Reverse factorial function

Question

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.

Answer 1

PowerShell, 34 bytes

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

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Answer 2

Keg, 10 bytes

&1{:¡⑻≠|1+

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Answer 3

05AB1E, 5 bytes

∞.Δ!Q

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Answer 4

JavaScript (V8), 29 bytes

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

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Answer 5

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.

Answer 6

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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Answer 7

HP‑41C series, 18 B

The \$x\$ in \$x! = n\$ can be determined by repeatedly dividing \$n\$ by an incrementing divisor (first \$1\$, then \$2\$, then \$3\$ and so forth) until the quotient equals the (next) divisor. The program works correctly up to and including \$x = 39\$.

01♦LBL "S"         5 Bytes    global label requires 4 + (length of string) Bytes
   NULL            1 Byte     invisible Null byte before numbers
02 1               1 Byte     place 1 on top of stack; X ≔ 1 ; Y ≔ 𝘯

03 LBL 00          2 Bytes    mark head of this loop with (local) label 00 ──┐
                                                                             │
04 X=Y?            1 Byte     if X = Y ──┐                                   │
05 RTN             1 Byte                └── then return                     │
                                                                             │
06 ∕               1 Byte     lastX ≔ X ;  X ≔ Y ∕ X                         │
                                                                             ↑
07 LASTX           1 Byte     put lastX back on top of stack again           │
   NULL            1 Byte     invisible Null byte before numbers             │
08 1               1 Byte     place 1 on top of stack                        │
09 +               1 Byte     sum X and Y, stack drops by one item           │
                                                                             │
10 GTO 00          2 Bytes    go to (local) label 00  ───────────────────────┘

Since this is pretty abstract, let’s consider a trace of the stack in the case \$n = 6 = 3! = x!\$. The step “sees” data from one row above, the result of a step follows in the same row. Empty cells are not relevant to the program.

step	X	Y	Z	T	LastX
	6
1	1	6
X=Y?	〃	〃
∕	6				1
LASTX	1	6			〃
1	1	1	6		〃
+	2	6			1
GTO 00	〃	〃			〃
X=Y?	〃	〃			〃
∕	3				2
LASTX	2	3			〃
1	1	2	3		〃
+	3	3			1
X=Y?	〃	〃			〃
RTN	〃	〃			〃

Note that GTO searches for a local label downward in program memory and wraps to the beginning of the currently executed program. To inhibit local label search, append an END command (3 Bytes). The above program presumes there are no other programs (following) in program memory, so the permanent (= always there) END is sufficient.

Answer 8

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

Answer 9

Vyxal 3 G, 5 bytes

K!=Þκ

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