Repetend length in 1/n

Question

This problem is based on non-terminating, repeating decimal points.

Let \$n\$ be any positive integer \$(n > 1 \text{ and } n < 10000)\$, say \$7\$. Then, \$1/n = 1/7 = 0.142857142857142857...\$

We see a pattern like, 0. 142857 142857 142857 ...
In this, the 142857 part is always repeating, which has length of \$6\$. Or, if \$n = 11\$, then \$1/n = 1/11 = 0.0909090909090909...\$

Here the length of the pattern is 2. So, here goes the problem!

Task

Given a positive integer \$n\$, \$(n > 1 \text{ and } n < 10000)\$, find the length of pattern in \$1/n\$, if it's repeating. Otherwise, return any non-positive integer (e.g., cases: \$1/5, 1/94, 1/22\$). Note that, the pattern should start just after the decimal point. Hint: \$1/22 = 0.04545454545454545454545\$.

Sample I/O

• 5 -> -1
• 13 -> 6
• 21 -> 6
• 27 -> 3
• 33 -> 2
• 37 -> 3
• 94 -> -1
• 22 -> -1
• 69 -> 22
• 197 -> 98
• 65 -> -1
• \$1/9979\$
• \$1/9967\$

This is a code-golf, so the fewest bytes will win!

Answer 1

Jelly, 6 bytes

R⁵*%i1

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Basically compute 10**[1..n] % n and get the 1-based index of 1.

Answer 2

JavaScript (Node.js), 39 bytes

Expects a BigInt. Returns \$0\$ if there's no repeating pattern.

f=(n,k=1n)=>10n**(k%=n)%n-1n?f(n,-~k):k

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Answer 3

05AB1E, 7 6 bytes

L.Δ°I%

Port of @Bubbler's Jelly answer, so make sure to upvote him!
-1 byte thanks to @ovs.

Outputs -1 if it's non-repeating.

Try it online or verify all test cases (times out for the final test case).

Explanation:

L       # Push a list in the range [1, (implicit) input-integer]
 .Δ     # Find the first value in this list which is truthy for:
        # (results in -1 if none are found)
   °    #  Take 10 to the power the current integer
    I%  #  Modulo the input-integer
        #  (Note: Only 1 is truthy in 05AB1E)
        # (after which the result is output implicitly as result)

Answer 4

K (ngn/k), 16 23 19 bytes

{1+(x(x!10*)\10)?1}

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As suggested by @traws (returns INT_MIN+1 if there's no repeating pattern), adapted to work with longer patterns that would otherwise overflow 64-bit integers.

• x(...)\10 set up a monadic-do-scan, run for x iterations and seeded with 10
• (x!10*) multiply the current value by 10, modding by the original x input
• (...)?1 get the index of the first 1 showing up in the result (returns 0N if no 1 is present, i.e. if there is no repeating pattern)
• 1+ add one to that result; converts 0N to -9223372036854775807

Answer 5

Husk, 8 bytes

€1m%¹↑İ⁰

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Same approach as Bubbler's answer.

€1m%¹↑İ⁰
€1          # index of first '1' in 
  m         # list of results of applying 
   %¹       # MOD n 
     ↑      # to first n elements of 
      İ⁰    # series of powers of 10 (starting at 10)

Answer 6

Perl 5 (-p -Mbigint), 31 bytes

$_=++$i<$_?9x$i*1%$_?redo:$i:-1

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Answer 7

GolfScript, 20 19 bytes

~:x,{10\?x%1=},0+1=

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~:x                   # Assign the input to x                  13
   ,                  # Make an array from 0 to x-1            [0 1 2 3 4 5 6 7 8 9 10 11 12]
    {        },       # Find all numbers that pass this test
     10\?x%           # (10^k)%x                               [1 10 9 12 3 4 1 10 9 12 3 4 1]
           1=         # Is it 1?                               [1  0 0  0 0 0 1  0 0  0 0 0 1]
                      # Only the index of the 1s are kept      [0 6 12]
               0+     # Append 0                               [0 6 12 0]
                 1=   # Get the second number                  6

If it doesn't repeat, the array will be [0 0] and 0 will be outputted.

Answer 8

Jelly, 21 bytes

R⁵*’ḍ@¹TṂȧ@⁸g10¤’¬¤o-

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This is probably very suboptimal (edit yes it is, i don't know why i didn't think to combine the two checks together cuz that would've given Bubbler's algorithm which is pretty smart).

Answer 9

C (gcc), 55 52 bytes

Saved 3 bytes thanks to the man himself Arnauld!!!

i;m;f(n){for(i=m=1;(m*=10)%n&&m%n-1;++i);m=m%n?i:0;}

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Returns either the length of repeated pattern of \$\frac{1}{\space n \space}\$ or \$0\$ for no repeated pattern.

Answer 10

Python 3, 50 bytes

f=lambda n,i=1:i*(10**i%n==1)or~(i<n and~f(n,i+1))

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Answer 11

Wolfram Language (Mathematica), 45 41 39 bytes

n_:>Lookup[Mod[10^#,n]->#&~Array~n,1,0]

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Defined as a delayed rule that can be applied to any integer.

Answer 12

Wolfram Language (Mathematica), 29 bytes

0&@@10~MultiplicativeOrder~#&

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Answer 13

Retina, 68 bytes

.+[1379]$
_,$&*_,;
\d+
0
{`;
;_
\G_
10*
+`(_+,)\1
,$1
^_,.+;(_+)
$.1

Try it online! Link includes faster test cases. Explanation:

.+[1379]$
_,$&*_,;

If the number is coprime to 10, then create a work area with the values p=1, n and k=0 (in unary).

\d+
0

But if it is not coprime to 10 then set the answer to 0 immediately.

{`

Repeat until the answer is found.

;
;_

Increment k.

\G_
10*

Multiply p by 10.

+`(_+,)\1
,$1

Reduce p modulo n.

^_,.+;(_+)
$.1

If p=1, then set the answer to k converted to decimal, which allows the loop to exit.

Answer 14

Charcoal, 13 bytes

ＮθＩ⊕⌕﹪Ｘχ…¹θθ¹

Try it online! Link is to verbose version of code. Basically a port of @Bubbler's answer, except that my range goes from 1 to n-1. Explanation:

Ｎθ              Take input as a number
        …       Exclusive range
         ¹      From literal `1`
          θ     To input number
      Ｘ         Vectorised raise to power
       χ        Predefined variable `10`
     ﹪          Vectorised reduce modulo
           θ    Input number
    ⌕           Find index of
            ¹   Literal `1`
   ⊕            Increment
  Ｉ             Cast to string
                Implicitly print

Answer 15

Batch, 91 bytes

@set/ap=1,k=0
:g
@set/ak=-~k%%%1,p=p*10%%%1
@if %k% neq 0 if %p% neq 1 goto g
@echo %k%

Explanation: Repeatedly increments the answer and multiplies the power by 10 until (modulo the input) the answer wraps around to zero or the power reduces to 1.