[Question]
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Your challenge, should you choose to accept it, is to create a function or a program that outputs "yes" if a given number is divisible by 13 and outputs "no" if it isn't.
**Rules:**  
 - You're not allowed to use the number 13 anywhere.  
 - No cop-out synonyms for 13 either (like using 15 - 2).  
 - Bonus points will be awarded for not using modulus, additional bonus for not using division.
**Scoring:**  
 - Your score will be the number of bytes in your code (whitespace *not* included) multiplied by your bonus.  
 - If you didn't use modulus, that bonus is 0.90; if you didn't use division, that bonus is 0.90.  
 - If you didn't use either, that bonus is 0.80.  
 - The lower your score, the better.
The input will always be an integer greater than 0 and less than 2^32.  
Your output should be a simple "yes" or "no".
**Clarifications:**  
 - Using some roundabout method of generating the number 13 for use is acceptable. Simple arithmetic synonyms like (10 + 3) are not allowed.  
 - The function or program must *literally output* "yes" or "no" for if the given number is divisible by 13.  
 - As always, clever solutions are recommended, but not required.
  
[Answer]
## Java (score 60.8 59.2)
```
void t(int n){System.out.print(Math.cos(.483321946706122*n)>.9?"yes":"no");}
```
Score: (76 - 2 whitespace) chars \* 0.8 = 59.2
[Answer]
**ASM - 16 bit x86 on WinXP command shell**
executable - 55 bytes \* 0.8 = 44
source - 288 characters \* 0.8 = 230.4
The number 13 doesn't even appear in the assembled .com file.
Assemble using A86.
```
    mov si,82h
    xor ax,ax
    xor cx,cx
a:  imul cx,10
    add cx,ax
    lodsb
    sub al,48
    jnc a
    inc cx
h:  mov dl,a and 255
c:  loop g
    sub dl,a and 255
    jz e
    mov dl,4
e:  add dl,k and 255
    mov dh,1
    mov ah,9
    int 21h
    ret
g:  inc dl
    cmp dl,c and 255
    jne c
    jmp h
k:  db 'yes$no$'
```
[Answer]
# Python 3.x: 54 \* 0.8 = 43.2
It may be a cop-out to have a string of length 13, but here it goes:
```
print('no' if any((' ' * int(input())).split('             ')) else 'yes')
```
It works by building a string of n spaces (the choice of delimiter is arbitrary, but I chose space for obvious reasons), and splitting away 13-space substrings until you're left with a string containing n%13 spaces.
[Answer]
## GolfScript, 32 chars
```
~){.14base{+}*.@<}do('no''yes'if
```
I wanted to try something different from everyone else, so my solution calculates the base 14 [digital root](http://en.wikipedia.org/wiki/Digital_root) of the number, by repeatedly converting the number to base **14** and summing the digits until the result no longer gets any smaller. This is essentially the same as calculating the remainder modulo 13, except that the result will be in the range 1 to 13 instead of 0 to 12.
Since checking whether the digital root equals 13 would be difficult without using the number 13 itself (or some lame workaround like 12+1), what I actually do is I increment the input number by one before the loop and decrement the result afterwards. That way, the result for numbers divisible by 13 will in fact be zero, which is much easier to check for.
Here's a commented version of the program:
```
~              # evaluate the input, turning it from a string to a number
)              # increment by one
{              # start of do-loop 
    .          # make a copy of the previous number, so we can tell when we're done
    14 base    # convert the number to base 14
    { + } *    # sum the digits
    . @ <      # check if the new number is less than the previous number...
} do           # ...and repeat the loop if so
(              # decrement the result by one
'no' 'yes' if  # output 'no' if the result is non-zero, 'yes' if it's zero
```
This program will actually handle *any* non-negative integer inputs, since GolfScript uses bignum arithmetic. Of course, extremely large inputs may consume excessive time and/or memory.
The code does not use either modulos or division directly, although it does use GolfScipt's base conversion operator, which almost certainly does some division and remainder-taking internally. I'll leave it for GigaWatt to decide whether this qualifies me for the bonus or not.
[Answer]
## C, 68 \* 0.8 = 54.4
After 24 answers, no one came up with this obvious algorithm yet:  
```
f(x){puts("no\0yes"+3*((x*330382100LL>>32)-(~-x*330382100LL>>32)));}
```
[Answer]
## JavaScript (27.9)
Current version (31 characters \* 0.90 bonus = 27.9).
```
alert(prompt()*2%26?'no':'yes')
```
Demo: <http://jsfiddle.net/9GQ9m/2/>
**Edit 1:** Forgo second bonus by using modulus to lower score considerably and avoid `for` loop. Also eliminate `~~` and save two chars (thanks `@copy`).
---
Older version (48 characters \* 0.80 bonus = 38.4)
```
for(n=~~prompt()*2;n-=26>0;);alert(n?'no':'yes')​
```
[Answer]
## BrainFuck
Score: 200 \* 0.8 = 160
```
>++++++[>++++++++<-]>>,[<[-<+>>-<]<[->+<]>>>[->++++++++++<]>[-<+>]<<[->+<],]++++
+++++++++>[>+<-<-[>>>]>>[[-<<+>>]>>>]<<<<]>[<<<[-<++>]<++++++++++++++.+.>]<<[[-<
++++++<++++++++>>]<-----.<---.>------.>]
```
Reads froms stdin. Probably not the most clever solution, but getting anything that works in BF is nice. It's quite compact though.
[Answer]
**Scala (38 \* 0.9 = 34.2)**
Similar to `0xD`(hex) or `015`(oct).
**ASCII** value of `CR` is 13.
```
def t(n:Int)=if(n%'\r'==0)"yes"else"no"
```
[Answer]
## Haskell, 28 \* 0.8 = 22.4
```
f x|gcd 26x>2="yes"|1<3="no"
```
[Answer]
Python:
```
f=lambda n:1==pow(8,n,79)
```
E.g. 
```
[i for i in range(100) if f(i)]
```
gives
```
[0, 13, 26, 39, 52, 65, 78, 91]
```
[Answer]
## C, 54.4 == 68 \* .8   80 \* .8
```
char*f(c){char*s=" yes\0\rno";while(c&&*s++);return c>0?f(c-*s):++s;}
```
[Answer]
### ECMAScript 6, 25 × 0.9 = 22.5
Yeah, it's a boring way of getting 13.
```
n => n % '             '.length ? 'no' : 'yes'
```
[Answer]
## APL ((21 - 1) × 0.8 = 16)
```
'yes' 'no'[1=⎕∨⌊⍟9*6]
```
`⎕IO` should be set to 0 for this to work properly in Dyalog APL. To generate 13, we take the floor (`⌊`) of the natural logarithm (`⍟`) of 9 to the power of 6 (`9*6`). After that, we find the GCD (`∨`) of our input (`⎕`) and 13, and we then test if that equals 1. This is used to index (`[...]`) the vector of answers.
If anyone wants to be pedantic about the mention of *bytes* in the scoring specification, the score for the UTF-8 encoded version of this is `(29 - 1) × 0.8 = 22.4`. :)
[Answer]
## C, 88
Fibonacci trick.
```
f(n){return n<2?n:f(n-1)+f(n-2);}main(x){printf("%s",x%f(7)?"No":"Yes",scanf("%d",&x));}
```
[Answer]
## Perl - 44 × 0.8 = 35.2
```
#!perl -p
map$_+=4*chop,($_)x10;$_=chop^$_*3?'no':yes
```
Counting the shebang as one byte.
I'm a bit late to the game, but I thought I'd share the algorithm, as no other posts to this point have used it.
This works under the observation that if *n* is divisible by *13*, then ⌊*n/10*⌋+*n%10\*4* is also divisible by *13*. The values *13*, *26* and *39* cycle onto themselves. All other multiples of *13* will eventually reach one of these values in no more than *log10 n* iterations.
---
**In Other Bases**
Admittedly, `chop` is a bit of a cop-out. With a base 10 representation, it's equivalent to `divmod`. But the algorithm works prefectly well in other bases, for example base 4, or 8.
Python style pseudo-code of the above algorithm (base 10):
```
def div13(n):
    while n > 40:
        q, r = n // 10, n % 10
        n = q + 4*r
    return n in [13, 26, 39]
```
In base 2:
```
def div13(n):
    while n > 40:
        q, r = n >> 1, n & 1
        n = q + 7*r
    return n in [13, 26, 39]
```
In base 4:
```
def div13(n):
    while n > 40:
        q, r = n >> 2, n & 3
        n = q + 10*r
    return n in [13, 26, 39]
```
In base 8:
```
def div13(n):
    while n > 40:
        q, r = n >> 3, n & 7
        n = q + 5*r
    return n in [13, 26, 39]
```
etc. Any base smaller than 13 works equally well.
[Answer]
## Javascript: 59\*0.8 = 47.2 (?)
[fiddle](http://jsfiddle.net/9aeCU/2/):
```
function r(n){
  for(c=0;n>c;n-=12,c++);
  return n==c?'yes':'no';
}
```
---
Including mellamokb's improvement (57\*0.8 = 45.6):
```
function r(n){
  for(c=0;n>c;n-=12,c++);
  return n-c?'no':'yes'
}
```
[Answer]
**Perl: (51-4 spaces)\*0.9 = 42.3**
```
say+<>%(scalar reverse int 40*atan2 1,1)?'no':'yes'
```
> 
> 40\*atan2(1,1) -> 31.41592 (PI\*10)
> 
> 
> 
[Answer]
# Perl (19.8)
21 bytes \* .9
```
say2*<>%26?"no":"yes"
```
*note:* My first Perl program ever. Weakly typed is good for golf i guess.
[Answer]
**in C (K&R): 47 \* 0.8 = 37.6** 
```
f(i){for(;i>0;i-=__LINE__);puts(i?"no":"yes");}
```
EDIT1: okay removed all dependencies on external functions, the above will work as long as you put this line on the 13th line of the file! :) If `__LINE__` is okay to be replaced by say `0xd` then can save a further 5 characters (score: 33.6)
[Answer]
## J - 22.4 = 28 \* 0.8
Based on [mxmul's](https://codegolf.stackexchange.com/users/2289/mxmul) clever [cyclic method](https://codegolf.stackexchange.com/a/4956/46).
```
f=:<:{('yes',~12 3$'no ')$~]
```
Examples:
```
   f 13
yes
   f 23
no
   f 13*513
yes
   f 123456789
no
```
[Answer]
# JavaScript (108 less 0 for whitespace) => 108, x 0.8 (no modulus, no division) = 86.4
`b=b=>{a=z,a=a+"";return+a.slice(0,-1)+4*+a.slice(-1)};z=prompt();for(i=99;i--;)z=b();alert(b()-z?"no":"yes")`
This method uses the following algorithm:
 1. Take the last digit, multiply it by four, add it to the rest of
 the truncated number.
 2. Repeat step 1 for 99 iterations...
 3. Test it one more time using step 1, if the resulting number is itself, you've found a multiple of 13.
Previous update, removed `var`, and reversed logic at the alert to remove more chars by using subtraction-false conditional.
Technically, the end result is that you'll eventually reach a two digit number like 13, 26, or 39 which when run through step 1 again will give 13, 26, or 39 respectively. So testing for iteration 100 being the same will confirm the divisibility.
[Answer]
# Cheddar, 20 bytes (noncompeting)
### Score is 20 \* 0.9 = 18
```
n->n*2%26?'no':'yes'
```
A straightforward answer.
[Answer]
# Common Lisp (71 bytes \* 0.8) = 56.8
Simple recursion, really.
`(defun w(x)(if(> x 14)(w(- x 13))(if(> 14 x 12)(print'yes)(print'no))))`
Ungolfed:
```
(defun w (x)
  (if (> x 14)
      (w (- x 13))
      (if (> 14 x 12)
          (print 'yes)
          (print 'no))))
```
[Answer]
## Ruby (50 48 \* 0.9 = 43.2)
Smart way to use `eval`
```
eval x="p gets.to_i*3%x.length == 0? 'yes':'no'"
```
[Answer]
## D 56 chars .80 bonus = 44.8
```
bool d(double i){
    return modf(i*0,0769230769,i)<1e-3;
}
```
this might have been a cop-out with using 1/13 and a double can store any 32 bit number exactly
edit: this works by multiplying with 1/13 and checking the fractional part if it's different from 0 (allowing for rounding errors) or in other words it check the fractional part of i/13
[Answer]
## Python 2.7
(20 - 1 whitespace) \* 0.9 (no division) = **17.1**
```
print input()%015==0
```
yes/no instead of true/false: 31 \* 0.9 (no division) = **27.9**
```
print'yneos'[input()%015!=0::2]
```
takes advantage of python's `int` to convert other bases from strings into base 10 integers. you can see in both versions they use a different (yet same character length) base
edit: 1 char save in yes/no version
edit2: another 2 chars shaved!
edit3: thanks again to comments! even more characters shaved off by using python's builtin octal representations (`015` == `13`...) instead of int's base translation
[Answer]
## Perl, 95 \* 0.8 = 76
```
$_=<>;
while($_>0){
$q=7*chop;
$d=3*($m=chop$q);
chop$d;
$_-=$d+$m}
if($_){print"no"}
else{print"yes"}
```
The line breaks were added for clarity. I could have probably made this answer a lot shorter, but I feel that this answer represents a unique way of approaching the problem.
[Answer]
# Python - score 27.9
**(31 characters \* 0.90)** -- forgoes some bonus for shorter code.
```
print'yneos'[2*input()%26>0::2]
```
**older version:** (47 characters \* 0.80) -- complete rip-off of mellamokb's Javascript answer, but in Python.
```
n=2*input()
while n>0:n-=26
print'yneos'[n<0::2]
```
**older version:** (60 characters \* 0.80)
```
n=input()
while n>12:
 for _ in'x'*12+'!':n-=1
print'yneos'[n>0::2]
```
**older version:** (105 characters \* 0.80)
```
n=abs(input())
while n>12:n=abs(sum(int(x)*y for x,y in zip(`n`[::-1],n*(1,-3,-4,-1,3,4))))
print'yneos'[n>0::2]
```
[Answer]
## In Q:
```
d:{$[0=x mod "I"$((string 6h$"q")[1 2]);`yes;`no]}
50*.9=45
```
[Answer]
# **Right Linear Grammar - ∞ points**
```
S->ε
S->1A
S->0S
S->9I
S->3C
S->5E
S->4D
S->2B
S->7G
S->6F
S->8H
F->3K
K->0F
A->2L
K->1G
A->5B
A->0J
B->7A
J->5A
G->6K
G->8S
H->9K
F->5S
K->2H
I->6E
I->5D
J->4S
D->8I
B->6S
K->9B
F->6A
G->9A
K->6L
K->4J
C->1E
L->8K
E->5C
B->4K
C->0D
J->2K
D->2C
A->9F
J->7C
C->6J
C->8L
E->0K
L->0C
B->9C
E->2S
L->6I
I->0L
J->0I
B->2I
I->3B
H->1C
I->7F
C->4H
F->1I
G->4I
I->0G
C->3G
F->8C
D->0A
E->3A
I->9H
A->7D
C->2F
H->7I
A->8E
F->9D
E->8F
A->6C
D->6G
G->0E
D->5F
E->9G
H->2D
D->7H
H->3E
I->2A
K->3I
C->9S
C->7K
E->4B
D->1B
L->1D
J->9E
I->1S
E->1L
J->8D
D->9J
L->2E
J->3L
B->5L
B->8B
L->7J
L->9L
G->1F
A->4A
K->5K
B->3J
H->6H
E->7E
J->1J
D->4E
G->2G
J->6B
D->3D
E->6D
H->4F
I->4C
C->5I
F->0H
H->5G
K->7S
G->3H
L->5H
H->8J
A->3S
H->0B
B->1H
G->7L
K->8A
F->2J
F->7B
L->4G
F->4L
A->1K
B->0G
G->5J
L->3F
```
Then depending on how you choose to 'run' it, it will output 'yes' or 'no'.
Not a serious entry, just some fun ;)
**EDIT:**
Perhaps I should explain a bit.
A [grammar](http://en.wikipedia.org/wiki/Formal_grammar) is a set of rules (productions) which define a [language](http://en.wikipedia.org/wiki/Formal_language). A language can be thought of as all of the possible strings formed by an alphabet, that conform to the rules of it's grammar.
Here the alphabet is the set of all decimal digits. The grammar's rules are that all strings must form decimal integers that are divisible by 13.
We can use the grammar above to test whether a string belongs to our language.
The rules of the grammar contain terminal symbols (which are elements in the language) as well as non-terminal symbols which are replaced recursively.
It's easier to explain what's going on with an example:
Lets say for example that the string we are testing is 71955.
There is always a start symbol (which is non-terminal), in the case of the grammar above this is 'S'. At this point we have not read any characters from our string:
```
current pattern                    symbol read
S                                  ε
```
Now, we read the first symbol in our string which is '7', then we look for a rule in the grammar which has any of the non-terminals in our current pattern in the left hand side of the '->' and that has our symbol in the right hand side of the '->'.
Luckily there is one (S->7G), so we replace the non-terminal symbols in our current pattern with the right hand side of the new rule:
```
current pattern                    symbol read
7G                                 7
```
Now we have the non-terminal 'G' in our pattern, and the next symbol to be read is '1', So we look for a rule in our grammar that begins with 'G->1". We find there is one (G->1F), so we replace the non terminal with the RHS of our new rule:
```
current pattern                    symbol read
71F                                1
```
Keep repeating this process:
Next rule: F->9D
```
current pattern                    symbol read
719D                               9
```
Next rule: D->5F
```
current pattern                    symbol read
7195F                              5
```
Next rule: F->5S
```
current pattern                    symbol read
71955S                             5
```
At this point we have no more symbols in our string, but we have another non-terminal symbol in there. We see from the first rule in the grammar that we can replace 'S' with the empty string (ε): S->ε
Doing so gives us the current patter: 71955ε which is the equivalent to 71955.
We have read all of the symbols in our string, and the pattern contains no non-terminal symbols. Which means that the string belongs to the language and therefore 71955 is in fact divisible by 13.
I.e. the goal is to have pattern = string.
If you are left with any non-terminal symbols, after reading all of the symbols in your string, the string doesnt belong to the language. Likewise, if you still have more symbols in your string to read, but there are no rules in the grammar allowing you to go forward, then the string does not belong to the language.
]