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A ​Note ​on ​N!

Pseudo-cyclic Number EvaluatorFind the remaining side of the tangential quadrilateralSelf Referential PolynomialsClosest Woodall PrimeTriple-balanced numbersprepend,append-SequenceReconstruct an arithmetic sequenceFour Spiraling AxesExact Partial Sum of Harmonic SeriesRyley's Theorem

5

$begingroup$

J. E. Maxfield proved following theorem (see DOI: 10.2307/2688966):

If A is any positive integer having m digits, there exists a positive integer N such that the first m digits of N! constitute the integer A.

Challenge

Your challenge is given some A find a corresponding N (in base 10).

Details

• Your submission should work for arbitrary A given enough time and memory. Just using e.g. 32-bit types to represent integers is not sufficient.


• You don't necessarily need to output the least possible N.

Examples

A            N

1            1

2            2

3            9

4            8

5            7

6            3

7            6

9           96

12           5

16          89

17          69

18          76

19          63

24           4

72           6

841      12745

206591378  314

The least possible N for each A can be found in https://oeis.org/A076219

code-golf math number integer factorial

share|improve this question

asked 1 hour ago

flawrflawr

27.4k668193

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I... why did he prove that theorem? Did he just wake up one day and say "I shall solve this!" or did it serve a purpose?
  $endgroup$
  – Magic Octopus Urn
  52 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By "any A" do you mean that like using numbers in Python or long longs in C++ is invalid?
  $endgroup$
  – HyperNeutrino
  24 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are programs allowed to throw stackoverflow exceptions if the input is too large?
  $endgroup$
  – Embodiment of Ignorance
  24 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can we return 0 for input 1? Lynn's answer currently does.
  $endgroup$
  – Erik the Outgolfer
  18 mins ago
  
  
  

  
  
  
  

add a comment | 

5

$begingroup$

J. E. Maxfield proved following theorem (see DOI: 10.2307/2688966):

If A is any positive integer having m digits, there exists a positive integer N such that the first m digits of N! constitute the integer A.

Challenge

Your challenge is given some A find a corresponding N (in base 10).

Details

• Your submission should work for arbitrary A given enough time and memory. Just using e.g. 32-bit types to represent integers is not sufficient.


• You don't necessarily need to output the least possible N.

Examples

A            N

1            1

2            2

3            9

4            8

5            7

6            3

7            6

9           96

12           5

16          89

17          69

18          76

19          63

24           4

72           6

841      12745

206591378  314

The least possible N for each A can be found in https://oeis.org/A076219

code-golf math number integer factorial

share|improve this question

asked 1 hour ago

flawrflawr

27.4k668193

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I... why did he prove that theorem? Did he just wake up one day and say "I shall solve this!" or did it serve a purpose?
  $endgroup$
  – Magic Octopus Urn
  52 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By "any A" do you mean that like using numbers in Python or long longs in C++ is invalid?
  $endgroup$
  – HyperNeutrino
  24 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are programs allowed to throw stackoverflow exceptions if the input is too large?
  $endgroup$
  – Embodiment of Ignorance
  24 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can we return 0 for input 1? Lynn's answer currently does.
  $endgroup$
  – Erik the Outgolfer
  18 mins ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

J. E. Maxfield proved following theorem (see DOI: 10.2307/2688966):

If A is any positive integer having m digits, there exists a positive integer N such that the first m digits of N! constitute the integer A.

Challenge

Your challenge is given some A find a corresponding N (in base 10).

Details

• Your submission should work for arbitrary A given enough time and memory. Just using e.g. 32-bit types to represent integers is not sufficient.


• You don't necessarily need to output the least possible N.

Examples

A            N

1            1

2            2

3            9

4            8

5            7

6            3

7            6

9           96

12           5

16          89

17          69

18          76

19          63

24           4

72           6

841      12745

206591378  314

The least possible N for each A can be found in https://oeis.org/A076219

code-golf math number integer factorial

share|improve this question

asked 1 hour ago

flawrflawr

27.4k668193

$endgroup$

A            N

1            1

2            2

3            9

4            8

5            7

6            3

7            6

9           96

12           5

16          89

17          69

18          76

19          63

24           4

72           6

841      12745

206591378  314

share|improve this question

asked 1 hour ago

flawrflawr

27.4k668193

share|improve this question

asked 1 hour ago

flawrflawr

27.4k668193

asked 1 hour ago

flawrflawr

27.4k668193

asked 1 hour ago

flawrflawr

27.4k668193

6 Answers
6

active

oldest

votes

5

$begingroup$

Python 2, 47 bytes

f=lambda a,n=1,p=1:`p`.find(a)and-~f(a,n+1,p*n)

Try it online!

Takes a string as input, like f('18').

The trick here is that x.find(y) == 0 precisely when x.startswith(y).

The and-expression will short circuit at `p`.find(a) with result 0 as soon as `p` starts with a; otherwise, it will evaluate to -~f(a,n+1,p*n), id est 1 + f(a,n+1,p*n).

The end result is 1 + (1 + (1 + (... + 0))), n layers deep, so n.

share|improve this answer

edited 24 mins ago

answered 35 mins ago

LynnLynn

51.4k899234

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Returns 0 for 1.
  $endgroup$
  – Shaggy
  14 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Shaggy I've asked about that in the comments, because 0 is a pretty sensible answer if it must just be non-negative (a fix could cost up to 8 bytes).
  $endgroup$
  – Erik the Outgolfer
  13 mins ago
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Python 3, 115 bytes

lambda x,a=0:str(F(a)).startswith(str(x))and a or f(x,a+Number(1))

F=lambda x:x and x*F(x-1)or 1

from sympy import*

Try it online!

Recursive so it dies quite quickly. 88 bytes if you don't require infinitely large numbers.

share|improve this answer

edited 41 mins ago

answered 49 mins ago

HyperNeutrinoHyperNeutrino

19.1k437148

$endgroup$

add a comment | 

0

$begingroup$

M, 19 bytes

®!Dḣ³L¤⁼³µ’

®‘$©$Ç¿

Try it online!

M is just Jelly but with Sympy. Explanation to come. This program is longer than it should be because M is a very old version of Jelly. I will add the golfed Jelly version later.

share|improve this answer

answered 23 mins ago

HyperNeutrinoHyperNeutrino

19.1k437148

$endgroup$

add a comment | 

0

$begingroup$

Jelly, 16 bytes

‘ɼ!³;D®ß⁼Lḣ@¥¥/?

Try it online!

Explanation

‘ɼ                | Increment the register (initially 0)

  !               | Factorial

   ³;             | Prepend the input

     D            | Convert to decimal digits

        ⁼   ¥¥/?  | If the input diguts are equal to...

         Lḣ@      | The same number of diguts from the head of the factorial

      ®           | Return the register

       ß          | Otherwise run the link again

share|improve this answer

edited 7 mins ago

answered 14 mins ago

Nick KennedyNick Kennedy

1,90149

$endgroup$

add a comment | 

0

$begingroup$

JavaScript, 47 bytes

Port of Lynn's Python solution. If 0 can be output for 1 then the last 3 bytes can be removed.

Takes input as a BigInt.

n=>(g=x=>`${x}`.search(n)&&1+g(x*i++))(i=1n)||1

Try it online

share

answered 4 mins ago

ShaggyShaggy

19.2k21768

$endgroup$

add a comment | 

0

$begingroup$

Jelly, 9 bytes

1!DḌƤċʋ1#

Try it online!

Takes an integer and returns a singleton.

share

answered 29 secs ago

Erik the OutgolferErik the Outgolfer

33.2k429106

$endgroup$

add a comment | 

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5

$begingroup$

Python 2, 47 bytes

f=lambda a,n=1,p=1:`p`.find(a)and-~f(a,n+1,p*n)

Try it online!

Takes a string as input, like f('18').

The trick here is that x.find(y) == 0 precisely when x.startswith(y).

The and-expression will short circuit at `p`.find(a) with result 0 as soon as `p` starts with a; otherwise, it will evaluate to -~f(a,n+1,p*n), id est 1 + f(a,n+1,p*n).

The end result is 1 + (1 + (1 + (... + 0))), n layers deep, so n.

share|improve this answer

edited 24 mins ago

answered 35 mins ago

LynnLynn

51.4k899234

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Returns 0 for 1.
  $endgroup$
  – Shaggy
  14 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Shaggy I've asked about that in the comments, because 0 is a pretty sensible answer if it must just be non-negative (a fix could cost up to 8 bytes).
  $endgroup$
  – Erik the Outgolfer
  13 mins ago
  
  
  

  
  
  
  

add a comment | 

5

$begingroup$

Python 2, 47 bytes

f=lambda a,n=1,p=1:`p`.find(a)and-~f(a,n+1,p*n)

Try it online!

Takes a string as input, like f('18').

The trick here is that x.find(y) == 0 precisely when x.startswith(y).

The and-expression will short circuit at `p`.find(a) with result 0 as soon as `p` starts with a; otherwise, it will evaluate to -~f(a,n+1,p*n), id est 1 + f(a,n+1,p*n).

The end result is 1 + (1 + (1 + (... + 0))), n layers deep, so n.

share|improve this answer

edited 24 mins ago

answered 35 mins ago

LynnLynn

51.4k899234

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Returns 0 for 1.
  $endgroup$
  – Shaggy
  14 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Shaggy I've asked about that in the comments, because 0 is a pretty sensible answer if it must just be non-negative (a fix could cost up to 8 bytes).
  $endgroup$
  – Erik the Outgolfer
  13 mins ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Python 2, 47 bytes

f=lambda a,n=1,p=1:`p`.find(a)and-~f(a,n+1,p*n)

Try it online!

Takes a string as input, like f('18').

The trick here is that x.find(y) == 0 precisely when x.startswith(y).

The and-expression will short circuit at `p`.find(a) with result 0 as soon as `p` starts with a; otherwise, it will evaluate to -~f(a,n+1,p*n), id est 1 + f(a,n+1,p*n).

The end result is 1 + (1 + (1 + (... + 0))), n layers deep, so n.

share|improve this answer

edited 24 mins ago

answered 35 mins ago

LynnLynn

51.4k899234

$endgroup$

f=lambda a,n=1,p=1:`p`.find(a)and-~f(a,n+1,p*n)

share|improve this answer

edited 24 mins ago

answered 35 mins ago

LynnLynn

51.4k899234

answered 35 mins ago

LynnLynn

51.4k899234

answered 35 mins ago

LynnLynn

51.4k899234

0

$begingroup$

Python 3, 115 bytes

lambda x,a=0:str(F(a)).startswith(str(x))and a or f(x,a+Number(1))

F=lambda x:x and x*F(x-1)or 1

from sympy import*

Try it online!

Recursive so it dies quite quickly. 88 bytes if you don't require infinitely large numbers.

share|improve this answer

edited 41 mins ago

answered 49 mins ago

HyperNeutrinoHyperNeutrino

19.1k437148

$endgroup$

add a comment | 

0

$begingroup$

Python 3, 115 bytes

lambda x,a=0:str(F(a)).startswith(str(x))and a or f(x,a+Number(1))

F=lambda x:x and x*F(x-1)or 1

from sympy import*

Try it online!

Recursive so it dies quite quickly. 88 bytes if you don't require infinitely large numbers.

share|improve this answer

edited 41 mins ago

answered 49 mins ago

HyperNeutrinoHyperNeutrino

19.1k437148

$endgroup$

add a comment | 

$begingroup$

Python 3, 115 bytes

lambda x,a=0:str(F(a)).startswith(str(x))and a or f(x,a+Number(1))

F=lambda x:x and x*F(x-1)or 1

from sympy import*

Try it online!

Recursive so it dies quite quickly. 88 bytes if you don't require infinitely large numbers.

share|improve this answer

edited 41 mins ago

answered 49 mins ago

HyperNeutrinoHyperNeutrino

19.1k437148

$endgroup$

lambda x,a=0:str(F(a)).startswith(str(x))and a or f(x,a+Number(1))

F=lambda x:x and x*F(x-1)or 1

from sympy import*

share|improve this answer

edited 41 mins ago

answered 49 mins ago

HyperNeutrinoHyperNeutrino

19.1k437148

answered 49 mins ago

HyperNeutrinoHyperNeutrino

19.1k437148

answered 49 mins ago

HyperNeutrinoHyperNeutrino

19.1k437148

0

$begingroup$

M, 19 bytes

®!Dḣ³L¤⁼³µ’

®‘$©$Ç¿

Try it online!

M is just Jelly but with Sympy. Explanation to come. This program is longer than it should be because M is a very old version of Jelly. I will add the golfed Jelly version later.

share|improve this answer

answered 23 mins ago

HyperNeutrinoHyperNeutrino

19.1k437148

$endgroup$

add a comment | 

0

$begingroup$

M, 19 bytes

®!Dḣ³L¤⁼³µ’

®‘$©$Ç¿

Try it online!

M is just Jelly but with Sympy. Explanation to come. This program is longer than it should be because M is a very old version of Jelly. I will add the golfed Jelly version later.

share|improve this answer

answered 23 mins ago

HyperNeutrinoHyperNeutrino

19.1k437148

$endgroup$

add a comment | 

$begingroup$

M, 19 bytes

®!Dḣ³L¤⁼³µ’

®‘$©$Ç¿

Try it online!

M is just Jelly but with Sympy. Explanation to come. This program is longer than it should be because M is a very old version of Jelly. I will add the golfed Jelly version later.

share|improve this answer

answered 23 mins ago

HyperNeutrinoHyperNeutrino

19.1k437148

$endgroup$

®!Dḣ³L¤⁼³µ’

®‘$©$Ç¿

share|improve this answer

answered 23 mins ago

HyperNeutrinoHyperNeutrino

19.1k437148

answered 23 mins ago

HyperNeutrinoHyperNeutrino

19.1k437148

answered 23 mins ago

HyperNeutrinoHyperNeutrino

19.1k437148

0

$begingroup$

Jelly, 16 bytes

‘ɼ!³;D®ß⁼Lḣ@¥¥/?

Try it online!

Explanation

‘ɼ                | Increment the register (initially 0)

  !               | Factorial

   ³;             | Prepend the input

     D            | Convert to decimal digits

        ⁼   ¥¥/?  | If the input diguts are equal to...

         Lḣ@      | The same number of diguts from the head of the factorial

      ®           | Return the register

       ß          | Otherwise run the link again

share|improve this answer

edited 7 mins ago

answered 14 mins ago

Nick KennedyNick Kennedy

1,90149

$endgroup$

add a comment | 

0

$begingroup$

Jelly, 16 bytes

‘ɼ!³;D®ß⁼Lḣ@¥¥/?

Try it online!

Explanation

‘ɼ                | Increment the register (initially 0)

  !               | Factorial

   ³;             | Prepend the input

     D            | Convert to decimal digits

        ⁼   ¥¥/?  | If the input diguts are equal to...

         Lḣ@      | The same number of diguts from the head of the factorial

      ®           | Return the register

       ß          | Otherwise run the link again

share|improve this answer

edited 7 mins ago

answered 14 mins ago

Nick KennedyNick Kennedy

1,90149

$endgroup$

add a comment | 

$begingroup$

Jelly, 16 bytes

‘ɼ!³;D®ß⁼Lḣ@¥¥/?

Try it online!

Explanation

‘ɼ                | Increment the register (initially 0)

  !               | Factorial

   ³;             | Prepend the input

     D            | Convert to decimal digits

        ⁼   ¥¥/?  | If the input diguts are equal to...

         Lḣ@      | The same number of diguts from the head of the factorial

      ®           | Return the register

       ß          | Otherwise run the link again

share|improve this answer

edited 7 mins ago

answered 14 mins ago

Nick KennedyNick Kennedy

1,90149

$endgroup$

‘ɼ!³;D®ß⁼Lḣ@¥¥/?

‘ɼ                | Increment the register (initially 0)

  !               | Factorial

   ³;             | Prepend the input

     D            | Convert to decimal digits

        ⁼   ¥¥/?  | If the input diguts are equal to...

         Lḣ@      | The same number of diguts from the head of the factorial

      ®           | Return the register

       ß          | Otherwise run the link again

share|improve this answer

edited 7 mins ago

answered 14 mins ago

Nick KennedyNick Kennedy

1,90149

answered 14 mins ago

Nick KennedyNick Kennedy

1,90149

answered 14 mins ago

Nick KennedyNick Kennedy

1,90149

0

$begingroup$

JavaScript, 47 bytes

Port of Lynn's Python solution. If 0 can be output for 1 then the last 3 bytes can be removed.

Takes input as a BigInt.

n=>(g=x=>`${x}`.search(n)&&1+g(x*i++))(i=1n)||1

Try it online

share

answered 4 mins ago

ShaggyShaggy

19.2k21768

$endgroup$

add a comment | 

0

$begingroup$

JavaScript, 47 bytes

Port of Lynn's Python solution. If 0 can be output for 1 then the last 3 bytes can be removed.

Takes input as a BigInt.

n=>(g=x=>`${x}`.search(n)&&1+g(x*i++))(i=1n)||1

Try it online

share

answered 4 mins ago

ShaggyShaggy

19.2k21768

$endgroup$

add a comment | 

$begingroup$

JavaScript, 47 bytes

Port of Lynn's Python solution. If 0 can be output for 1 then the last 3 bytes can be removed.

Takes input as a BigInt.

n=>(g=x=>`${x}`.search(n)&&1+g(x*i++))(i=1n)||1

Try it online

share

answered 4 mins ago

ShaggyShaggy

19.2k21768

$endgroup$

n=>(g=x=>`${x}`.search(n)&&1+g(x*i++))(i=1n)||1

share

answered 4 mins ago

ShaggyShaggy

19.2k21768

answered 4 mins ago

ShaggyShaggy

19.2k21768

answered 4 mins ago

ShaggyShaggy

19.2k21768

0

$begingroup$

Jelly, 9 bytes

1!DḌƤċʋ1#

Try it online!

Takes an integer and returns a singleton.

share

answered 29 secs ago

Erik the OutgolferErik the Outgolfer

33.2k429106

$endgroup$

add a comment | 

0

$begingroup$

Jelly, 9 bytes

1!DḌƤċʋ1#

Try it online!

Takes an integer and returns a singleton.

share

answered 29 secs ago

Erik the OutgolferErik the Outgolfer

33.2k429106

$endgroup$

add a comment | 

$begingroup$

Jelly, 9 bytes

1!DḌƤċʋ1#

Try it online!

Takes an integer and returns a singleton.

share

answered 29 secs ago

Erik the OutgolferErik the Outgolfer

33.2k429106

$endgroup$

1!DḌƤċʋ1#

share

answered 29 secs ago

Erik the OutgolferErik the Outgolfer

33.2k429106

answered 29 secs ago

Erik the OutgolferErik the Outgolfer

33.2k429106

answered 29 secs ago

Erik the OutgolferErik the Outgolfer

33.2k429106