This is one possible solution path. We’ll use the notation of the form “1ac = 1(23)45” to refer to a full clue answer containing its two-digit prime, and “[1ac] = 145” for the resulting grid entry. There are 21 two-digit primes, namely {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}, and 21 clues, so all of them are used as primes removed from answers.

5ac and 17ac are two-digit squares,
containing primes C and M which are reverses of each other.
The same applies to 3dn and 15dn with Q and Y.
The reversible prime pairs are
{(13, 31), (17, 71), (37, 73), (79, 97)}.
The entry [11dn] = G + W, so it’s even,
which means [17ac] starts with one of {2, 4, 6, 8}
and its reverse [3dn] ends with the same.
The square of a number with digits *ab* is
(10*a* + *b*)^{2} =
100*a*^{2} + 20*ab* + *b*^{2}.
If it’s odd, *b*^{2} is one of {1, 9, 25, 49, 81};
20*ab* is 2*ab* (even) followed by 0,
and 100*a*^{2} ends in 00,
so the tens digit of an odd square is always even.
Thus the reversible primes, which don’t contain any even digits,
can never be the final digits of the squares.

We know the last digit of [3dn],
which must be the last digit of the square 3dn, is even
(and non-zero, because it’s also the first digit of [17ac]),
ie it’s 4 or 6 (as no squares end in 2 or 8).
The square of an even number is a multiple of 4
(because (2*n*)^{2} = 4*n*^{2}),
so its last two digits must be divisible by 4
(because any multiple of 100 is clearly a multiple of 4).
If 3dn ends in 4 then its final digits are one of {04, 24, 44, 64, 84},
so the prime must be the first two digits,
and the only square starting with one of the reversible primes
and ending in 4 is (17)64, which would need 15dn to contain 71.
There’s no square matching (71)__,
as 7056 and 7225 are consecutive squares.
A square matching 1(71)_ would have its square root between
√1710 ≈ 41.35 and √1719 ≈ 41.46,
but there are no integers in that range;
similarly, between √2710 ≈ 52.06 and √2719 ≈ 52.14
there are no integers, and so on up to
√9710 ≈ 98.54 and √9719 ≈ 98.58.
Therefore 3dn doesn’t end in 4, it ends in 6,
and is the square of a number ending in 4 or 6.

The 14 candidates for 3dn are now
{1156, 1296, 1936, 2116, 2916, 3136, 4096, 4356, 5476, 5776, 7056, 7396, 8836, 9216}.
From a visual inspection, the only ones containing reversible primes are
{(31)36, 3(13)6, (73)96},
so [3dn] is one of {36, 96}
and [17ac] is the corresponding one of {63, 69}.
But we know the last digit of [17ac] is the last digit of the square 17ac,
so it can’t be 3,
which makes **[17ac] = 69**,
**[3dn] = 96**,
**3dn = (73)96**,
**Q = 73** and
**Y = 37**.

Now 15dn needs to contain 37.
If it’s the middle two digits,
the square root has to be in one of the ranges
[√1370 ≈ 37.01, √1379 ≈ 37.13],
[√2370 ≈ 48.68, √2379 ≈ 48.77]
and so on up to [√9370 ≈ 96.80, √9379 ≈ 96.85],
but none of those ranges contains an integer.
Therefore 37 is at the start, giving (37)21,
so **[15dn] = 21**.

Returning to 17ac, we know the square ends in 9,
so the penultimate digit is even,
which means the prime can’t be in the middle, ie 6(__)9,
so it’s (__)69, where the first two digits
are from {13, 31, 17, 71, 79, 97}.
The only matching square is (13)69,
so **M = 13** and **C = 31**.
Now 5ac needs to contain 31.
If it’s the middle two digits,
the square root has to be in one of the ranges
[√1310 ≈ 36.19, √1319 ≈ 36.32],
[√2310 ≈ 48.06, √2319 ≈ 48.16]
and so on up to [√9310 ≈ 96.49, √9319 ≈ 96.53],
but none of those ranges contains an integer.
Therefore 31 is at the start, giving (31)36,
so **[5ac] = 36**.

15ac is a five-digit cube of a two-digit prime,
ie one of {12167, 24389, 29791, 50653, 68921, 79507}.
The entry [15ac] starts with 2 and from the 6dn clue
we know [15ac] = S^{2};
the only three-digit squares starting with 2 are
{15^{2} = 225, 16^{2} = 256, 17^{2} = 289},
so 15ac = 2(43)89,
**L = 43**,
**D = 29**,
**[15ac] = 289** and
**S = 17**.

Now 6dn is a palindrome containing 17 and the entry [6dn] = 6__ = KT,
which must be odd; the only possibility is (17)671,
so **[6dn] = 671**
and {K, T} = {11, 61} in some order.
8dn has descending digits, with no repeats, and contains T,
so **T = 61** and **K = 11**.
The entry [8dn] can’t end in 0,
so the only possibility for 8dn is 987(61),
with **[8dn] = 987**.

For [7ac] we have _96, which is a multiple of 29;
the only matching number is **[7ac] = 696**;
the full answer is also a multiple of 29 and contains 29,
so it’s either (29)696 or 696(29),
but we don’t need to know which.

From the grid, [11dn] is __6;
it’s equal to G + W,
so it can’t be greater than 89 + 97 = 186;
thus it starts with 1,
making **[11ac] = 18**.
But the 11ac clue tells us [11dn] is a multiple of [11ac],
so **[11dn] = 126** = 7×18.
11ac with ascending digits including G must be one of
{1(23)8, 1(47)8, 1(67)8}
(not 1(37)8 because 37 is already assigned to Y).
Since G + W = 126, W is one of {103 (too big), 79, 59};
11dn has ascending digits, so it must be 11dn = 126(79),
with **W = 79** and **G = 47**.

Entries [18ac] and [12dn] have the same digits,
and their full answers including N and X respectively have the same digit sum,
so N and X must have the same digit sum.
Of the primes still unassigned,
the only pairs with the same digit sum are
{23, 41} and {53, 71}.
But N has ascending digits,
so **N = 23** and **X = 41**.
In the grid, [18ac] = _1_
and it’s a multiple of 79, ie one of {316, 711}.
[12dn] has the same digits,
so it’s one of {163, 613, 117},
which means [14ac] is one of {71, 76},
so the possibilities for the triangular number 14ac, including 11,
are {1171, 7111, 1176, 7116, 7611}.
Any triangular number *t* = *n*(*n* + 1)/2,
with *n* and *n* + 1 being
the integers either side of √(2*t*).
For 7611, we have √(2×7611) ≈ 123.38,
with 123×124/2 = 7626, not a match.
Both √(2×7111) ≈ 119.26
and √(2×7116) ≈ 119.30
give 119×120/2 = 7140, not a match.
Both √(2×1171) ≈ 48.39
and √(2×1176) ≈ 48.50
give 48×49/2 = 1176, so 14ac = (11)76,
**[14ac] = 76**,
**[12dn] = 163** and
**[18ac] = 316**.
The full answer 12dn could be any of
{(41)163, 1(41)63, 16(41)3, 163(41)}
but it doesn’t matter which.

In the grid we have [10dn] = __8,
and 10dn including V has ascending digits.
Of the remaining unassigned primes with ascending digits,
{19, 59, 67, 89}, only **V = 67**
can be inserted to keep 10dn’s digits in order, being __(67)8.
Of the ten possibilities
{12678, 13678, 14678, 15678, 23678, 24678, 25678, 34678, 35678, 45678},
only 24678 is divisible by the sum of its digits,
so 10dn = 24(67)8 and **[10dn] = 248**,
making **[13ac] = 41**.

From the grid, [16dn] = 9_ and [1dn] = _6,
which can’t be any of {36, 76, 96}
because they appear elsewhere in the grid.
From 16dn’s clue, the sum of the two entries is a square.
If [1dn] is 16, the sum is in the range [16 + 90 = 106, 16 + 99 = 115],
but there are no squares in that range;
similarly, if [1dn] is any of {56, 66, 86}
there are no squares in the range of sums;
so the two entries are either 26 + 95 = 121 or 46 + 98 = 144.
The unassigned primes are {19, 53, 59, 71, 83, 89, 97}.
16dn contains Z and is divisible by Z.
If it has Z in the middle then it’s one of
{9(19)5, 9(53)5, 9(59)5, 9(71)5, 9(83)5, 9(89)5, 9(97)5} or
{9(19)8, 9(53)8, 9(59)8, 9(71)8, 9(83)8, 9(89)8, 9(97)8},
but none of those numbers is divisible by its central prime.
So the prime is either at the start or the end,
which means 16dn is either 100Z + [16dn]
or 100×[16dn] + Z (it doesn’t matter which);
either way, the entry [16dn] is also a multiple of Z.
The only available prime that’s a factor of 95 or 98 is
**Z = 19**,
with **[16dn] = 95**,
16dn = (19)95 or 95(19),
making **[1dn] = 26**.

P + Z is a triangular number;
adding 19 to the remaining primes gives
{72, 78, 90, 102, 108, 116}.
72 = 12×12/2, so can’t fit the triangular formula
of *n*(*n* + 1)/2.
Of the rest, 78 = 12×13/2 is triangular
but the others aren’t;
so **P = 59**.
We don’t need to know what 1dn is,
but of the three possibilities {(59)26, 2(59)6, 26(59)},
only 2596 is divisible by 59.

18ac is one of {(23)316, 3(23)16, 31(23)6, 316(23)}
and is a multiple of H,
which is one of {53, 71, 83, 89, 97}.
Of the 20 combinations,
only 31623/83 gives an integer,
so **H = 83**.
The triangular entry [12ac] starts with 1,
so it must be one of {10, 15}.
12ac is also triangular and contains H,
so it’s one of
{10(83), 15(83), 1(83)0, 1(83)5, (83)10, (83)15}.
Checking the integers either side of √(2*t*) as above,
the only one that gives a matching triangular value is 1830,
so **[12ac] = 10**.

[9ac] = _2 is the reverse of another entry;
the only entries matching 2_ are [15dn] = 21 and [1dn] = 26,
so [9ac] is one of {12, 62}.
9ac contains F and is a multiple of R,
which are both from {53, 71, 89, 97}.
If F is at the end, 9ac is 12(__) or 62(__) and odd;
the matching multiples of 53 are {1219, 6201},
for 71 there’s only {1207},
89 has no matching odd multiples
and 97 only has {1261},
none of which ends in one of the remaining primes.
If F is at the start, the possibilities are
{(53)12, (53)62, (71)12, (71)62, (89)12, (89)62, (97)12, (97)62},
none of which is divisible by one of the other primes.
So F is in the middle, and of the possibilities
{1(53)2, 6(53)2, 1(71)2, 6(71)2, 1(89)2, 6(89)2, 1(97)2, 6(97)2}
only 6(53)2 = 71×92 works,
so **[9ac] = 62**,
**F = 53** and
**R = 71**.

The remaining primes are {B, J} = {89, 97} in some order.
13ac is a prime formed from the entry
[13ac] = 41 with J inserted somewhere.
If J is 97 then the digit sum of 13ac would be 21,
so 13ac would be a multiple of 3;
therefore **J = 89** and **B = 97**.
We don’t need to know what 13ac is,
but 41(89) = 59×71 and 4(89)1 = 67×73,
so it’s the prime (89)41.

[4dn] = _60 and 4dn is a multiple of 97, containing 71,
one of {(71)_60, _(71)60, _6(71)0, _60(71)}.
Of the 36 combinations,
only 2(71)60 = 97×280 fits,
so **[4dn] = 260**.
Finally, 2ac contains 97 and is a multiple of its entry [2ac] = _92.
Of the 36 possible combinations,
only 5(97)92 = 592×101 fits,
so **[2ac] = 592** and the grid is complete.