The two factors of 12ac are 5839 = 6 × 973 + 1 and 6827 = 6 × 1138 − 1, the latter appearing in column 6, so the grid entry is 5839988.

This is one possible solution path. We’ll use the notation p:d for grid entries, to show which digits represent the lower prime (p) and which are the difference (d). To start with, it will be useful to produce a table of n values such that p = 6n±1, for the primes less than 100:

p | 5 | 7 | 11 | 13 | 17 | 19 | 23 | 29 | 31 | 37 | 41 | 43 | 47 | 53 | 59 | 61 | 67 | 71 | 73 | 79 | 83 | 89 | 97 |

n | 1 | 1 | 2 | 2 | 3 | 3 | 4 | 5 | 5 | 6 | 7 | 7 | 8 | 9 | 10 | 10 | 11 | 12 | 12 | 13 | 14 | 15 | 16 |

For 4dn and 19dn, the sum of 2 means
both factors have n = 1 and can only be 5 or 7.
The only combinations are 5×5 (entered as 5:0),
5×7 (5:2) and 7×7 (7:0).
The middle digit in 8ac can’t be 0
(because _0:_ would imply a prime ending in 0,
while _:0_ would imply a difference starting with 0),
so **4dn = 5:2** and 19d is 5:0 or 7:0.

Now 21ac ends in 0.
If it’s __:0, ie, a square,
the prime factor must have n = 12 ÷ 2 = 6, ie, p = 37;
but 37×37 = 1369, which has too many digits.
So, the entry is _:_0.
The single-digit factor must be 5 or 7, with n = 1,
so the other factor has n = 12 − 1 = 11, which can only be 67.
For the difference to end in 0,
the lower factor is 7, so **21ac = 7:60**.

For 18ac, the product has 4 digits.
If the entry is _:__, the product is less than 7×(7 + 99) = 742,
which is too small, so the entry is __:_.
From 19dn we know the middle digit is 5 or 7;
it can’t be the former because _5 is a multiple of 5, not a prime,
so the lower factor is a prime ending in 7, and we’ve resolved **19dn = 7:0**.
The n-sum for 18ac is 13 and the two primes are at most 8 apart (the difference has 1 digit);
consulting the table, the only options are 37×41 and 37×43,
the entry being 37:4 or 37:6.

Then 16dn has a middle digit of 4 or 6,
which can’t be the last digit of a two-digit prime,
so the initial prime must be 5 or 7 and the difference is 4_ or 6_.
The n-sum is 13, so the second factor has n = 13 − 1 = 12, ie 71 or 73.
The entry options are 5:66, 5:69, 7:64 and 7:66,
which fixes the last digit of **18ac = 37:6**.

For 15dn we have the entry _36 and n-sum 5.
If the first factor is a single digit, then the second factor has n = 5 − 1 = 4,
ie 23, but that’s too small for a difference of 36.
So, 15dn is _3:6 and the factors end in 3 and 9;
the only pair with an n-sum of 5 are 13 and 19, so **15dn = 13:6**.

Now 14ac is _1_ and its product has 4 digits.
An entry of _:1_ can only manage 3 digits, so it must be _1:_;
thus, we’re looking for a lower factor ending in 1 and a higher factor no more than 8 away,
with their n values summing to 11.
The only option is 31×37, so **14ac = 31:6**.

For 2dn, the n-sum is 9, so both prime factors are ≤ 47.
Neither the lower prime nor the difference can have 3 digits,
so the entry is __:__ and we’re looking for two 2-digit primes at least 10 apart.
The options are 11×41, 11×43, 13×41, 13×43, 17×37 and 19×37,
so we know the entry starts with 1 and the difference is in the range [18, 32].
That makes 10ac start with 1, 2 or 3, not a single-digit prime.
The n-sum for 10ac is 5, so the lower prime (with 2 digits) must have n = 2,
ie, 10ac starts with 11 or 13.
The difference in 2dn now starts with 1;
the only option with a difference < 20 was 19×37, so **2dn = 19:18**.

Now 6ac is _9_ with an n-sum of 7.
If the lower prime is a single digit (n = 1), the upper prime must be 37 (n = 6),
but that doesn’t give a difference starting with 9.
So, we’re looking for a lower prime ending in 9 with n ≤ 3.
The only one is 19, with 23 for the other factor, so **6ac = 19:4**.

For 1dn, we have _1_ and an n-sum of 21.
A single-digit prime for the lower factor can’t produce a 4-digit product,
so we’re looking for a lower prime ending in 1 with n ≤ 10
and an upper prime no more than 8 away.
The only option is 61×67, so **1dn = 61:6**.

For 1ac, we now have 61__ and an n-sum of 26.
The lower prime can’t be 6 (not a prime) or 61_ (giving a 6-digit product),
so it’s 61 (n = 10) and the upper prime is 97 (n = 16);
so **1ac = 61:36**.

For 3dn, we have 6_:_ and an n-sum of 23. If the lower prime is 61 (n = 10), the upper one must be 79 (n = 13), but the difference would be more than 1 digit. So the lower prime is 67 (n = 11) and the upper one is 71 or 73 (n = 12), the entry being 67:4 or 67:6. Then 8ac is 7:2_, with an n-sum of 6, so it’s 7×29 (7:22) or 7×31 (7:24).

4ac starts with 5 and has an n-sum of 15.
The lower prime must have n ≤ 7, so it can’t be 53 or 59,
leaving only 5 (n = 1).
Then the upper prime has n = 14, so it’s 83, and we have **4ac = 5:78**.
Now 5dn (n-sum = 10) starts with 7 and the middle digit is 2 or 4,
so the lower prime must be 7 (n = 1) and the upper one is 53 (n = 9),
giving the entry **5dn = 7:46**, which resolves **8ac = 7:24**.

For both 11ac and 16dn, we have _6_ and an n-sum of 13.
The lower prime must be 5 or 7 (n = 1) and the upper one is 71 or 73 (n = 12),
so the possible entries are 5:66, 5:68, 7:64, 7:66.
For 9dn (n-sum = 10) the middle digit is now 4, 6 or 8,
so its lower factor is 5 or 7 (n = 1) and the upper one is 53 (n = 9);
the entry can’t be 7:46 (already used in 5dn),
so **9dn = 5:48**, which fixes **11ac = 7:64**.

Now 22ac starts with 6 or 8 and has n-sum = 24.
The lower prime can’t be 6 or 8 (not prime)
or a 3-digit number (giving a product of more than 4 digits),
so it’s a 2-digit prime with n ≤ 12, ie, 61 (n = 10) or 67 (n = 11).
The options are 61×83 and 67×79,
with corresponding entries 61:22 and 67:12.
But 13dn must end with an even digit, so **22ac = 61:22**.

For 13dn to have a 6-digit product, the lower prime must have 3 digits,
and the difference must be the 2 that we got from 22ac.
So, we’re looking for two primes with the same n value;
the n-sum is 276, so n = 138 and the primes are
6×138 − 1 = 827 and 6×138 + 1 = 829 and **13dn = 827:2**.

For 16ac we have 5_2 or 7_2.
The n-sum is 6, so the lower prime has n ≤ 3 and can only be 5 or 7 (n = 1),
making the upper prime 29 or 31 (n = 5).
The only combination with a difference ending in 2 is 7×29,
so **16ac = 7:22**, which fixes **16dn = 7:66**.

For 20ac we have _7_; the product has 4 digits,
so the lower factor is a 2-digit prime ending in 7
and the higher factor is more than 8 away.
The n-sum is 17, so the lower prime has n ≤ 8
and the two factors must be 47 (n = 8) and 53 (n = 9),
giving **20ac = 47:6**.
That gives _:62 for 17dn,
where the lower factor can’t be 7 because 7 + 62 = 69 isn’t a prime,
so **16dn = 5:62**.

For 11dn we have 7_24, a 4-digit product and an n-sum of 30.
If the lower prime is 7 (n = 1), the upper one must be 6×29 ± 1,
but 6×29 + 1 = 175 (not a prime, obviously) and 7×125 = 875,
which doesn’t have enough digits.
The lower prime can’t be 7_2 (an even number),
so it’s 7_ with a difference of 24.
Of the three options, 71 + 24 = 95 (not a prime),
and 73 + 24 = 97 gives an n-sum of 12 + 16 = 28,
so the factors must be 79 and 79 + 24 = 103
(which is 6×17 + 1, giving the correct n-sum of 13 + 17 = 30)
and the entry is **11dn = 79:24**.

We established earlier that 10ac starts with 11 or 13,
so for 7dn we have 4__6 with 1 or 3 for the second digit.
The lower factor can’t be 4 (not a prime) or 4__ (giving a 6-digit product),
so it’s 41 or 43 (n = 7) and the upper factor has n = 13,
ie, it’s 79.
For the difference to end in 6, we must have 43×79,
so **7dn = 43:36**.
Then 10ac starts with 13, and it ends with 4 or 6 from 3dn;
it can’t be 13:6 (already used at 15dn),
so **10ac = 13:4**, which makes **3dn = 67:4**.

For 12dn we have _3_ and an n-sum of 7,
so the lower factor has n ≤ 3.
If it’s 13 (n = 2), the upper factor is 29 or 31 (n = 5),
but the difference would be 16 or 18, which have too many digits.
So the lower factor is 5 or 7 (n = 1) and the upper one is 37 (n = 6),
the entry being 5:32 or 7:30.
Then for 12ac we have _83_988, starting with 5 or 7.
If the lower factor is a single digit,
the maximum product is 7×(7 + 839988) = 5879965,
which doesn’t have enough digits;
it can’t be 58 or 78 (not a prime);
if it’s 3 digits,
the maximum is 783×(783 + 9988) = 8433693,
still not having enough digits;
if it’s 5 digits,
the product would have more then 8 digits;
so we must have a 4-digit prime matching _83_ and a difference of 988.
The higher factor must then also have 4 digits,
and the only odd 4-digit number appearing in column 6 of the grid is 6827,
so the lower factor must be 6827 − 988 = 5839
and **12ac = 5839:988**.
To verify, 6827 = 6×1138 − 1 and 5839 = 6×973 + 1,
so the n-sum is 1138 + 973 = 2111.
And 12dn is now resolved as **12dn = 5:32**, completing the grid.