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Given a number n, find x such that x! = n, where both x and n are positive integers. Assume the input n will always be the factorial of a positive integer, so something like n=23 will not be given as input.

Examples: n=1 -> x=1 (0 is not a positive integer), n=24 -> x=4

Shortest code wins.

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    \$\begingroup\$ Btw, we once had a similar challenge but 1) it's pretty old 2) it's over real numbers, not just positive integers 3) it bans factorial-related built-ins, which isn't quite good for our current standards. So I think the challenge itself is fine and not a dupe. \$\endgroup\$
    – Bubbler
    May 20, 2020 at 23:54
  • \$\begingroup\$ Is there an upper bound on the possible inputs? \$\endgroup\$
    – xnor
    May 21, 2020 at 8:14
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    \$\begingroup\$ What are the bounds on the input? Can we assume that it is no more than 19! (largest factorial that can be fully represented in a 53+11 double precision floating point), no more than 23! (largest factorial that can be accurately represented in a double precision 53+11 floating point), or no more than 170! (largest factorial whose magnitude is less than the maximum of a double precision floating point ~= 10^308) \$\endgroup\$
    – JDL
    May 21, 2020 at 10:29
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    \$\begingroup\$ @JDL I'm not the challenge author, but I'd say "up to the highest number that your language's number type supports (without loss of precision), but the underlying algorithm should work for higher numbers". Related standard loophole. \$\endgroup\$
    – Bubbler
    May 22, 2020 at 4:33
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    \$\begingroup\$ @Bubbler: the question does not limit the input number in any way except that x and n are positive integers. As such 646077305624121491462330357080396430806673805704796612248389053020040737981389397373513335318926846519441974218777961448245634895440330929720840926954349439434654453860427703550673839109903970520283495061590634864022312082259902655711571689179112428197039756156051147969300077437438615382409042832551650139224687809841080780412598454920634889005911333104355143592477664451230317936640000000000000000000000000000000000000000000000000000000 is a perfectly valid value for n :-) \$\endgroup\$
    – Bob Jarvis - Слава Україні
    May 23, 2020 at 18:01

37 Answers 37

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PowerShell, 34 bytes

param($n)for(;++$x-$n;$n/=$x){};$x

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  • \$\begingroup\$ nice. PS allows you to skip the ; after the operators for, if, param and etc. Try it online! \$\endgroup\$
    – mazzy
    Apr 1, 2021 at 10:35
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Keg, 10 bytes

&1{:¡⑻≠|1+

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1
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05AB1E, 5 bytes

∞.Δ!Q

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JavaScript (V8), 29 bytes

f=(n,i=1)=>n/i^1?f(n/i,++i):i

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3
  • \$\begingroup\$ yes, your are right \$\endgroup\$
    – Yaroslav Gaponov
    May 21, 2020 at 10:20
  • \$\begingroup\$ @Arnauld f(1) required to be 1, your function output false. \$\endgroup\$
    – tsh
    May 22, 2020 at 3:20
  • \$\begingroup\$ @tsh My bad. Here is another 28 that works for n=1. \$\endgroup\$
    – Arnauld
    May 22, 2020 at 8:58
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Prolog (SWI), 62 55 bytes

X-N-N:-A is N/D,A==1,X is D;X-N/D-(D+1).
f(X,N):-X-N-1.

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If N equals D, sets X to D, otherwise, it calls itself with N/D and D+1.

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1
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Python 2, 73 bytes

lambda x:[n for n in range(1,x)if reduce(lambda a,b:a*b,range(1,n+1))==x]

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Add++, 23 bytes

D,f,@,R¦*BK=
L,dBkRÞ{f}

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This is what happens when there's no "find index" or "first n where" or auto-vectorisation.

Explained

D,f,@,R¦*BK= # Helper function f
      R      #    range 1...n
       ¦*    #    reduce by multiplication
         BK= #    does that equal the global register?
L,dBkRÞ{f}   # main lambda
  dBk        #    place the input into the global register.
     R       #    range 1...input
      Þ{f}   #    filter by helper function f
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