Sub-prime More-guess Relief by Ruslan

Puzzle explanation

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.

Puzzle solution process

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.

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