# Divisibility Streak

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.

### Rules

• You can assume the input is greater than 1.

### Scoring

:The submission with the lowest score wins.

• I suggest changing "smallest positive integer" to "smallest nonnegative integer". It doesn't change the challenge at all, but with the current description, it implies we don't need to check for divisibility by 1 (which we technically shouldn't need to). Either that, or you could remove the divisibility by 1 checks from the description. – TehPers Aug 14 '17 at 15:11
• The smallest positive integer is 1, and k + 1 is 2, where k is the smallest positive integer. Sorry for the nitpick. – TehPers Aug 14 '17 at 15:33
• Isn't this the same as finding the smallest k which doesn't divide n-1? – Paŭlo Ebermann Aug 14 '17 at 22:28
• @PaŭloEbermann Take n=7 where k=3: n-1 is divisible by k. – Oliver Aug 15 '17 at 0:19
• Ah, I missed the +1. – Paŭlo Ebermann Aug 15 '17 at 16:57

# Pyth, 6 5 bytes

f%tQh


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# Java 8, 444241 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

• 42 bytes – Leaky Nun Aug 14 '17 at 15:02
• 41 bytes Just shaved a byte from LeakyNun's suggestion. – TheLethalCoder Aug 14 '17 at 15:08
• 39 bytes – Nevay Aug 14 '17 at 15:23

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


# Husk, 7 bytes

ḟ§%→+⁰N


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# 05AB1E, 7 6 bytes

Ý+āÖ0k


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Alternate 7 byte solutions:
<DLÖγнg
Ls<ÑKн<

# 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>

# 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&


10

1until$_++%++$\}{$\--  Try it online! # Python 2, 35 bytes f=lambda n,x=1:n%x<1and-~f(n+1,x+1)  Try it online! # Cubix, 17 bytes )uUqI1%?;)qUO(;/@  Try it online! 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 # Python 2, 43 41 bytes Saved 2 bytes thanks to Leaky Nun! i=input();k=1 while~-i%-~k<1:k+=1 print k  Try it online! # Python 2, 40 bytes f=lambda i,k=1:~-i%-~k<1and f(i,k+1)or k  Try it online! # 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)  Try it online! # 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)}  Try it here. # 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!

# 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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# 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.

# 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.


# 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)
]   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


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.

# Japt, 7 bytes

õ b!%UÉ


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# 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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# PHP, 34 bytes

for(;$argv++%++$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.

# SOGL V0.12, 8 bytes

]e.-ē⁴I\


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Not bad for a language which is made for a completely different type of challenges.

Explanation:

]         do .. while top of the stack is truthy
e          push the variable E contents, by default user input
.-        subtract the input from it
ē       push the value of the variable E and then increase the variable
⁴      duplicate the item below one in the stack
I     increase it
\    test if divides
if it does divide, then the loop restarts, if not, outputs POP which is e-input


# Mathematica, 40 bytes

Min@Complement[Range@#,Divisors[#-1]-1]&


Try it online! (Mathics)

Mathematical approach, n+k is divisible by k+1 if and only if n-1 is divisible by k+1. And n-1 is not divisible by n, so Range@# is enough numbers.

Originally I intend to use Min@Complement[Range@#,Divisors[#-1]]-1&, but this also work.

• Why does the captcha appear when I use the submission from tio? – user202729 Aug 14 '17 at 14:57
• Because you typed (copied-and-pasted) it too quickly. It isn't about TIO. – Leaky Nun Aug 14 '17 at 14:58

## Julia 0.6.0 (47 bytes) (38 bytes)

n->(i=1;while isinteger(n/i) i+=1;n+=1 end;i-1)

n->(i=1;while n%i<1 i+=1;n+=1end;i-1)

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9 bytes were cut thanks to Mr.Xcoder

• Normally a "Try it online" link allows people to actually try the code by defining some combination of header, footer, and arguments which mean that pressing the play button gives output. – Peter Taylor Aug 14 '17 at 15:43
• @PeterTaylor By a pure guess, I tried running it as such, and to my surprise it worked. I recommend the OP to edit in with the testable version. – Mr. Xcoder Aug 14 '17 at 15:55
• 46 bytes (removing one space): n->(i=1;while isinteger(n/i) i+=1;n+=1end;i-1) – Mr. Xcoder Aug 14 '17 at 15:59
• Another pure guess allowed be to golf it down to 38 bytes: n->(i=1;while n%i<1 i+=1;n+=1end;i-1) – Mr. Xcoder Aug 14 '17 at 16:03
• @PeterTaylor Sorry forgot it! – Goysa Aug 14 '17 at 17:08

# C (gcc), 34 bytes

i;f(n){for(i=1;++n%++i<1;);n=i-1;}


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• Suggest i-- instead of n=i-1 – ceilingcat Dec 31 '18 at 6:02

## Batch, 70 bytes

@set/an=%1-1,i=0
:l
@set/ai+=1,r=n%%~i
@if %r%==0 goto l
@echo %i%


All this is doing is finding the largest i such that LCM(1..i) divides n-1.

# R, 43 bytes

function(n,k=0:n)which((n+k)%%(k+1)>0)-1


Anonymous function.

Verify all test cases!

# Aceto, 28 27 bytes

[;%
I)@]
iIk2I(D(
rk[(&Xpu


I could save one byte if I don't have to exit.

### Explanation:

We use three stacks: The left stack holds a counter starting at 2, the right one holds the given number (or its increments), the center stack is used for doing the modulo operations. We could, of course, do everything in one stack, but this way we can set the outer stacks to be "sticky" (values that are popped aren't really removed) and save ourselves many duplication operations. Here's the method in detail:

Read an integer, increment it, make the current stack sticky, and "move" it (and ourselves) to the stack to the left:

iI
rk[


Go one more stack to the left, push a literal 2, make this stack sticky, too. Remember this position in the code (@), and "move" a value and ourselves to the center stack again.

  @]
k2
(


Now we test: Is the modulo of the top two numbers not 0? If so, jump to the end, otherwise go one stack to the right, increment, and push the value and us to the middle. Then go to the left stack, increment it, too, and jump back to the mark that we set before.

[;%
I)
I(
&


When the result of the modulo was not zero, we invert the position the IP is moving, go one stack to the left (where our counter lives), decrement it, and print the value, then exit.

      D(
Xpu


# Ruby, 3432 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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## F#, 86 bytes 84 bytes

let s n =
let rec c n1 d r=if n1%d=0 then c(n1+1)(d+1)(r+1)else r
c n 1 0


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Edit: -2 characters from Oliver

• Welcome to PPCG! Does your program take stdin? You can use TIO, which has an online F# interpreter. Also, can remove the whitespace in r = if? – Oliver Aug 16 '17 at 17:41
• @Oliver Thank you, I changed the link to TIO, so now you can actually pass the argument to test it. :) – Vladislav Khapin Aug 21 '17 at 6:52

# Befunge, 19 bytes

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


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