Say, that we have N-1 lines, with N-1 primes. Each N digits. What we now need is an N digit prime number to put it below.
Its most significant digit may be 1, 3, 7 or 9. Otherwise, the leftmost vertical number wouldn’t be prime. If the sum of all N-1 other rightmost digits is X, then:
If X mod 3 = 0, then just 1 and 7 are possible, otherwise the leftmost vertical would be divisible by 3.
If X mod 3 = 1, then 1, 3, 7 and 9 are possible.
If X mod 3 = 2, then just 3 and 9 are possible, otherwise the leftmost vertical would be divisible by 3.
The probability is (1/3)*(((1+2+1)/5))=4/15 that the first digit fits. (4/15)^N, that all N digit fit.
Actually, we must consider the probability of divisibility by 11, which is roughly 1⁄11, which further reduces 4⁄15 per number to 40⁄165. And with 7 … and so on.
For the divisibility with 3, we render out not only one permutation of N-1 primes but all of them. For the divisibilty with 11, some of them.
It is indeed quite complicated. But if you handwavily estimate the results of all that complexity—the probabilities of divisibility by various things—then the estimate you get is the one cousin_it gave earlier, because the Prime Number Theorem is what you get when you estimate the density of prime numbers by treating divisibility-by-a-prime as a random event. (Which for many purposes works very well.)
There are 143 primes between 100 and 999. We can, therefore, make 2,924,207 3x3 different squares with 3 horizontal primes. 50,621 of them have all three vertical numbers prime. About 1.7%.
There are 1061 primes between 1000 and 9999. We can, therefore, make 1,267,247,769,841 4x4 different squares with 4 horizontal primes. 406,721,511 of them have all four vertical numbers prime. About 0.032%.
I strongly suspect that this goes to 0, quite rapidly.
How many Sudokus can you get with 9 digit primes horizontally and vertically?
Not a single one. Which is quite obvious when you consider that you can’t have a 2, 4, 6, or 8 in the bottom row. But you have to, to have a Sudoku, by the definition.
I’m sure you’re right that the fraction of all-horizontals-prime grids that have all verticals prime tends to 0 as the size increases. But the number of such grids increases rapidly too.
Say, that we have N-1 lines, with N-1 primes. Each N digits. What we now need is an N digit prime number to put it below.
Its most significant digit may be 1, 3, 7 or 9. Otherwise, the leftmost vertical number wouldn’t be prime. If the sum of all N-1 other rightmost digits is X, then:
If X mod 3 = 0, then just 1 and 7 are possible, otherwise the leftmost vertical would be divisible by 3. If X mod 3 = 1, then 1, 3, 7 and 9 are possible. If X mod 3 = 2, then just 3 and 9 are possible, otherwise the leftmost vertical would be divisible by 3.
The probability is (1/3)*(((1+2+1)/5))=4/15 that the first digit fits. (4/15)^N, that all N digit fit.
Actually, we must consider the probability of divisibility by 11, which is roughly 1⁄11, which further reduces 4⁄15 per number to 40⁄165. And with 7 … and so on.
For the divisibility with 3, we render out not only one permutation of N-1 primes but all of them. For the divisibilty with 11, some of them.
It’s quite complicated.
It is indeed quite complicated. But if you handwavily estimate the results of all that complexity—the probabilities of divisibility by various things—then the estimate you get is the one cousin_it gave earlier, because the Prime Number Theorem is what you get when you estimate the density of prime numbers by treating divisibility-by-a-prime as a random event. (Which for many purposes works very well.)
There are 143 primes between 100 and 999. We can, therefore, make 2,924,207 3x3 different squares with 3 horizontal primes. 50,621 of them have all three vertical numbers prime. About 1.7%.
There are 1061 primes between 1000 and 9999. We can, therefore, make 1,267,247,769,841 4x4 different squares with 4 horizontal primes. 406,721,511 of them have all four vertical numbers prime. About 0.032%.
I strongly suspect that this goes to 0, quite rapidly.
How many Sudokus can you get with 9 digit primes horizontally and vertically?
Not a single one. Which is quite obvious when you consider that you can’t have a 2, 4, 6, or 8 in the bottom row. But you have to, to have a Sudoku, by the definition.
It’s a bit analogous situation here.
I’m sure you’re right that the fraction of all-horizontals-prime grids that have all verticals prime tends to 0 as the size increases. But the number of such grids increases rapidly too.