254, 6674}, the only one that fits is 4ac = 5254. For 11ac “4 : 11” we have _4_0 in the grid, so one of its factors is 5, ie it’s 2×5×p×13 where p×13 matches _4_ and p > 140/13 ≈ 10.8; the only primes in range are {11, 13}, giving {1430, 1690} respectively, and the one that fits is 11ac = 1430.

In the southeast corner, 23ac “6 : 4” matches __61 or __11 (since 21dn is 16 or 81) and therefore doesn’t have a factor of 5. If the smallest factor is 7, the minimum product is 7⁵×11 = 184877, too big; therefore it’s 3^r×7^{(6 − r)}, ie one of {3⁵×7 = 1701, 3⁴×7² = 3969, 3³×7³ = 9261, 3²×7⁴ = 21609, 3×7⁵ = 50421}, of which only 23ac = 9261 fits, making 21dn = 16.

For 15dn “4 : 8” we have _2_5 in the grid, so one of its prime factors is 5. It’s extreme prime factors are 8 apart (both odd), so they must be either {3, 11} or {5, 13}. In the first case we have 3×5×p×11, where p is one of {3, 5, 7, 11}; the corresponding products are {495, 825, 1155, 1815}, none of which fits the grid. So 15dn is 5×pq×13 where p and q are in {5, 7, 11, 13}; of the 10 combinations, the only ones that match _2_5 are {5×5×7×13 = 2275, 5×5×13×13 = 4225, 5×11×13×13 = 9295}

16dn “4 : 29” has extreme prime factors of 2 and 31 and has to match ___2 in the grid, ie it’s 2×pq×31 where p and q are primes from 2 to 31 (not 5, which would make the product end in 0) and pq ends in 1 (if they’re both odd) or 6 (if one of them is 2); let’s say p is the lower one. If p is 13 then q must end in 7 and has to be ≥ 17, but 2×13×17×31 = 13702 is too big. If p is 11 then q ends in 1 and is one of {11, 31}, but 2×11×31×31 = 21142 is too big. Considering the similar options for p in {2, 3, 7}, we get seven possibilities for 16dn: {2×2×13×31 = 1612, 2×2×23×31 = 2852, 2×3×7×31 = 1302, 2×3×17×31 = 3162, 2×7×13×31 = 5642, 2×7×23×31 = 9982, 2×11×11×31 = 7502}.

19ac “7 : 9” has extreme prime factors of 2 and 11 and has to match ___4 in the grid, so it can’t have a factor of 5. If 11 is used three times, the minimum product is 2⁴×11³ = 21296, too big. If there’s only one factor of 2, the minimum product is 2×3⁵×11 = 5346 (not ending in 4) and the next is 2×3⁴×7×11 = 12474 (too big); so 19ac is 2²×pqrs×11, where pqrs ends in 1 or 6 and is between 1004/2/2/11 ≈ 22.8 and 9994/2/2/11 ≈ 227.1. Taking p  <= q  <= r  <= s, if s is 11 then pqr is between 2³ = 8 and 9994/2²/11² ≈ 20.6, ie one of {11, 16}; but neither of those is the product of three primes, so p, q, r and s are all in {2, 3, 7}. Of the nine combinations that make pqrs end in 1 or 6, the ones with a four-digit total product are {2⁵×7×11 = 2464, 2⁴×3²×11 = 1584, 2⁴×7²×11 = 8624, 2³×3²×7×11 = 5544, 2²×3⁴×11 = 3564}. The second digit of 19ac has to be the same as the second digit of 16dn, so it’s either 5, with 16dn being 7502 and 19ac in {1584, 3564, 5544}, or 6, with 16dn in {1612, 5642} and 19ac being 8624.

We now have to consider the second version of the grid, with all the entries anagrammed, which we’ll indicate like this. To make all the new entries distinct from the old ones, all the two-digit answers must be reversed. That makes 1ac 3375 → _5__; its first digit has to be from 1dn’s 3249 and the third digit is from 3dn’s 7776, so the only possibility is 1ac = 3573. For 4173 at 10ac, the first digit is 2|4|9 and the third digit is 6|7, so the 4 and 7 are fixed; to be different it must be 10ac = 4371. Now 3dn is 7_7_ and to be different from 7776 it must be 3dn = 7677. Similarly, 1dn is 3_4_ and we need 3249 → 1dn = 3942.

For the anagram of 4ac 5254, the second and fourth digits must be 2 and 4 (from the crossings with 5dn and 6dn), and to be different from 4ac we need 4ac = 5452. The second digit of 11ac is from 028 (5dn) and 1430 (11ac), ie it’s 0. At 6dn we have 4207 → 27__ and the third digit is from 143 (11ac), so 6dn = 2740.

At 13dn we have 5132 → __2_; the second digit is from 18ac 1024, so it’s 1; the last digit is from 22ac 2858, so it’s 5; thus 13dn = 3125. Now 22ac is 2858 → 5___, but 15dn is one of {2275, 4225, 9295} and doesn’t contain an 8, so 22ac = 5828 and 15dn ends in 2. The second digit of 15dn is 2 or 4 (from 18ac’s 024), which reduces 15dn to one of {2275, 4225}.

For 23ac we have 9261 → __1_ and the last digit is from 2401 (17dn), ie it’s 2; the second digit is from 16dn, which is in {1612, 5642, 7502}, ie it’s 6, making 23ac = 9621 and limiting 16dn to {1612, 5642} and making 19ac = 8624. For 17dn we have 2401 → ___2 and the second digit is from 8624 (19ac), ie 4; the 0 can’t go at the start, so 17dn = 1402. Now 16dn is ___6 and the second digit is from 862 (19ac), which can only be 2.

We now have the initial and second grids partially completed as shown. For each of the unclued entries, the number by which the initial value is multiplied to get its anagram must be in the range [2, 9], because anything higher would give more than six digits. Their first digits are from {1, 2, 3, 4}, so that multiplying them by at least 2 doesn’t similarly overflow.

For 12ac we have _6__8_ → _7__2_ or _7__8_. The 7 in the anagram is the ones digit of m×6 + (any digit carried from m×(the third digit)), where m is the multiplier and the carry is in the range [0, m − 1]. If m is 2 then the anagram’s second digit is from 2×6 + {0, 1} = {2, 3}; if it’s 3, the digit is 3×6 + {0, 1, 2} = {8, 9, 0}; so the multiplier is m ≥ 4, 12ac must start with 16 (because 4×26 = 104 would produce more than six digits), and m  <= 6 (because 7×16 = 112 would produce more than six digits). In that range, the only ways to get 7 for the second digit are {4×16 + 3 = 67, 6×16 + 1 = 97}. The only ways to multiply the fifth digit by 4 or 6 to convert 8 to 2 or 8 are {4×8 + 0 = 32, 6×8 + 0 = 48, 6×8 + 4 = 52}; to get those carried digits, we need one of {4×80 = 320, 4×81 = 324, 4×82 = 326, 6×80 = 480, 6×81 = 486, 6×87 = 522, 6×88 = 528}. If m is 6, the options are {6×16__80 = 97__80, 6×16__81 = 97__86, 6×16__87 = 97__22, 6×16__88 = 97__28}. We can rule out 16__87 → 97__22 and because the first number has no place for the 2, 2 and 9; similarly, 16__88 and 97__28 can’t be anagrams. Also, 16__80 → 97__80 would consist of the digits 016789 in some order and 16__81 → 97__86 would use 116789; but neither set of digits adds up to a multiple of 3, as the digits of any number 6n must. So the multiplier is 4 and the options are {4×16__80 = 67__20, 4×16__81 = 67__24, 4×16__82 = 67__28}; 16__81 and 67__24 can’t be anagrams, which leaves 12ac as one of {162780, 167280, 16_782, 167_82}. Multiplying the first two by 4 gives {651120, 669120}, not anagrams, so 12ac is 16__82, with a 7 in it, and the third digit has to be at least 7, to give a carry of 3 to add to 4×16 = 64 to give 67; and 12ac is 67__28, which makes 5dn = 4802.

For 7dn we have _3__4_ → _1____, with the 4 becoming one of {0, 2, 4}. The multiplier can’t be > 7 because 8×13 = 104 would add an extra digit at the front. To get the 1, we need multipliers and carries from {3×3 + 2, 6×3 + 3, 7×3 + 0}. If m is 6, the last two digits multiply from 6×40 = 240 to 6×49 = 294, none of which has 0, 1 or 2 as the penultimate digit. If m is 7, 7dn starts with 13 and 7dn starts with 91 and the options are {7×13__43 = 91__01, 7×13__44 = 91__08, 7×13__45 = 91__15, 7×13__46 = 91__22, 7×13__47 = 91__29}. We can rule out 13__43 → 91__01 because the first number has no place for the 0, 9 and another 1; similarly, none of the other pairs can be anagrams, so the multiplier is 3. The first digit can’t be ≥ 3, because 3×33 + 2 (carried) = 101, adding an extra digit at the front. The options are {13__40 → 41__20, 13__41 → 41__23, 13__42 → 41__26, 13__43 → 41__29, 13__47 → 41__41, 13__48 → 41__44, 13__49 → 41__47, 23__40 → 71__20, 23__41 → 71__23, 23__42 → 71__26, 23__43 → 71__29, 23__47 → 71__41, 23__48 → 71__44, 23__49 → 71__47}, of which we can rule out many because they can’t be anagrams, or because the six digits used don’t make a multiple of 3. From 12ac, the third digit must be ≥ 7, but 13__48 → 41__44 would need 134448 and 23__47 → 71__41 would need 231147, so they’re ruled out, leaving {13__40 → 41__20, 13__41 → 41__23, 13__42 → 41__26, 23__41 → 71__23}. For 13__40 → 41__20, the known digits are 01234 and the other one must be from {7, 8, 9}, and to make the digit sum a multiple of 3 it must be 8, giving 138240, but 3×138240 = 414720 isn’t an anagram; for 13__41 → 41__23, to the known digits 11234 we need to add 7, giving 3×137241 = 411723 as a possibility; for 13__42 → 41__26 we need 12346 and 8, but 3×138642 = 415926 isn’t an anagram; finally, 23__41 → 71__23 needs 12347 and 8, but 3×237841 = 713523 and 3×238741 = 716223 aren’t anagrams. The only combination that works is 7dn = 137241 and 7dn = 411723.

Now 12ac is 167_82 → 671_28. From 4×82 = 328, a 3 is carried into the hundreds column, and the digit there needs to be n such that 4n + 3 = n + 30 (so the same digit appears in the anagram and 3 is carried to the thousands column), which solves as n = 9, ie 12ac = 167982 and 12ac = 671928. 18ac is 1__2, an anagram of 1024, and can’t be 1402, which is the anagram at 17dn, so 18ac = 1042. Now we know that 15dn contains a 4, so 15dn = 4225.

14ac is now _42____ → _27___ or _57___. The fifth digit of 14ac is one of {1, 4, 5} from the options at 16dn; the multiplier can’t be ≥ 8 because 8×142 = 1136 would add an extra digit at the front. The multipliers and carries that can convert 2 to 7 are {3×2 + 1, 6×2 + 5, 7×2 + 3}, with a carry of 0 or 1 to the next column; those that can convert 4 to 2 are {3×4 + 0, 5×4 + 2, 7×4 + 4} and from 4 to 5 there’s only {6×4 + 1}, so the only options using the available carry of 0 or 1 are with a multiplier of 3 or 6 and in either case the numbers are divisible by 3. If the multiplier is 3, the starts of 14ac and 14ac are from {3×142 + 1 → 427, 3×242 + 1 → 727}; if it’s 6, there’s only {6×142 + 5 → 857}. If 14ac is 242___ then the multiplier is 3, giving 727___; their digits are 22477 and one of 145, of which only 5 gives a digit sum divisible by 3. The fourth digit of 14ac needs to give a carry of 1 when multiplied by 3, so it’s not 7 and is 5, making 242577, but 3×242577 = 727731 isn’t an anagram; therefore 14ac starts with 142.

For 8dn we have _19_8_, and the anagram is __9___ with the first digit being > 1, the second digit being 1 or 3 and the fifth digit being 6 or 8. The multiplier can’t be 9 because the minimum of 9×119 = 1071 would overflow. The multiplier and carry options to convert 9 to 9 are {2×9 + 1, 3×9 + 2, …, 8×9 + 7}: generalised, they’re m×9 + (m − 1), and the result carries another (m − 1) to the next column (eg 3×9 + 2 = 29, carrying the 2), where the calculation will be m×1 + (m − 1) = 2m − 1; since we know the result of that ends in 1 or 3, 2m ends in 2 or 4 and the multiplier m is one of {2, 6, 7}. The options to convert 8 to 6 are then {2×8 + 0, 7×8 + 0}; those converting 8 to 8 are {6×8 + 0, 7×8 + 2}; to convert 1 → 1 there’s {6×1 + 5, 7×1 + 4}; for 1 → 3 there’s {2×1 + 1, 7×1 + 6}. If the multiplier is 2, we have _19_8_ → _39_6_ and the 1 in the anagram can’t be the last digit (because it’s even), so the anagram is _39168 or 83916_; whatever the first digit is, _39168/2 = __9584, but there’s no place for the anagram to contain both a 4 and a 5; whatever the last digit is, 83916_/2 = 41958_, but again both a 4 and a 5 are needed. If the multiplier is 6, we have _19_8_ → _19_8_; the last digit of 8dn has to be 1, to avoid a carry into the fifth digit, ie _19_81 → _19_86; then the 6 in 8dn has to be in fourth place and the second 1 in the anagram has to be in fourth place, ie _19681 → _19186; but that means the first digit is the same in both numbers, which is impossible if multiplying by 6. Therefore the multiplier is 7, the first digit of 8dn is 1 and we have 7×119_8_ = 839_6_ or 839_8_, which makes 11ac = 3014.

If the fifth digit of 8dn is 6 then the last digit of 8dn has to be 1, to avoid a carry, ie 119_81 → 839_67, but no anagrams are possible from those digits. To convert the 8 to 8, we need a carry of 2 from the last digit, which must be 3 or 4; if it’s 4 then 7×119_84 ends in 88 and 8dn needs another 8, but 7×119884 = 839188 isn’t an anagram. Therefore 8dn is 119_83 and 8dn is 839_81; we need two 8s in 8dn, so it’s 8dn = 119883 and 8dn = 839181, which makes 19ac = 8264.

For 14ac we now have 1428__ (with the fifth digit one of 145) → __71__, so 14ac is one of {142817, 142847, 142857}. Multiplying by 3 gives {428451, 428541, 428571}, none of which fit in the grid, so the multiplier is 6 and the anagram is one of {856902, 857082, 857142}; the pair that fits is 14ac = 142857 and 14ac = 857142. That completes the initial grid with 16dn = 5642, and the anagram grid with 16dn = 4256 and 15dn = 5422.