How many times, are they multiples?

Question

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...

Answer 1

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

Answer 2

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?

Answer 3

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

Answer 4

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

Answer 5

MATL, 5 bytes

:i\~z

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

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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

Answer 6

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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Answer 7

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).

Answer 8

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)!!!

Answer 9

Ruby, 41 bytes

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

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Answer 10

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=⎕∘.|⎕

Answer 11

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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Answer 12

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)

Answer 13

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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Answer 14

APL (Dyalog Extended), 9 bytes^SBCS

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

Answer 15

Bash, 40 bytes

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

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Answer 16

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

Answer 17

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.

Answer 18

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.

Answer 19

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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Answer 20

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

Answer 21

Python 54 bytes

lambda K,M,H:sum(n%l<1for n in range(K,M+1)for l in H)

Answer 22

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

Answer 23

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

Answer 24

Perl 5 (-ap), 33 bytes

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

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Answer 25

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

Answer 26

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

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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.

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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.

Answer 27

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]]

Answer 28

Kotlin, 69 bytes

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

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