Divisibility Streak [duplicate]

Question

We can define the Divisibility Streak k of a number n by finding the smallest non-negative integer k such that n+k is not divisible by k+1.

Challenge

In your language of choice, write a program or function that outputs or returns the Divisibility Streak of your input.

Examples:

n=13:
13 is divisible by 1 
14 is divisible by 2 
15 is divisible by 3 
16 is divisible by 4 
17 is not divisible by 5

The Divisibilty Streak of 13 is 4

n=120:
120 is divisible by 1 
121 is not divisible by 2 

The Divisibilty Streak of 120 is 1

Test Cases:

n      DS
2      1
3      2
4      1
5      2
6      1
7      3
8      1
9      2
10     1
2521   10

More test cases can be found here.

Notes

• Based on Project Euler Problem 601
• This sequence can be found on OEIS, shifted down by 1.

Rules

• You can assume the input is greater than 1.

Scoring

code-golf:The submission with the lowest score wins.

Answer 1

Java 8, 44 42 41 39 bytes

^{Crossed out 44 is still regular 44 ;(}

n->{int r=0;for(;~-n%--r<1;);return~r;}

-2 bytes thanks to @LeakyNun.
-1 byte thanks to @TheLethalCoder.
-2 bytes thanks to @Nevay.

Explanation:

Try it here.

n->{                 // Method with integer as parameter and return-type
  int r=0;           //  Result-integer (starting at 0)
  for(;~-n%--r<1;);  //  Loop as long as `n-1` is divisible by `r-1`
                     //   (after we've first decreased `r` by 1 every iteration)
  return~r;          //  Return `-r-1` as result integer
}                    // End of method

Answer 2

Haskell, 35 bytes

f n=[k|k<-[1..],rem(n+k)(k+1)>0]!!0

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Using until is also 35 bytes

f n=until(\k->rem(n+k)(k+1)>0)(+1)1

Answer 3

Pyth, 6 5 bytes

f%tQh

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Answer 4

Husk, 7 bytes

ḟ§%→+⁰N

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Answer 5

05AB1E, 7 6 bytes

Ý+āÖ0k

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Alternate 7 byte solutions:
<DLÖγнg
Ls<ÑKн<

Answer 6

APL (Dyalog Unicode), 16 14 13 bytes

⊃∘⍸⍳(×+∘×|+)⊢

-1b thanks to Adám

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Previous solution made tacit (i.e. using trains)

---

{⊃⍸×(1+⍳⍵)|⍵+⍳⍵}

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Only my second APL answer so I'm sure it can be golfed further!

Justification for only searching up to 2n is due to Bertrand's Postulate which states that for any n>3, there exists a prime number between n and 2^n-2. Since by definition a prime only has factors of itself and 1, it will always terminate the divisibility streak.

Explanation

{⊃⍸×(1+⍳⍵)|⍵+⍳⍵}  ⍝
{              }  ⍝ dfn, in this case taking a single argument (e.g. 13) mapped to ⍵
             ⍳⍵   ⍝ computes array of digits 0,1,2,3,4...,12
           ⍵+     ⍝ adds the original argument to each element of this array - 13,14,15,16...25
    (1+⍳⍵)        ⍝ computes a new array from 1...13
          |       ⍝ applies modulo to elements at the same index across the two arrays, e.g. 13%1,14%2,15%3,16%4,17%5...26%14 = 0 0 0 0 2...12
   ×              ⍝ converts any number >1 to 1, e.g. 0 0 0 0 1...1
  ⍸               ⍝ find the index of all 'true's. e.g. 4 6 7 8 9 10 12 13
 ⊃                ⍝ take the first, e.g. 4

Answer 7

JavaScript (ES6), 28 bytes

n=>g=(x=2)=>++n%x?--x:g(++x)

---

Test it

o.innerText=(f=

n=>g=(x=2)=>++n%x?--x:g(++x)

)(i.value=2521)();oninput=_=>o.innerText=f(+i.value)()

<input id=i><pre id=o>

Answer 8

Mathematica, 30 27 bytes

0//.i_/;(i+1)∣(#+i):>i+1&

An unnamed function that takes an integer argument.

Try it on Wolfram Sandbox

Usage:

0//.i_/;(i+1)∣(#+i):>i+1&[2521]

10

Answer 9

Perl 5, 23 21 + 1 (-p) = 22 bytes

1until$_++%++$\}{$\--

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Answer 10

Python 2, 35 bytes

f=lambda n,x=1:n%x<1and-~f(n+1,x+1)

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Answer 11

K (oK), 23 15 bytes

#1_(~!/)(1+)\1,

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Explanation:

(~!/)(1+)\1, / the solution
          1, / prepend 1 to input
     (1+)\   / iterate, increasing by 1
(~!/)        / until the modulo (!) is not (~) zero

• -8 bytes thanks to @Traws!

Answer 12

Cubix, 17 bytes

)uUqI1%?;)qUO(;/@

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Cubified

    ) u
    U q
I 1 % ? ; ) q U
O ( ; / @ . . .
    . .
    . .

• I1 setup the stack with input and divisor
• %? do mod and test
  • ;)qU)uqU if 0 remove result and increment input and divisor. Bit of a round about path to get back to %
  • /;(O@ if not 0, drop result, decrement divisor, output and exit

Watch it run

Answer 13

Python 2, 43 41 bytes

Saved 2 bytes thanks to Leaky Nun!

i=input();k=1
while~-i%-~k<1:k+=1
print k

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Python 2, 40 bytes

f=lambda i,k=1:~-i%-~k<1and f(i,k+1)or k

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Answer 14

Python 2, 44 40 bytes

-4 bytes thanks to Leaky Nun.

f=lambda n,x=1:~-n%-~x and x or f(n,x+1)

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Answer 15

Swift 4, 56 bytes

This is a full function f, with an integer parameter i that prints the output.

func f(i:Int){var k=0;while(i-1)%(k+1)<1{k+=1};print(k)}

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Swift 4, 56 bytes

This is an anonymous function, that returns the result.

{var k=0;while($0-1)%(k+1)<1{k+=1};return k}as(Int)->Int

Try it here.

Check out the Test Suite!

Answer 16

C# (Mono), 41 39 bytes

n=>{int r=0;while(~-n%--r<1);return~r;}

Essentially a port of @Kevin Cruijssen's Java 8 answer with further golfing.

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Answer 17

dc, 28 bytes

1si[1+dli1+dsi%0=M]dsMxli1-p

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This feels really suboptimal, with the incrementing and the final decrement, but I can't really see a way to improve on it. Basically we just increment a counter i and our starting value as long as value mod i continues to be zero, and once that's not true we subtract one from i and print.

Answer 18

Gaia, 8 bytes

@1Ė₌)†↺(

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Explanation

@         Push input (call it n).
 1        Push 1 (call it i).
      ↺   While...
  Ė₌       n is divisible by i:
    )†     Increment both n and i.
       (  Decrement the value of i that failed this test and print.

Answer 19

J, 17 bytes

[:{.@I.>:@i.|i.+]

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I think there's still room for golfing here.

Explanation (ungolfed)

[: {.@I. >:@i. | i. + ]
                 i. + ]  Range [n,2n)
                 i.       Range [0,n)
                    +     Added to each
                      ]   n
         >:@i. | i. + ]  Divisibility test
         >:@i.            Range [1,n+1)
               |          Modulo (in J, the arguments are reversed)
                 i. + ]   Range [n,2n)
    {.@I.                Get the index of the first non-divisible
       I.                 Indices of non-zero values
    {.                    Head

The cap ([:) is there to make sure that J doesn't treat the last verb ({.@I.) as part of a hook.

The only sort of weird thing about this answer is that I. actually duplicates the index of each non-zero number as many times as that number's value. e.g.

   I. 0 1 0 2 3
1 3 3 4 4 4

But it doesn't matter since we want the first index anyways (and since i. gives an ascending range, we know the first index will be the smallest value).

Finally, here's a very short proof that it is valid to check division only up to n.

We start checking divisibility with 1 | n, so assuming the streak goes that far, once we get to checking divisibility by n we have n | 2n - 1 which will never be true (2n - 1 ≡ n - 1 (mod n)). Therefore, the streak will end there.

Answer 20

Japt, 7 bytes

õ b!%UÉ

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Answer 21

Ruby, 34 32 31 bytes

f=->n,d=1{n%d<1?1+f[n+1,d+1]:0}

A recursive lambda. Still new to Ruby, so suggestions are welcome!

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Answer 22

x86 Machine Code, 16 bytes

49                 dec    ecx        ; decrement argument
31 FF              xor    edi, edi   ; zero counter

                Loop:
47                 inc    edi        ; increment counter
89 C8              mov    eax, ecx   ; copy argument to EAX for division
99                 cdq               ; use 1-byte CDQ with unsigned to zero EDX
F7 FF              idiv   edi        ; EDX:EAX / counter
85 D2              test   edx, edx   ; test remainder
74 F6              jz     Loop       ; keep looping if remainder == 0

4F                 dec    edi        ; decrement counter
97                 xchg   eax, edi   ; move counter into EAX for return
C3                 ret               ;  (use 1-byte XCHG instead of 2-byte MOV)

The above function takes a single parameter, n, in the ECX register. It computes its divisibility streak, k, and returns that via the EAX register. It conforms to the 32-bit fastcall calling convention, so it is easily callable from C code using either Microsoft or Gnu compilers.

The logic is pretty simple: it just does an iterative test starting from 1. It's functionally identical to most of the other answers here, but hand-optimized for size. Lots of nice 1-byte instructions there, including INC, DEC, CDQ, and XCHG. The hard-coded operands for division hurt us a bit, but not terribly so.

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Answer 23

PHP, 34 bytes

for(;$argv[1]++%++$r<1;);echo$r-1;

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Simple enough. Checks the remainder of division (mod) each loop while incrementing each value, outputs when the number isn't divisible anymore.

Answer 24

Befunge, 19 bytes

#v_1+::&+1-\%
1<@.-

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Answer 25

Jelly, 7 bytes

Ḷ+%RTḢ’

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Explanation:

Ḷ+%RTḢ’  main link, argument: n
Ḷ+       [n, 2n-1]
   R     [1, n]
  %      modulo
    TḢ   index of first truthy value (first value where k+1 does not divide n+k), 1-indexed
      ’  decrement

Answer 26

Keg, 17 bytes

®n0&{©n⑻+⑹⑻%0=|}⑺

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Answer 27

K4, 27 25 bytes

Solution:

*(1+)/[{y=x*_y%x}. 1+;]1,

Explanation:

Pretty naive. Start with list of [1, x] and increment both until no longer divisible.

*(1+)/[{y=x*_y%x}. 1+;]1, / the solution
                       1, / prepend input with 1
 (  )\[              ;]   / (loop)\[condition;start]
       {        }.        / lambda taking implicit x & y
                   1+     / add 1 (vectorised) [1, 13] => [2, 14]
             y%x          / divide y by x
            _             / floor
          y*              / multiply by x
        y=                / equals y?
  1+                      / add 1 (vectorised) [1, 13] => [2, 14]
*                         / take first [2, 14] => 2

Answer 28

Japt -æ, 9 8 7 5 bytes

N°u°U

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Original (w/o flag), 7 bytes

@°uXÄ}a

Test it

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Explanation

Implicit input of integer U.

@     }a

Return the first integer that returns a truthy value (a number >0) when passed through a function (with X being the current integer) ...

°

That postfix increments U ...

uXÄ

And modulos it (u) by X+1.

Answer 29

PowerShell Core, 30 bytes

for(;!($args[0]++%++$i)){}$i-1

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Answer 30

Vyxal, 7 bytes

?+n›%)Ṅ

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Explanation:

     )   # Lambda f from beginning of program
      Ṅ  # First non-negative integer n such that f(n) is truthy
?        # Push the input
 +       # Add n to it
    %    # Modulo
  n›     # N incremented