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 = x^{2},
K = x^{6},
S = x^{4},
for some values of x.
K has two digits,
so it can only be 2^{6}
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
2^{4} = 16 or 3^{4} = 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 x^{20}
or x^{6}y^{2};
we can rule out the former because
2^{20} = 1048576 has far more than m’s three digits.
Trying primes for the latter,
2^{6}×3^{2} = 576
but all other combinations are too big
(2^{6}×5^{2} = 1600,
3^{6}×2^{2} = 2916 etc),
so **m = 576**.

Similarly, p has four digits and is clued by 35 = 7×5;
no value of x^{34} fits,
so it must be x^{6}y^{4}.
Trying primes,
2^{6}×3^{4} = 5184 but the rest are too big
(2^{6}×5^{4} = 40000,
3^{6}×2^{4} = 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 x^{8}
or x^{2}y^{2}
and one of its prime factors must be 2 (because j is even);
2^{8} = 256 doesn’t fit,
so j = 4y^{2} for some (odd) prime y;
for j to end in 6,
y^{2} 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;
2^{13} = 8192 is too big,
so u = x^{6}y.
If x is 3,
3^{6}×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 {x^{15},
x^{7}y,
x^{3}y^{3},
x^{3}yz,
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 x^{15} (only one prime factor)
and x^{3}y^{3}
(2^{3}×3^{3} = 216);
and 5 is not a factor.
Repeatedly dividing the four candidates by 2 and then by 3,
we get {348 = 2^{2}×3×29,
384 = 2^{7}×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 {x^{11},
x^{5}y,
x^{3}y^{2},
x^{2}yz}.
For the first,
2^{11} = 2048 is too big;
for the third,
2^{3}×3^{2} = 72 clashes with the 7 already in the row and
3^{3}×2^{2} = 108 is too big;
for the fourth (avoiding having both 2 and 5 as factors),
2^{2}×3×7 = 84 clashes with the 8 in the row,
3^{2}×2×7 = 126 is too big.
So we’re left with x^{5}y,
and 3^{5} = 243 is too big,
so the only value that fits is
**B = 96** (= 2^{5}×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
{x^{11},
x^{5}y,
x^{3}y^{2},
x^{2}yz}.
We can eliminate x^{11},
as 2^{11} = 2048 is too big.
For x^{5}y,
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 5^{5} = 3125 is too big;
so J isn’t x^{5}y.
For x^{3}y^{2},
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 11^{3} = 1331 is too big,
so J isn’t x^{3}y^{2}
and must be x^{2}yz.
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
7^{2}×11×13 = 7007, too big;
579 is not a multiple of 2 or 5 but 579 = 3×193,
so we’d need 193 = x^{2}y with primes greater than 5,
but 7^{2}×11 = 539 is too big;
578 is not a multiple of 3 or 5 but
578 = 2×289 and 289 is recognisable as 17^{2},
so 578 = 17^{2}×2,
which doesn’t match x^{2}yz.
Therefore **J = 572**
(which we can verify as 2^{2}×11×13).

H is clued by 4,
which can only be broken down as 2×2,
so its prime factorisation is either x^{3} 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
{x^{17},
x^{8}y,
x^{5}y^{2},
x^{2}y^{2}z}.
2^{17} = 131072 is too big;
2^{8}×3 = 768 clashes with the 6 already in the row,
and 2^{8}×7 = 1792
and 3^{8} = 6561 are too big.
For the fourth option,
3^{2}×5^{2}×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
{2^{2}×3^{2}×7 = 252,
2^{2}×3^{2}×11 = 396,
2^{2}×3^{2}×13 = 468,
2^{2}×3^{2}×17 = 612,
2^{2}×3^{2}×19 = 684,
2^{2}×3^{2}×23 = 828,
2^{2}×7^{2}×3 = 588,
3^{2}×7^{2}×2 = 882},
all of which either have a repeated digit or clash with the 6 in the row.
So N has to be x^{5}y^{2};
3^{5}×5^{2} = 6075 is too big,
so one of the prime factors has to be 2, ruling out 5;
the only possibilities with three digits are
2^{5}×3^{2} = 288 with a repeated digit and
3^{5}×2^{2} = 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
{x^{19},
x^{9}y,
x^{4}y^{3},
x^{4}yz} with 2 to a power greater than 1 and no factor of 5.
For the first option,
2^{19} = 524288 is too big.
For the second,
2^{9} = 512,
so to give an answer ending in 6 the prime y would have to end in 3,
but 2^{9}×3 = 1536
and 2^{9}×13 = 6656 don’t fit
and 2^{9}×23 = 11776 is too big.
For the third,
neither of {2^{4}×3^{3} = 432,
2^{4}×7^{3} = 5488} fits
and 2^{4}×11^{3} = 21296 is too big,
while 3^{4}×2^{3} = 648 is too small
and 7^{4}×2^{3} = 19208 is too big.
So M is 2^{4}xy, 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
{x^{9}y,
x^{4}y^{3},
x^{4}yz}
(x^{19} being too big).
For the first option,
consecutive values are
2^{9}×13 = 6656,
2^{9}×17 = 8704,
neither of which fits,
and 3^{9} = 19683 is too big.
For the second option,
with x = 2 consecutive values are
2^{4}×7^{3} = 5488,
2^{4}×11^{3} = 21296;
with x = 3 consecutive values are
3^{4}×2^{3} = 648,
3^{4}×5^{3} = 10125;
with x = 5 consecutive values are
5^{4}×2^{3} = 5000,
5^{4}×3^{3} = 16875;
with x = 7 the minimum value is
7^{4}×2^{3} = 19208;
so there are no values that fit.
Thus n must be x^{4}yz.
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 23^{2},
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**
(= 2^{4}×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} = 100x^{2} + 10(8x) + 16 or
(10x + 6)^{2} = 100x^{2} + 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} = 100x^{2} + 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} = 100x^{2} + 10(2x) + 1 or
(10x + 9)^{2} = 100x^{2} + 10(18x) + 81,
so its penultimate digit is even;
similarly, T ending in 9 is either
(10x + 3)^{2} = 100x^{2} + 10(6x) + 9 or
(10x + 7)^{2} = 100x^{2} + 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**.