This is one possible solution path. No two entries in the same grid can end with the same digit, and all the non-final digits must be distinct, too. It will be useful to note that the nth triangular number is n(n + 1)/2, and the “triangular root” n of a number T is the nearest integer below √(2T).

Left grid

We’ll start with the left grid, where the preamble tells us 6ac is a number that when multiplied by 3ac gives 3dn, a number starting with the same digit. If 6ac starts with 2 and 3ac starts with the digit n, then the product is greater than (2×10)(n×100) = 2n×1000, whose first digit is at least 2n, which can’t be the same as n (unless n is 0, not allowed). The same applies if 6ac starts with any digit > 2, so 6ac must start with 1. If it ends with 0 then 3dn will also end with 0, but we’re not allowed two entries with the same last digit; if it ends with 1 then 3ac and 3dn will have the same last digit; therefore 6ac is at least 12.

The two non-final digits of 3ac must be different from 6ac’s 1 (and from each other), and its last digit can’t be 0 or 1 (which would make the last digit of 3dn the same as that of 3ac or 6ac respectively), so the minimum for 3ac is 232. 6ac can’t be 13, which is neither triangular nor even. If it’s 14 then 3dn is at least 14×232 = 3248, starting with a higher digit than 3ac, and the discrepancy would only be greater for larger values of 3ac or 6ac. Therefore 6ac = 12, not triangular but even.

Now if 3ac starts with 4 (or higher), the minimum is 421 (prime), but then 3dn would be 5052, starting with a different digit; therefore 3ac starts with 2 or 3. If it starts with 2 it can’t be a palindrome (because 6ac ends with 2) and must be prime; if it starts with 3, it ends in either 3 as a palindrome or {3, 7, 9} as a prime; either way, 3ac ends with one of {3, 7, 9}. If it starts with 2, 3dn is ≤ 2998, so 3ac ≤ 2998/12 ≈ 249.8, ie one of {233, 237, 239, 243, 247, 249}; 237, 243 and 249 are all multiples of 3 and 247 = 13×19, which leaves the primes {233, 239}, for which 3dn would be {2796, 2868} respectively. But 2868 would make both 5ac and 3dn end in 8, so we’d have 233×12 = 2796. If 3ac starts with 3, it’s ≤ 3998/12 ≈ 333.2, ie one of {323, 327, 329}; 327 = 3×109 and 329 = 7×47 are neither primes nor palindromes, which leaves 323×12 = 3876. So 3ac is one of {233, 323} with 3dn in {2796, 3876} respectively, ie 3ac ends in 3, 3dn ends in 6 and 2 and 3 have both been used as non-final digits.

If 3ac is 233 and 3dn is 2796, making 5ac end in 7 and 8ac (square or triangular) start with 9. If 8ac is triangular, it’s one of {903 = 42×43/2, 946 = 43×44/2, 990 = 44×45/2}. We can rule out 903 because 3ac already ends in 3, 946 because 3dn ends in 6 and 990 because its two non-final digits are the same, so 8ac has to be square, one of {900, 961}, making 4dn (triangular or prime) either 310_ or 316_. If 4dn is triangular and matches 310_, its “root” is just below a number between √(2×3101) ≈ 78.8 and √(2×3109) ≈ 78.9, but the 78th triangular number is 78×79/2 = 3081, not in the range required; if it’s triangular and matches 316_, its root is just below a number between √(2×3161) ≈ 79.5 and √(2×3169) ≈ 79.6, but the 79th triangular number is 79×80/2 = 3160, also out of range. That leaves 4dn as a prime ending in 1 or 9 (not 3, which 3ac ends in, or 7, which 5ac ends in), ie one of {3101, 3109, 3161, 3169}; 3101 = 7×443 and 3161 = 29×109 aren’t primes, so 4dn has to end in 9. Now 10ac is 9_ and there are no such Fibonacci numbers (the only two-digit ones are {13, 21, 34, 55, 89}), so 10ac has to be prime, ie 97, but we already have 5ac ending in 7. This rules out 233 for 3ac, so 3ac = 323 and 3dn = 3876.

Now 8ac (triangular or square) is 7__. If it’s triangular, the root is roughly between √(2×700) ≈ 37.4 and √(2×799) ≈ 39.97, which gives triangular numbers {37×38/2 = 703, 38×39/2 = 741, 39×40 = 780}, but we can rule out 703 (as 3ac ends in 3). If it’s square, it’s one of {729, 784} but we can rule out 729 because 4dn already has 2 as a non-final digit. So 8ac is one of {741, 780, 784}. Now 4dn (prime or triangular) is 214_ or 218_. If it’s prime it has to end in one of {1, 7, 9}, and the only prime in {2141, 2147, 2149, 2181, 2187, 2189} is 2141. If it’s triangular, its root is roughly between √(2×2141) ≈ 65.4 and √(2×2189) ≈ 66.2, which gives triangular numbers {2145, 2211}. The only possibilities for 4dn are thus {2141, 2145}, which makes 8ac = 741 (triangular). Now 4dn can’t also end in 1, so 4dn = 2145 (triangular).

Now 10ac is 5_. If it’s a Fibonacci number it can only be 55, but then both 4dn and 10ac would end in 5, so it’s prime. It can’t be 53 because 3ac ends in 3, so 10ac = 59 and we have columns 3 to 5 of the left grid completed. The 0 that isn’t a final digit can’t be in column 1, because those cells are all the starts of entries, so it must be the second digit of 2dn. The other digits of 2dn are the ones not yet used as final digits, ie {0, 4, 7}. If 2dn is square, it can’t end in 7; it also can’t end in 0 because the square would then end in 00, giving three occurrences of 0 in the grid. If it’s a square ending in 4 then 9ac is _46, which can’t be the reverse of a square (there are no squares between 25² = 625 and 26² = 676). It also can’t be a square because a square ending in 6 has an odd tens digit (it’s either (10n + 4)² = 100n² + 80n + 16 or (10n + 6)² = 100n² + 120n + 36). Therefore 2dn can’t be a square at all, so it’s prime and must end in 7; it can’t start with 0, so 2dn = 4007.

For 7ac we now have _0, not odd, so it’s 2n², which can only be 7ac = 50 = 2×5². For 1ac we have _4, the reverse of either a square or a triangular number. The only triangular number matching 4_ is 45, but 1ac can’t start with 5 (because 7ac does), so the square is 49 and 1ac = 94. We’re left with 5ac = _08 and 9ac = _76 and the available digits are 6 and 8. 876 is neither a square nor the reverse of a square, so 9ac = 676 (fitting either clue) and 5ac = 808 (not triangular, the nearest being 39×40/2 = 780 and 40×41/2 = 861, so it uses the palindrome clue). We now have the left grid complete, with all the clues assigned except for 9ac, so we can work on the right grid with almost all the clues known.

Right grid

In the right grid, 7ac is odd, and multiplying it by 9ac gives 3dn, which ends in the same digit as 9ac. If 7ac ends with any of {3, 7, 9} then for 9ac and 3dn to end in the same digit, 9ac must end in 0 or 5. If 9ac ends in 0 then either it’s a square matching _00, which would mean 2dn also ends in 0 (not allowed), or 9ac is the reverse of a square starting with 0 (also not allowed). If it ends in 5 then it’s either a square matching _25 or the reverse of a square starting with 5, ie one of {925, 675}; but the square 2dn can’t end in 2 or 7, so 9ac doesn’t end in 0 or 5, so 7ac can’t end in {3, 7, 9}. If 7ac ends in 5 then the product 3dn has to end in 0 or 5; but 9ac can’t end in 0 (as above) and if it ends in 5 it has the same last digit as 7ac. Therefore 7ac ends in 1.

The prime 4dn now ends in one of {3, 7, 9} (not 1, which is the last digit of 7ac) and 10ac (Fibonacci) starts with that digit, so 10ac = 34. The square 2dn now only has {0, 5, 6, 9} available for its last digit. A square ending in 0 ends in 00 and a square ending in 5 ends in 25, both clashing with the 1 already in third place. A square ending in 9 is either (10n + 3)² = 100n² + 60n + 9 or (10n + 7)² = 100n² + 140n + 49, both of which have an even penultimate digit, so 2dn ends in 6. 9ac is now _6_, a square from {169, 361, 961} reading forwards or backwards; it can’t end in 1 (because 7ac does) or 3 (because 4dn does), so 9ac = 169. Note that in both grids 9ac (676 and 169) is a square reading in both directions, so we can’t decide which clue goes with which grid, but it doesn’t matter.

The maximum for 3dn = 7ac×169 is 9889, so 7ac < 9889/169 ≈ 58.5 and is one of {21, 31, 41, 51}, giving {3549, 5239, 6929, 8619} respectively for 3dn; 6929 or 8619 would make 5ac end in 9 or 6, but we already have them as final digits in 9ac and 2dn, so 3dn is in {3549, 5239} and 7ac is in {21, 31}. But 3dn makes either 3ac or 8ac start with 3, so 7ac can’t start with 3 too; thus 7ac = 21 and 3dn = 3549. The digits not used so far as final digits are {0, 2, 7, 8}, so the square 8ac must end in 0, ie 8ac = 400.

The prime 3ac can’t end in 2 or 8, so it ends in 7, leaving final digits {2, 8} for 1ac and 6ac. For 1ac, there are no triangular numbers between 12×13/2 = 78 and 13×14/2 = 91, so the triangular number to be reversed is one of {21, 28}. We already have two 1s in the grid (and 21 reversed would be 12, which is already in the left grid), so 1ac = 82 and 6ac ends in 8. For 6ac (triangular), the options are {28, 78} but we already have 7ac starting with 2, so 6ac = 78.

For 2dn (square) we have 2_16, with {5, 6, 9} available for the missing digit; the only square that fits is 2dn = 2916. Now 3ac can only be one of {357, 367} but 357 is a multiple of 3, so 3ac = 367 (prime), which completes 4dn = 6703 (also prime), and the remaining 5 goes at the start of 5ac = 595 (triangular, 34×35/2).