What is the smallest solution of $42!+k=P18cdot P18cdot P18 $?
What is the smallest positive integer $k$ , such that $$42!+k$$ splits into three primes with $18$ digits ?
My currently best result is $$k=31449145975909$$ http://factordb.com/index.php?query=42%21%2B31449145975909
I found this result by choosing random primes $p$ , then choosing $k$ minimal with $pmid 42!+k$ and factoring the rest. This gives better results than just dividing $42!$ by two random primes , but still the results are probably much too high. Is there a better way than brute force ?
number-theory elementary-number-theory prime-factorization
edited Nov 25 '18 at 13:04
Moo
5,53131020
asked Nov 25 '18 at 10:41
Peter
46.7k1039125
|
show 1 more comment
What is the smallest positive integer $k$ , such that $$42!+k$$ splits into three primes with $18$ digits ?
My currently best result is $$k=31449145975909$$ http://factordb.com/index.php?query=42%21%2B31449145975909
I found this result by choosing random primes $p$ , then choosing $k$ minimal with $pmid 42!+k$ and factoring the rest. This gives better results than just dividing $42!$ by two random primes , but still the results are probably much too high. Is there a better way than brute force ?
number-theory elementary-number-theory prime-factorization
edited Nov 25 '18 at 13:04
Moo
5,53131020
asked Nov 25 '18 at 10:41
Peter
46.7k1039125
1
If we just pick the first two primes randomly and then take the samllest suitable prime as third, it will typically be $approx frac12ln 10^{17}approx 19$ too big, thus producing an offset of typically $2cdot 10^{35}$. With this in mind, your $kapprox 3cdot 10^{13}$ seems to be exceptionally small already. Then again, we have a lot of $P18$ at our disposal for trying ...
– Hagen von Eitzen
Nov 25 '18 at 11:10
Enzo Creti found the solution : $k=279 557$. Probably, it is the optimal solution.
– Peter
Nov 25 '18 at 13:36
After seeing Enzo Creti's $k$, I started brute-forcing. $k=193$, $k=6211$, $k=12973$ could be called "near misses" (three primes, smallest $>10^{15}$), still a lot to go ...
– Hagen von Eitzen
Nov 25 '18 at 13:59
@HagenvonEitzen Thank you for your double-check.
– Peter
Nov 25 '18 at 14:12
2
Closing remark: Yep, $k=279557$ is the optimal solution
– Hagen von Eitzen
Nov 25 '18 at 16:24
|
show 1 more comment
What is the smallest positive integer $k$ , such that $$42!+k$$ splits into three primes with $18$ digits ?
My currently best result is $$k=31449145975909$$ http://factordb.com/index.php?query=42%21%2B31449145975909
I found this result by choosing random primes $p$ , then choosing $k$ minimal with $pmid 42!+k$ and factoring the rest. This gives better results than just dividing $42!$ by two random primes , but still the results are probably much too high. Is there a better way than brute force ?
number-theory elementary-number-theory prime-factorization
edited Nov 25 '18 at 13:04
Moo
5,53131020
asked Nov 25 '18 at 10:41
Peter
46.7k1039125
What is the smallest positive integer $k$ , such that $$42!+k$$ splits into three primes with $18$ digits ?
My currently best result is $$k=31449145975909$$ http://factordb.com/index.php?query=42%21%2B31449145975909
I found this result by choosing random primes $p$ , then choosing $k$ minimal with $pmid 42!+k$ and factoring the rest. This gives better results than just dividing $42!$ by two random primes , but still the results are probably much too high. Is there a better way than brute force ?
number-theory elementary-number-theory prime-factorization
number-theory elementary-number-theory prime-factorization
edited Nov 25 '18 at 13:04
Moo
5,53131020
asked Nov 25 '18 at 10:41
Peter
46.7k1039125
edited Nov 25 '18 at 13:04
Moo
5,53131020
asked Nov 25 '18 at 10:41
Peter
46.7k1039125
edited Nov 25 '18 at 13:04
Moo
5,53131020
edited Nov 25 '18 at 13:04
Moo
5,53131020
edited Nov 25 '18 at 13:04
Moo
5,53131020
5,53131020
asked Nov 25 '18 at 10:41
Peter
46.7k1039125
asked Nov 25 '18 at 10:41
Peter
46.7k1039125
asked Nov 25 '18 at 10:41
Peter
46.7k1039125
46.7k1039125
1
If we just pick the first two primes randomly and then take the samllest suitable prime as third, it will typically be $approx frac12ln 10^{17}approx 19$ too big, thus producing an offset of typically $2cdot 10^{35}$. With this in mind, your $kapprox 3cdot 10^{13}$ seems to be exceptionally small already. Then again, we have a lot of $P18$ at our disposal for trying ...
– Hagen von Eitzen
Nov 25 '18 at 11:10
Enzo Creti found the solution : $k=279 557$. Probably, it is the optimal solution.
– Peter
Nov 25 '18 at 13:36
After seeing Enzo Creti's $k$, I started brute-forcing. $k=193$, $k=6211$, $k=12973$ could be called "near misses" (three primes, smallest $>10^{15}$), still a lot to go ...
– Hagen von Eitzen
Nov 25 '18 at 13:59
@HagenvonEitzen Thank you for your double-check.
– Peter
Nov 25 '18 at 14:12
2
Closing remark: Yep, $k=279557$ is the optimal solution
– Hagen von Eitzen
Nov 25 '18 at 16:24
|
show 1 more comment
1
If we just pick the first two primes randomly and then take the samllest suitable prime as third, it will typically be $approx frac12ln 10^{17}approx 19$ too big, thus producing an offset of typically $2cdot 10^{35}$. With this in mind, your $kapprox 3cdot 10^{13}$ seems to be exceptionally small already. Then again, we have a lot of $P18$ at our disposal for trying ...
– Hagen von Eitzen
Nov 25 '18 at 11:10
Enzo Creti found the solution : $k=279 557$. Probably, it is the optimal solution.
– Peter
Nov 25 '18 at 13:36
After seeing Enzo Creti's $k$, I started brute-forcing. $k=193$, $k=6211$, $k=12973$ could be called "near misses" (three primes, smallest $>10^{15}$), still a lot to go ...
– Hagen von Eitzen
Nov 25 '18 at 13:59
@HagenvonEitzen Thank you for your double-check.
– Peter
Nov 25 '18 at 14:12
2
Closing remark: Yep, $k=279557$ is the optimal solution
– Hagen von Eitzen
Nov 25 '18 at 16:24
1
1
If we just pick the first two primes randomly and then take the samllest suitable prime as third, it will typically be $approx frac12ln 10^{17}approx 19$ too big, thus producing an offset of typically $2cdot 10^{35}$. With this in mind, your $kapprox 3cdot 10^{13}$ seems to be exceptionally small already. Then again, we have a lot of $P18$ at our disposal for trying ...
– Hagen von Eitzen
Nov 25 '18 at 11:10
If we just pick the first two primes randomly and then take the samllest suitable prime as third, it will typically be $approx frac12ln 10^{17}approx 19$ too big, thus producing an offset of typically $2cdot 10^{35}$. With this in mind, your $kapprox 3cdot 10^{13}$ seems to be exceptionally small already. Then again, we have a lot of $P18$ at our disposal for trying ...
– Hagen von Eitzen
Nov 25 '18 at 11:10
Enzo Creti found the solution : $k=279 557$. Probably, it is the optimal solution.
– Peter
Nov 25 '18 at 13:36
Enzo Creti found the solution : $k=279 557$. Probably, it is the optimal solution.
– Peter
Nov 25 '18 at 13:36
After seeing Enzo Creti's $k$, I started brute-forcing. $k=193$, $k=6211$, $k=12973$ could be called "near misses" (three primes, smallest $>10^{15}$), still a lot to go ...
– Hagen von Eitzen
Nov 25 '18 at 13:59
After seeing Enzo Creti's $k$, I started brute-forcing. $k=193$, $k=6211$, $k=12973$ could be called "near misses" (three primes, smallest $>10^{15}$), still a lot to go ...
– Hagen von Eitzen
Nov 25 '18 at 13:59
@HagenvonEitzen Thank you for your double-check.
– Peter
Nov 25 '18 at 14:12
@HagenvonEitzen Thank you for your double-check.
– Peter
Nov 25 '18 at 14:12
2
2
Closing remark: Yep, $k=279557$ is the optimal solution
– Hagen von Eitzen
Nov 25 '18 at 16:24
Closing remark: Yep, $k=279557$ is the optimal solution
– Hagen von Eitzen
Nov 25 '18 at 16:24
|
show 1 more comment
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1
If we just pick the first two primes randomly and then take the samllest suitable prime as third, it will typically be $approx frac12ln 10^{17}approx 19$ too big, thus producing an offset of typically $2cdot 10^{35}$. With this in mind, your $kapprox 3cdot 10^{13}$ seems to be exceptionally small already. Then again, we have a lot of $P18$ at our disposal for trying ...
– Hagen von Eitzen
Nov 25 '18 at 11:10
Enzo Creti found the solution : $k=279 557$. Probably, it is the optimal solution.
– Peter
Nov 25 '18 at 13:36
After seeing Enzo Creti's $k$, I started brute-forcing. $k=193$, $k=6211$, $k=12973$ could be called "near misses" (three primes, smallest $>10^{15}$), still a lot to go ...
– Hagen von Eitzen
Nov 25 '18 at 13:59
@HagenvonEitzen Thank you for your double-check.
– Peter
Nov 25 '18 at 14:12
2
Closing remark: Yep, $k=279557$ is the optimal solution
– Hagen von Eitzen
Nov 25 '18 at 16:24