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You are given three parameters: start(int), end(int) and list(of int);

Make a function that returns the amount of times all the numbers between start and end are multiples of the elements in the list. example:

start = 15; end = 18; list = [2, 4, 3]; 15 => 1 (is multiple of 3) 16 => 2 (is multiple of 2 and 4) 17 => 0 18 => 2 (is multiple of 2 and 3) result = 5

The function should accept two positive integer numbers and an array of integers as parameters, returning the total integer number. Assume that start is less <= end.

examples: 

Multiple(1, 10, [1, 2]); => 15 Multiple(1, 800, [7, 8]); => 214 Multiple(301, 5000,[13, 5]); => 1301

The shortest solution is the victor!!! May he odds be ever in your favor...

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    \$\begingroup\$ Nice challenge! Presumably the winning criterion is shortest code? In which case you should say so explicitly, and add the code-golf tag. \$\endgroup\$ Commented Dec 2, 2019 at 10:16
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    \$\begingroup\$ Can we assume that start will always be less than end? \$\endgroup\$ Commented Dec 2, 2019 at 11:49
  • \$\begingroup\$ You say integers. So start, end or the elements in list may be non-positive? \$\endgroup\$ Commented Dec 3, 2019 at 11:17
  • \$\begingroup\$ Thanks for the feedback, I was lacking that in the problem. \$\endgroup\$ Commented Dec 3, 2019 at 14:43
  • \$\begingroup\$ I'd like to ask a question (as I'm really new in this channel), how do I decide the winner? since each language has it's own limitations... Or is it simply a personal preference? \$\endgroup\$ Commented Dec 6, 2019 at 8:52

31 Answers 31

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05AB1E, 5 bytes

ŸÑ˜åO

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Explanation

Ÿ - the numbers between start and end Ñ - get their divisors ˜ - deep flatten this list å - find the instances of the elements in the list in these divisors O - Sum this
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10
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Haskell, 33 bytes

(a#b)x=sum[1|0<-mod<$>[a..b]<*>x]

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Explanation:

(a#b)x --take two values a and b and the list x [a..b] --generate the range a, a+1, ... , b mod<$>[a..b] --partially apply "mod" to each entry of the list mod<$>[a..b]<*>x --apply the partially applied functions to all values of the list x [1|0<-mod<$>[a..b]<*>x] --generate a new list with a one for every zero in the previously computed list (a#b)x=sum[1|0<-mod<$>[a..b]<*>x] --sum it all up
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  • \$\begingroup\$ Nice method. I found a variant that ties and tried to shorten it point-free but could tie at best -- maybe you have a better idea. \$\endgroup\$ Commented Dec 3, 2019 at 9:58
  • \$\begingroup\$ @xnor I've also tried to get rid of the x, but without success, the (0^) trick is neat, please consider posting it as an own answer! (So far I have had no success in golfing your suggestions further.) \$\endgroup\$ Commented Dec 3, 2019 at 15:45
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JavaScript (Node.js), 44 bytes

(a,x,y)=>a.map(n=>t+=y/n-(~-x/n|0)|0,t=0)&&t

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Python 2, 38 bytes

lambda a,x,y:sum(y/i-~-x/i for i in a)

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\$ \sum _{i \in list} \lfloor \frac{end}{i} \rfloor - \lfloor \frac{start-1}{i} \rfloor \$

The answer seems trivial and naive. So am I misunderstand something?

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  • \$\begingroup\$ Your Python answer is right -- I was going to post the same. Congrats on figuring out the simplification. I'm disappointed nobody else had found the floor-division idea earlier, at least on my quick glance of the answers. \$\endgroup\$ Commented Dec 3, 2019 at 9:50
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    \$\begingroup\$ @xnor the formula comes up with my mind just after I read the question. But since the question is posted ~1d ago (and no one using it), I can't quite believe it could be correct. After posting the JavaScript version, I realized that by using Python I could save many bytes.. \$\endgroup\$ Commented Dec 3, 2019 at 9:55
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Perl 6, 21 bytes

{sum $^a..$^b X%%@^c}

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Explanation

{ } # Anonymous codeblock returning sum # The sum of $^a..$^b # How many numbers in the range a to b X%% # Are divisible by any of @^c # The third parameter
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MATL, 5 bytes

:i\~z

Inputs are a cell array containing the start and end values, and then a column vector defining list.

Try it online! Or verify all test cases.

Explanation

: % Range, with implicit input: cell array {start, end}. This gives the range % from start to end i % Input: list of numbers in the form of a column vector \ % Modulo, with boradcast. This gives a matrix with each number in the range % modulo each number in the list ~ % Logical negation z % Number of nonzero entries. Implicit display
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Python 3, 54 bytes

lambda s,e,L:sum(n%l<1for n in range(s,e+1)for l in L)

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5
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APL+WIN, 20 15 bytes

5 bytes saved thanks to Adam.

Prompts for end integer, start integer and then list:

+/,0=⎕∘.|⎕↓0,⍳⎕

Try it online! Courtesy of Dyalog Classic

Could be reduced to 9 bytes if we could enter two lists and not have to generate the first:

+/,0=⎕∘.|⎕
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  • \$\begingroup\$ +/+/ → +/, \$\endgroup\$ Commented Dec 2, 2019 at 11:50
  • \$\begingroup\$ s+⍳1+⎕-s←⎕ → ⎕↓0,⍳⎕ \$\endgroup\$ Commented Dec 2, 2019 at 11:53
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R, 35 bytes

function(a,b,l)sum(b%/%l-(a-1)%/%l)

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Implementation of tsh's method by Giuseppe. 3 bytes shorter than the previous version:

R, 45 38 bytes

-7 bytes thanks to Nick Kennedy and Xi'an.

function(a,b,l)sum(!outer(a:b,l,`%%`))

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Boils down to summing the values of !(x %% y) for x in a:b and y in l (%% is \$mod\$ in R, so x %% y == 0 iff x is a multiple of y).

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  • \$\begingroup\$ Though it removes your first use of Negate, tio.run/##K/… is shorter. \$\endgroup\$ Commented Dec 2, 2019 at 11:08
  • \$\begingroup\$ @NickKennedy Yes, of course, thanks! Xian just told me as much… Don't know why I didn't think of it! I guess I was too eager to use Negate. \$\endgroup\$ Commented Dec 2, 2019 at 12:11
  • \$\begingroup\$ 35 using tsh's method. \$\endgroup\$ Commented Dec 3, 2019 at 21:08
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C++ (clang), 113 110 109 108 106 100 93 99 91 89 bytes

using I=int;I s;void f(I a,I b,I*l,I z,I&s){for(s=0;a<=b;++a)for(I i=z;i--;)s+=a%l[i]<1;}

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Saved 3 bytes thanks to @AZTECCO!!!
Saved 6 13 bytes thanks to @bznein!!!
Added 6 Saved 8 bytes thanks to @AZTECCO (and now it's playing by the rules)!!!

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  • \$\begingroup\$ You can use #import<vector> to save 1 \$\endgroup\$ Commented Dec 2, 2019 at 15:23
  • \$\begingroup\$ @AZTECCO Thanks, didn't know about #import. :-) \$\endgroup\$ Commented Dec 2, 2019 at 17:42
  • \$\begingroup\$ #include is standard.. See stackoverflow.com/a/172264/7699953 btw you can save 2 by using s+=a%i<1; instead of s+=a%i?0:1; \$\endgroup\$ Commented Dec 2, 2019 at 20:58
  • \$\begingroup\$ @AZTECCO Ah, good to see Microsoft doing something useful for the code-golfing community. Thanks for the -2! :-) \$\endgroup\$ Commented Dec 2, 2019 at 21:12
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    \$\begingroup\$ @AZTECCO So many ints! Managed to save 2 bytes using using. :-) \$\endgroup\$ Commented Dec 8, 2019 at 16:09
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Ruby, 41 bytes

->a,e,l{l.sum{|x|(a..e).count{|r|r%x<1}}}

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4
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JavaScript (ES6), 47 bytes

Takes input as (list, start)(end).

(a,x,s=0)=>g=y=>y<x?s:g(y-1,a.map(c=>y%c||s++))

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4
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K (oK), 16 bytes

{+//~z!\:x_!1+y}

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Explanation:

{+//~z!\:x_!1+y} - function with 3 arguments [x;y;z] <- (start; end; list) !1+y - a list 0 to end (inclusive) x_ - drop the first start numbers z!\: - find the modulo of each number in the range start..end with each of list ~ - negate (0 becomes 1, non-zero becomes 0) +// - reduce by addition and converge (repeat until result stops changing)

J, 22 bytes

1#.1#.0=[|/(}.i.,])/@]

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Explanation:

The verbs (functions) in J can take one (right) or two (left and right) arguments. The left argument is the list, the right one - a list of the start and end numbers. 

 ( )/@ insert the verb in () between the elements of ] the right argument (start and end) , append i. a list 0..end-1 ] to the right argument (end) }. drop start elements |/ insert modulo verb between the above list and [ the left argument (the list) 0= compare each with 0 1#. sum by base 1 conversion 1#. sum by base 1 conversion (we do it twice, because the result of |/ is a table)
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    \$\begingroup\$ Some explanation would be good \$\endgroup\$ Commented Dec 2, 2019 at 21:29
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    \$\begingroup\$ @AZTECCO I added an explanation of the K answer. I'm going to add the J one soon. \$\endgroup\$ Commented Dec 3, 2019 at 8:03
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APL (Dyalog Extended), 9 bytesSBCS

Full program, prompting for end, start, list, from stdin.

≢⍸=⎕|\⎕…⎕

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⎕ prompt for end

⎕… prompt for start and generate progression vector

⎕|\ prompt for **list* and make division remainder table

= Boolean mask indicating where equal to zero

⍸ get the indices of the Trues

≢ tally them

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Python 2, 54 bytes

f=lambda n,m,l:m/n and sum(n%i<1for i in l)+f(n+1,m,l)

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  • \$\begingroup\$ Cool trick using the recursion and division! \$\endgroup\$ Commented Dec 3, 2019 at 7:51
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Bash, 40 bytes

eval set !\$[{$1..$2}%{$3}]+;echo $[$*0]

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3
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APL(NARS), chars 13, bytes 26

{+/∊0=⍺∣../⍵}

⍺ it is the list of divisors, ⍵ it is the range. Test:

 1 2{+/∊0=⍺∣../⍵}1 10 15 7 8{+/∊0=⍺∣../⍵}1 800 214 13 5{+/∊0=⍺∣../⍵}301 5000 1301 13 5 3 11{+/∊0=⍺∣../⍵}301 5000 3294
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2
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Jelly, 7 bytes

r/ḍ@€FS

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A dyadic link taking a list of [start, end] as the left argument and the list of possible divisors as the right argument. Returns an integer indicating the total number of possible divisions.

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    \$\begingroup\$ Nice answer! My version was also a 7 byter (tryadic link). \$\endgroup\$ Commented Dec 2, 2019 at 18:11
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Java (JDK), 115 67 bytes

(s,e,d)->{int t=0;for(;s<=e;s++)for(int j:d)t+=s%j<1?1:0;return t;}

48 bytes saved thanks to Olivier

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First time golfing in Java, so this is a new experience for me.

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    \$\begingroup\$ Well, you can declare lambda methods: (s,e,d)->{...}. Also, you can remove spaces and probably use a few golfing tips. Here's my take on golfing your answer. \$\endgroup\$ Commented Dec 2, 2019 at 22:30
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Icon, 59 bytes

procedure f(a,b,l) n:=0 (a to b)%!l<1&n+:=1&\z return n end

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1
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Python 54 bytes

lambda K,M,H:sum(n%l<1for n in range(K,M+1)for l in H)
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  • \$\begingroup\$ Welcome to the site! I've edited your answer to fit the standard format, feel free to edit in any explanations, or links to online testing environments such as Try it online! \$\endgroup\$ Commented Dec 2, 2019 at 20:31
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Japt -x, 7 6 bytes

rõ ïvV

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rõ ïvV :Implicit input of arrays U=[start,end] and V=list > e.g., U=[15,18] V=[2,4,3] r :Reduce U by õ : Inclusive range > [15,16,17,18] ï V :Cartesian product with V > [[15,2],[15,4],[15,3],[16,2],[16,4],[16,3],[17,2],[17,4],[17,3],[18,2],[18,4],[18,3]] v :Reduce each pair by testing the divisibility of the first by the second > [0,0,1,1,1,0,0,0,0,1,0,1] :Implicit output of sum of resulting array > 5
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1
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Kotlin, 69 bytes

{s:Int,e:Int,l:List<Int>->(s..e).toList().sumBy{v->l.count{v%it==0}}}

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1
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Husk, 6 bytes

#0×`%…

Try it online! Nothing very special.

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Vyxal d, 5 4 bytes

ṡ$vḊ

-1 byte thanks to @lyxal

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Explanation:

ṡ # Range from a to b $ # Swap (last input is now on top) vḊ # For each in the range, test for divisibility with the last input # Deep flatten and sum with the d flag
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  • \$\begingroup\$ Try it Online! for 4 bytes \$\endgroup\$ Commented Oct 10, 2022 at 2:49
  • \$\begingroup\$ @lyxal Nice, I suspected there would be a better way to get that last input \$\endgroup\$ Commented Oct 10, 2022 at 2:53
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Charcoal, 11 bytes

Ｉ⁻Σ÷ηζΣ÷⊖θζ

Try it online! Link is to verbose version of code. Explanation:

 η Second input (End) ζ Third input (List) ÷ Vectorised integer division Σ Sum θ First input (Start) ⊖ Decremented ζ Third input (List) ÷ Vectorised integer division Σ Sum ⁻ Subtract Ｉ Cast to string Implicitly print
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0
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Pyth, 15 bytes

/.nm%LdeQ}.*PQ0

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Takes input as start, end, list

Explanation:

 } # Inclusive range .* # Splat (list becomes list of arguments) PQ # All but the last element of the input # --> inclusive range from start to end m # map each number of that L eQ # to a map of each member of Q[-1] = list % d # remainder modulo member (2, 4, 3 in example) # --> List of lists of remainders .n # flatten / 0 # count zeroes
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  • \$\begingroup\$ -5 bytes using multiple inputs and pfns \$\endgroup\$ Commented Dec 4, 2019 at 23:12
0
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Perl 5 (-ap), 33 bytes

for$i(<>..<>){$i%$_||$\++for@F}}{

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0
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C# (Visual C# Interactive Compiler), 54 bytes

(a,b,l)=>l.Where(x=>a%x<1).Count()+(++a>b?0:f(a,b,l));

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Recursive approach

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    \$\begingroup\$ It's recursive, but you don't define f. You should define it like this: f=(a,b,l)=>... \$\endgroup\$ Commented Dec 2, 2019 at 22:39
  • \$\begingroup\$ @OlivierGrégoire If it is recursive, you have to include the whole function definition: Func<int,int,IEnumerable<int>,int>f=(a,b,l)=>..., since f=(a,b,l)=>... is not a valid expression (f does not exist yet). However, it would just be shorter to do int f(int a,int b,IEnumerable<int>l)=>... \$\endgroup\$ Commented Dec 3, 2019 at 4:12
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    \$\begingroup\$ if you take in int[] instead of IEnumerable<int> you can use Length instead of Count(), removing 1 byte \$\endgroup\$ Commented Dec 3, 2019 at 18:15
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SimpleTemplate, 86 bytes

Yeah, kinda long, but it works!
Receives 2 numbers (in any order) and an array/string of 1-digit numbers (e.g.: "243"), passed to the render() method of the compiler.

{@fori fromargv.0 toargv.1}{@eachargv.2}{@ifi is multiple_}{@incR}{@/}{@/}{@/}{@echoR}

Displays the result or nothing, if none is

Ungolfed

{@set count 0} {@for i from argv.0 to argv.1} {@each argv.2 as number} {@if i is multiple of number} {@inc by 1 count} {@/} {@/} {@/} {@echo count}

Should be mostly straightforward.

You can try it on http://sandbox.onlinephpfunctions.com/code/b26cd7f75e0c4676038573c4274180891a890895

Change line 986 to test the golfed and ungolfed versions, and line 988 to get the input.

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0
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Mathematica, 52 bytes

f[s_,e_,l_]:=(q=Quotient;Total[q[e,#]-q[s-1,#]&/@l])

This uses the simplification formula from @tsh

A less clever solution at 66 bytes but more directly following the OP:

f[s_,e_,l_]:=Total[Length@Intersection[Divisors@#,l]&/@Range[s,e]]
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