Is $k=553276187$ the smallest solution?
I search the smallest positive integer $k$, such that $40!+k$ splits into three primes having $16$ decimal digits. The smallest solution I found is
$$k=553 276 187$$
You can see the factorization here : http://factordb.com/index.php?id=1100000001197449481
According to my calculations, no $kle 2cdot 10^6$ does the job.
Is my $k$ the optimal solution ?
number-theory elementary-number-theory prime-factorization
asked Nov 27 at 8:44
Peter
46.6k1039125
|
show 2 more comments
I search the smallest positive integer $k$, such that $40!+k$ splits into three primes having $16$ decimal digits. The smallest solution I found is
$$k=553 276 187$$
You can see the factorization here : http://factordb.com/index.php?id=1100000001197449481
According to my calculations, no $kle 2cdot 10^6$ does the job.
Is my $k$ the optimal solution ?
number-theory elementary-number-theory prime-factorization
asked Nov 27 at 8:44
Peter
46.6k1039125
2
Thanks again for the unmotivated downvotes ! Or does someone have a good argument why this question should be downvoted ? I am curious.
– Peter
Nov 27 at 9:00
Same story here : math.stackexchange.com/questions/3012676/…
– Peter
Nov 27 at 9:10
Didn't downvote, but it might help if you could share your efforts? That is, how did you find this factorization? Or if you give some context to the question?
– Servaes
Nov 27 at 9:20
1
Welcome to the club of the victims of downvoters to good questions. $to +1$
– Claude Leibovici
Nov 27 at 9:24
2
Peter I'm not downvoting these, but if you ask for the downvoters' motivation, the same can, frankly, be asked about the question. What motivates this particular search? Why $40!$ and $42!$ as bases in particular? Are you really asking for general results about the density of products of three large primes? Et cetera. I dare guess that addressing questions like that rather than just posting a "random challenge search" would make for a better question.
– Jyrki Lahtonen
Nov 30 at 11:07
|
show 2 more comments
I search the smallest positive integer $k$, such that $40!+k$ splits into three primes having $16$ decimal digits. The smallest solution I found is
$$k=553 276 187$$
You can see the factorization here : http://factordb.com/index.php?id=1100000001197449481
According to my calculations, no $kle 2cdot 10^6$ does the job.
Is my $k$ the optimal solution ?
number-theory elementary-number-theory prime-factorization
asked Nov 27 at 8:44
Peter
46.6k1039125
I search the smallest positive integer $k$, such that $40!+k$ splits into three primes having $16$ decimal digits. The smallest solution I found is
$$k=553 276 187$$
You can see the factorization here : http://factordb.com/index.php?id=1100000001197449481
According to my calculations, no $kle 2cdot 10^6$ does the job.
Is my $k$ the optimal solution ?
number-theory elementary-number-theory prime-factorization
number-theory elementary-number-theory prime-factorization
asked Nov 27 at 8:44
Peter
46.6k1039125
asked Nov 27 at 8:44
Peter
46.6k1039125
asked Nov 27 at 8:44
Peter
46.6k1039125
asked Nov 27 at 8:44
Peter
46.6k1039125
asked Nov 27 at 8:44
Peter
46.6k1039125
46.6k1039125
2
Thanks again for the unmotivated downvotes ! Or does someone have a good argument why this question should be downvoted ? I am curious.
– Peter
Nov 27 at 9:00
Same story here : math.stackexchange.com/questions/3012676/…
– Peter
Nov 27 at 9:10
Didn't downvote, but it might help if you could share your efforts? That is, how did you find this factorization? Or if you give some context to the question?
– Servaes
Nov 27 at 9:20
1
Welcome to the club of the victims of downvoters to good questions. $to +1$
– Claude Leibovici
Nov 27 at 9:24
2
Peter I'm not downvoting these, but if you ask for the downvoters' motivation, the same can, frankly, be asked about the question. What motivates this particular search? Why $40!$ and $42!$ as bases in particular? Are you really asking for general results about the density of products of three large primes? Et cetera. I dare guess that addressing questions like that rather than just posting a "random challenge search" would make for a better question.
– Jyrki Lahtonen
Nov 30 at 11:07
|
show 2 more comments
2
Thanks again for the unmotivated downvotes ! Or does someone have a good argument why this question should be downvoted ? I am curious.
– Peter
Nov 27 at 9:00
Same story here : math.stackexchange.com/questions/3012676/…
– Peter
Nov 27 at 9:10
Didn't downvote, but it might help if you could share your efforts? That is, how did you find this factorization? Or if you give some context to the question?
– Servaes
Nov 27 at 9:20
1
Welcome to the club of the victims of downvoters to good questions. $to +1$
– Claude Leibovici
Nov 27 at 9:24
2
Peter I'm not downvoting these, but if you ask for the downvoters' motivation, the same can, frankly, be asked about the question. What motivates this particular search? Why $40!$ and $42!$ as bases in particular? Are you really asking for general results about the density of products of three large primes? Et cetera. I dare guess that addressing questions like that rather than just posting a "random challenge search" would make for a better question.
– Jyrki Lahtonen
Nov 30 at 11:07
2
2
Thanks again for the unmotivated downvotes ! Or does someone have a good argument why this question should be downvoted ? I am curious.
– Peter
Nov 27 at 9:00
Thanks again for the unmotivated downvotes ! Or does someone have a good argument why this question should be downvoted ? I am curious.
– Peter
Nov 27 at 9:00
Same story here : math.stackexchange.com/questions/3012676/…
– Peter
Nov 27 at 9:10
Same story here : math.stackexchange.com/questions/3012676/…
– Peter
Nov 27 at 9:10
Didn't downvote, but it might help if you could share your efforts? That is, how did you find this factorization? Or if you give some context to the question?
– Servaes
Nov 27 at 9:20
Didn't downvote, but it might help if you could share your efforts? That is, how did you find this factorization? Or if you give some context to the question?
– Servaes
Nov 27 at 9:20
1
1
Welcome to the club of the victims of downvoters to good questions. $to +1$
– Claude Leibovici
Nov 27 at 9:24
Welcome to the club of the victims of downvoters to good questions. $to +1$
– Claude Leibovici
Nov 27 at 9:24
2
2
Peter I'm not downvoting these, but if you ask for the downvoters' motivation, the same can, frankly, be asked about the question. What motivates this particular search? Why $40!$ and $42!$ as bases in particular? Are you really asking for general results about the density of products of three large primes? Et cetera. I dare guess that addressing questions like that rather than just posting a "random challenge search" would make for a better question.
– Jyrki Lahtonen
Nov 30 at 11:07
Peter I'm not downvoting these, but if you ask for the downvoters' motivation, the same can, frankly, be asked about the question. What motivates this particular search? Why $40!$ and $42!$ as bases in particular? Are you really asking for general results about the density of products of three large primes? Et cetera. I dare guess that addressing questions like that rather than just posting a "random challenge search" would make for a better question.
– Jyrki Lahtonen
Nov 30 at 11:07
|
show 2 more comments
1 Answer
1
active
oldest
votes
The answer to the question is no, here is a slightly smaller solution:
$$
494;804;473 .
$$
sage: factor( factorial(40) + 494804473 )
8912658466556113 * 9232286052422441 * 9915818101022081
It was obtained also in sage, the program was running one day, and this is in this intermediate situation a found solution. (It is still running. It was not written to touch the numbers one by one starting with $40!$ in order, so this may also not be the minimal number doing the job.)
Note: As i also said in the comment, the fact that $40!^{1/3}$ is so close to the number $999dots 9$ with $16$ digits makes it harder to get such decompositions with three factors of $16,16,16$ digits each. Because the range of the primes is not really between the $16$ digits numbers $1000dots 0$ and $999dots 9$. The computer was founding for instance also
k= 33377 40!+k = 1056334373719901 * 22413347482811693 * 34461718591042889
k= 42269 40!+k = 1859373333290887 * 2719883435120179 * 161334845671400153
k= 47189 40!+k = 1822739844251111 * 15851904710438411 * 28238324719193609
k= 55049 40!+k = 1101465899839957 * 3378301824756127 * 219268155834100091
with factors having $ge 16$ digits,
and also insisting that the smaller prime factor among all three of the three is bigger than $7000dots 0$ ($16$ digits)...
k=300722963 40!+k = 7073009263847467 * 8126384986908847 * 14195263159044887
k=349245887 40!+k = 7209029739646403 * 8613116168531887 * 13140380656971067
k= 77639477 40!+k = 7252468942583363 * 7992027613367573 * 14076744068584523
k=300731429 40!+k = 7321097013439289 * 7404841114350401 * 15050577417998861
k=339499973 40!+k = 7398940260351053 * 10008305130239719 * 11018310576981439
k= 96999373 40!+k = 7532964349652891 * 9232067350535819 * 11732219831109637
k= 48531323 40!+k = 7564931816003599 * 10225749107742383 * 10547387873068219
k=465611681 40!+k = 7866432514968563 * 7912332084219151 * 13108794049712437
k=465608309 40!+k = 8221566296360779 * 9835813811956651 * 10089745359967421
k=126205139 40!+k = 8278202201940731 * 9790182376796351 * 10067421590328119
k=388112077 40!+k = 8601556182258047 * 8687613919887803 * 10918614312712297
answered Nov 30 at 10:49
dan_fulea
6,2301312
Thank you for your search. I think, the question is clear enough, I accept only the case of three primes with $16$ digits. I also noticed that the fact that $40!$ is "near" $10^{48}$ makes the search longer. I tried to construct good solutions, but the constructions are far worse than the number I accidently found. I found plenty of "near-misses"
– Peter
Nov 30 at 12:25
The used sage code is posted now on ask.sagemath.org/question/44581/sage-program-for-40k-project/…
– dan_fulea
Dec 10 at 15:25
add a comment |
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1 Answer
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The answer to the question is no, here is a slightly smaller solution:
$$
494;804;473 .
$$
sage: factor( factorial(40) + 494804473 )
8912658466556113 * 9232286052422441 * 9915818101022081
It was obtained also in sage, the program was running one day, and this is in this intermediate situation a found solution. (It is still running. It was not written to touch the numbers one by one starting with $40!$ in order, so this may also not be the minimal number doing the job.)
Note: As i also said in the comment, the fact that $40!^{1/3}$ is so close to the number $999dots 9$ with $16$ digits makes it harder to get such decompositions with three factors of $16,16,16$ digits each. Because the range of the primes is not really between the $16$ digits numbers $1000dots 0$ and $999dots 9$. The computer was founding for instance also
k= 33377 40!+k = 1056334373719901 * 22413347482811693 * 34461718591042889
k= 42269 40!+k = 1859373333290887 * 2719883435120179 * 161334845671400153
k= 47189 40!+k = 1822739844251111 * 15851904710438411 * 28238324719193609
k= 55049 40!+k = 1101465899839957 * 3378301824756127 * 219268155834100091
with factors having $ge 16$ digits,
and also insisting that the smaller prime factor among all three of the three is bigger than $7000dots 0$ ($16$ digits)...
k=300722963 40!+k = 7073009263847467 * 8126384986908847 * 14195263159044887
k=349245887 40!+k = 7209029739646403 * 8613116168531887 * 13140380656971067
k= 77639477 40!+k = 7252468942583363 * 7992027613367573 * 14076744068584523
k=300731429 40!+k = 7321097013439289 * 7404841114350401 * 15050577417998861
k=339499973 40!+k = 7398940260351053 * 10008305130239719 * 11018310576981439
k= 96999373 40!+k = 7532964349652891 * 9232067350535819 * 11732219831109637
k= 48531323 40!+k = 7564931816003599 * 10225749107742383 * 10547387873068219
k=465611681 40!+k = 7866432514968563 * 7912332084219151 * 13108794049712437
k=465608309 40!+k = 8221566296360779 * 9835813811956651 * 10089745359967421
k=126205139 40!+k = 8278202201940731 * 9790182376796351 * 10067421590328119
k=388112077 40!+k = 8601556182258047 * 8687613919887803 * 10918614312712297
answered Nov 30 at 10:49
dan_fulea
6,2301312
Thank you for your search. I think, the question is clear enough, I accept only the case of three primes with $16$ digits. I also noticed that the fact that $40!$ is "near" $10^{48}$ makes the search longer. I tried to construct good solutions, but the constructions are far worse than the number I accidently found. I found plenty of "near-misses"
– Peter
Nov 30 at 12:25
The used sage code is posted now on ask.sagemath.org/question/44581/sage-program-for-40k-project/…
– dan_fulea
Dec 10 at 15:25
add a comment |
The answer to the question is no, here is a slightly smaller solution:
$$
494;804;473 .
$$
sage: factor( factorial(40) + 494804473 )
8912658466556113 * 9232286052422441 * 9915818101022081
It was obtained also in sage, the program was running one day, and this is in this intermediate situation a found solution. (It is still running. It was not written to touch the numbers one by one starting with $40!$ in order, so this may also not be the minimal number doing the job.)
Note: As i also said in the comment, the fact that $40!^{1/3}$ is so close to the number $999dots 9$ with $16$ digits makes it harder to get such decompositions with three factors of $16,16,16$ digits each. Because the range of the primes is not really between the $16$ digits numbers $1000dots 0$ and $999dots 9$. The computer was founding for instance also
k= 33377 40!+k = 1056334373719901 * 22413347482811693 * 34461718591042889
k= 42269 40!+k = 1859373333290887 * 2719883435120179 * 161334845671400153
k= 47189 40!+k = 1822739844251111 * 15851904710438411 * 28238324719193609
k= 55049 40!+k = 1101465899839957 * 3378301824756127 * 219268155834100091
with factors having $ge 16$ digits,
and also insisting that the smaller prime factor among all three of the three is bigger than $7000dots 0$ ($16$ digits)...
k=300722963 40!+k = 7073009263847467 * 8126384986908847 * 14195263159044887
k=349245887 40!+k = 7209029739646403 * 8613116168531887 * 13140380656971067
k= 77639477 40!+k = 7252468942583363 * 7992027613367573 * 14076744068584523
k=300731429 40!+k = 7321097013439289 * 7404841114350401 * 15050577417998861
k=339499973 40!+k = 7398940260351053 * 10008305130239719 * 11018310576981439
k= 96999373 40!+k = 7532964349652891 * 9232067350535819 * 11732219831109637
k= 48531323 40!+k = 7564931816003599 * 10225749107742383 * 10547387873068219
k=465611681 40!+k = 7866432514968563 * 7912332084219151 * 13108794049712437
k=465608309 40!+k = 8221566296360779 * 9835813811956651 * 10089745359967421
k=126205139 40!+k = 8278202201940731 * 9790182376796351 * 10067421590328119
k=388112077 40!+k = 8601556182258047 * 8687613919887803 * 10918614312712297
answered Nov 30 at 10:49
dan_fulea
6,2301312
Thank you for your search. I think, the question is clear enough, I accept only the case of three primes with $16$ digits. I also noticed that the fact that $40!$ is "near" $10^{48}$ makes the search longer. I tried to construct good solutions, but the constructions are far worse than the number I accidently found. I found plenty of "near-misses"
– Peter
Nov 30 at 12:25
The used sage code is posted now on ask.sagemath.org/question/44581/sage-program-for-40k-project/…
– dan_fulea
Dec 10 at 15:25
add a comment |
The answer to the question is no, here is a slightly smaller solution:
$$
494;804;473 .
$$
sage: factor( factorial(40) + 494804473 )
8912658466556113 * 9232286052422441 * 9915818101022081
It was obtained also in sage, the program was running one day, and this is in this intermediate situation a found solution. (It is still running. It was not written to touch the numbers one by one starting with $40!$ in order, so this may also not be the minimal number doing the job.)
Note: As i also said in the comment, the fact that $40!^{1/3}$ is so close to the number $999dots 9$ with $16$ digits makes it harder to get such decompositions with three factors of $16,16,16$ digits each. Because the range of the primes is not really between the $16$ digits numbers $1000dots 0$ and $999dots 9$. The computer was founding for instance also
k= 33377 40!+k = 1056334373719901 * 22413347482811693 * 34461718591042889
k= 42269 40!+k = 1859373333290887 * 2719883435120179 * 161334845671400153
k= 47189 40!+k = 1822739844251111 * 15851904710438411 * 28238324719193609
k= 55049 40!+k = 1101465899839957 * 3378301824756127 * 219268155834100091
with factors having $ge 16$ digits,
and also insisting that the smaller prime factor among all three of the three is bigger than $7000dots 0$ ($16$ digits)...
k=300722963 40!+k = 7073009263847467 * 8126384986908847 * 14195263159044887
k=349245887 40!+k = 7209029739646403 * 8613116168531887 * 13140380656971067
k= 77639477 40!+k = 7252468942583363 * 7992027613367573 * 14076744068584523
k=300731429 40!+k = 7321097013439289 * 7404841114350401 * 15050577417998861
k=339499973 40!+k = 7398940260351053 * 10008305130239719 * 11018310576981439
k= 96999373 40!+k = 7532964349652891 * 9232067350535819 * 11732219831109637
k= 48531323 40!+k = 7564931816003599 * 10225749107742383 * 10547387873068219
k=465611681 40!+k = 7866432514968563 * 7912332084219151 * 13108794049712437
k=465608309 40!+k = 8221566296360779 * 9835813811956651 * 10089745359967421
k=126205139 40!+k = 8278202201940731 * 9790182376796351 * 10067421590328119
k=388112077 40!+k = 8601556182258047 * 8687613919887803 * 10918614312712297
answered Nov 30 at 10:49
dan_fulea
6,2301312
The answer to the question is no, here is a slightly smaller solution:
$$
494;804;473 .
$$
sage: factor( factorial(40) + 494804473 )
8912658466556113 * 9232286052422441 * 9915818101022081
It was obtained also in sage, the program was running one day, and this is in this intermediate situation a found solution. (It is still running. It was not written to touch the numbers one by one starting with $40!$ in order, so this may also not be the minimal number doing the job.)
Note: As i also said in the comment, the fact that $40!^{1/3}$ is so close to the number $999dots 9$ with $16$ digits makes it harder to get such decompositions with three factors of $16,16,16$ digits each. Because the range of the primes is not really between the $16$ digits numbers $1000dots 0$ and $999dots 9$. The computer was founding for instance also
k= 33377 40!+k = 1056334373719901 * 22413347482811693 * 34461718591042889
k= 42269 40!+k = 1859373333290887 * 2719883435120179 * 161334845671400153
k= 47189 40!+k = 1822739844251111 * 15851904710438411 * 28238324719193609
k= 55049 40!+k = 1101465899839957 * 3378301824756127 * 219268155834100091
with factors having $ge 16$ digits,
and also insisting that the smaller prime factor among all three of the three is bigger than $7000dots 0$ ($16$ digits)...
k=300722963 40!+k = 7073009263847467 * 8126384986908847 * 14195263159044887
k=349245887 40!+k = 7209029739646403 * 8613116168531887 * 13140380656971067
k= 77639477 40!+k = 7252468942583363 * 7992027613367573 * 14076744068584523
k=300731429 40!+k = 7321097013439289 * 7404841114350401 * 15050577417998861
k=339499973 40!+k = 7398940260351053 * 10008305130239719 * 11018310576981439
k= 96999373 40!+k = 7532964349652891 * 9232067350535819 * 11732219831109637
k= 48531323 40!+k = 7564931816003599 * 10225749107742383 * 10547387873068219
k=465611681 40!+k = 7866432514968563 * 7912332084219151 * 13108794049712437
k=465608309 40!+k = 8221566296360779 * 9835813811956651 * 10089745359967421
k=126205139 40!+k = 8278202201940731 * 9790182376796351 * 10067421590328119
k=388112077 40!+k = 8601556182258047 * 8687613919887803 * 10918614312712297
answered Nov 30 at 10:49
dan_fulea
6,2301312
answered Nov 30 at 10:49
dan_fulea
6,2301312
answered Nov 30 at 10:49
dan_fulea
6,2301312
answered Nov 30 at 10:49
dan_fulea
6,2301312
6,2301312
Thank you for your search. I think, the question is clear enough, I accept only the case of three primes with $16$ digits. I also noticed that the fact that $40!$ is "near" $10^{48}$ makes the search longer. I tried to construct good solutions, but the constructions are far worse than the number I accidently found. I found plenty of "near-misses"
– Peter
Nov 30 at 12:25
The used sage code is posted now on ask.sagemath.org/question/44581/sage-program-for-40k-project/…
– dan_fulea
Dec 10 at 15:25
add a comment |
Thank you for your search. I think, the question is clear enough, I accept only the case of three primes with $16$ digits. I also noticed that the fact that $40!$ is "near" $10^{48}$ makes the search longer. I tried to construct good solutions, but the constructions are far worse than the number I accidently found. I found plenty of "near-misses"
– Peter
Nov 30 at 12:25
The used sage code is posted now on ask.sagemath.org/question/44581/sage-program-for-40k-project/…
– dan_fulea
Dec 10 at 15:25
Thank you for your search. I think, the question is clear enough, I accept only the case of three primes with $16$ digits. I also noticed that the fact that $40!$ is "near" $10^{48}$ makes the search longer. I tried to construct good solutions, but the constructions are far worse than the number I accidently found. I found plenty of "near-misses"
– Peter
Nov 30 at 12:25
Thank you for your search. I think, the question is clear enough, I accept only the case of three primes with $16$ digits. I also noticed that the fact that $40!$ is "near" $10^{48}$ makes the search longer. I tried to construct good solutions, but the constructions are far worse than the number I accidently found. I found plenty of "near-misses"
– Peter
Nov 30 at 12:25
The used sage code is posted now on ask.sagemath.org/question/44581/sage-program-for-40k-project/…
– dan_fulea
Dec 10 at 15:25
The used sage code is posted now on ask.sagemath.org/question/44581/sage-program-for-40k-project/…
– dan_fulea
Dec 10 at 15:25
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2
Thanks again for the unmotivated downvotes ! Or does someone have a good argument why this question should be downvoted ? I am curious.
– Peter
Nov 27 at 9:00
Same story here : math.stackexchange.com/questions/3012676/…
– Peter
Nov 27 at 9:10
Didn't downvote, but it might help if you could share your efforts? That is, how did you find this factorization? Or if you give some context to the question?
– Servaes
Nov 27 at 9:20
1
Welcome to the club of the victims of downvoters to good questions. $to +1$
– Claude Leibovici
Nov 27 at 9:24
2
Peter I'm not downvoting these, but if you ask for the downvoters' motivation, the same can, frankly, be asked about the question. What motivates this particular search? Why $40!$ and $42!$ as bases in particular? Are you really asking for general results about the density of products of three large primes? Et cetera. I dare guess that addressing questions like that rather than just posting a "random challenge search" would make for a better question.
– Jyrki Lahtonen
Nov 30 at 11:07