Find the sum of all numbers below n that are a multiple of some set of numbers

Question

Almost equivalent to Project Euler's first question:

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

Challenge:

Given a positive integer N and a set of at least one positive integer A, output the sum of all positive integers less than N that are multiples of at least one member of A.

For example, for the Project Euler case, the input would be:

1000
3
5

Test cases:

Input : 50, [2]
Output: 600

Input : 10, [3, 5]
Output: 23

Input : 28, [4, 2]
Output: 182

Input : 19, [7, 5]
Output: 51

Input : 50, [2, 3, 5]
Output: 857

1) Do we count numbers that are multiples of both twice? 2) Can we only get two other numbers? or any amount say one or 3? — Wheat Wizard, Commented Dec 19, 2016 at 3:48
Can you give some test cases? Obviously don't post the answer to the PE one, but what about other examples? — Riker, Commented Dec 19, 2016 at 3:55
@WheatWizard: The word "or" implies that each number is counted only once, at most. I agree that the question needs to make it clear how many "numbers to check for multiples of" arguments must be supported, though. Exactly two? One or more? Zero or more? — smls, Commented Dec 19, 2016 at 3:58
Can we take "numbers equal to or below 10", or take 9 as input instead of 10? — Stewie Griffin, Commented Dec 19, 2016 at 7:41
"and a set of at least one positive integer A" how big can the set be? — betseg, Commented Dec 19, 2016 at 7:41

coredump · Accepted Answer · 2016-12-19 13:05:45Z

1

Common Lisp, 77

(lambda(n x)(loop for i below n when(some(lambda(u)(zerop(mod i u)))x)sum i))

Ungolfed

(lambda (limit seeds)
  (loop for i below limit
        when (some (lambda (u) (zerop (mod i u))) seeds)
          sum i))

answered Dec 19, 2016 at 13:05

coredump

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AdmBorkBork · Accepted Answer · 2016-12-19 16:15:54Z

PowerShell, 57 bytes

param($a,$b)(1..--$a|?{$i=$_;$b|?{!($i%$_)}})-join'+'|iex

Try it online!

Iterative solution. Takes input as a number $a and as a literal array $b. Loops from 1 up to one below $a (via --$a), using a Where-Object operator |?{...} with a clause to select certain numbers.

The clause sets $i to be the current number before sending input array $b into another |?{...}, here picking out those items where the current number is evenly divided by at least one of the numbers in $b. Those elements of $b that do divide evenly are left on the pipeline.

Thus, if there is at least one element from $b, the pipeline contains an element, so the outer Where is $true and the current number is left on the pipeline. Otherwise, with no elements from $b on the pipeline, the outer Where is $false, so the current number is not placed on the pipeline.

Those numbers are all gathered up in parens, -joined together with + signs, and piped to |iex (short for Invoke-Expression and similar to eval). The summation result is left on the pipeline, and output is implicit.

Community · Accepted Answer · 2020-06-17 09:04:33Z

1

PHP, 78 76 74 bytes

for(;++$i<$argv[$f=$k=1];$s+=$i*!$f)for(;$v=$argv[++$k];)$f*=$i%$v;echo$s;

The outer loop runs $i from 1 to below first argument and adds $i to $s if $f is not set.
The inner loop multiplies $f with ($i modulo argument) for all subsequent arguments, setting $f to 0 if $i is the multiple of any of them.

Run with -r.

edited Jun 17, 2020 at 9:04

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1

answered Dec 19, 2016 at 17:00

Titus

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Scala, 47 bytes

n=>1.to(n(0)-1).filter(i=>n.exists(i%_==0)).sum

n is a List which contains of a first argument N, the rest are elements of A

Works by filtering out numbers where there doesn't exist at least one A of which i is a multiple, then summing. Strictly speaking we should use n.tail.exists inside the closure, but as i is always less than N and therefore never a multiple of N the solution is still complete without this.

edited Jun 17, 2020 at 9:04

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1

answered Dec 20, 2016 at 11:31

sprague44

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NonlinearFruit · Accepted Answer · 2016-12-20 15:52:58Z

1

Java 8, 75 bytes

(N,A)->IntStream.range(1,N).filter(x->A.stream().anyMatch(y->x%y==0)).sum()

The method signature for this is int f(int N, List<Integer> A)

answered Dec 20, 2016 at 15:52

NonlinearFruit

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Community · Accepted Answer · 2020-06-17 09:04:33Z

C11, 177 bytes

#include"object.h"
#define S size_t
S g(S m,array_t*d){S s,i,l,j;for(s=i=0;i<m;i++){for(l=1,j=0;j<d->idx+1;l*=i%(S)(*array_get_ref(d,j++,NULL))->fwi->value);s+=l?0:i;}return s;}

Requires this set of headers in the same folder, and the fnv-hash library found there as well. Compile like gcc 1.c ../fnv-hash/libfnv.a -o 1 -DNODEBUG

Test Program:

#include "../calc/object/object.h"
#include <stdio.h>

size_t f (const size_t max, const size_t a, const size_t b);
size_t f2 (const size_t max, const array_t* const divs);
size_t g (size_t max, array_t* divs);

define_array_new_fromctype(size_t);

int main(void) {
  printf("%zu\n", f(10, 3, 5));
  static const size_t a[] = {
    3, 5
  };
  array_t* b = array_new_from_size_t_lit(a, 2, t_realuint);
  printf("%zu\n", f2(10, b));
  printf("%zu\n", g(10, b));
  array_destruct(b);
  return 0;
}

size_t f (const size_t max, const size_t a, const size_t b) {
  size_t sum = 0;
  for (size_t i = 0; i < max; i++) {
    sum += (i % a * i % b) ? 0 : i;
  }
  return sum;
}

size_t f2 (const size_t max, const array_t* const divs) {
  size_t sum = 0;
  const size_t len = array_length(divs);

  for (size_t i = 0; i < max; i++) {
    size_t mul = 1;
    for (size_t j = 0; j < len; j++) {
      object_t** this = array_get_ref(divs, j, NULL);

      fixwid_t*   num = (*this)->fwi;

      mul *= i % (size_t) num->value;
    }
    sum += mul ? 0 : i;
  }
  return sum;
}

#define S size_t
S g(S m,array_t*d){S s,i,l,j;for(s=i=0;i<m;i++){for(l=1,j=0;j<d->idx+1;l*=i%(S)(*array_get_ref(d,j++,NULL))->fwi->value);s+=l?0:i;}return s;}

outputs

23
23
23

mkst · Accepted Answer · 2018-11-11 22:16:47Z

1

K (oK), 15 14 bytes

Solution:

{+/&|/~y!\:!x}

Try it online!

Examples:

{+/&|/~y!\:!x}[50;,2]
600
{+/&|/~y!\:!x}[10;3 5]
23

Explanation:

{+/&|/~y!\:!x} / the solution
{            } / lambda taking implicit x and y
           !x  / range 0..x-1
       y!\:    / modulo (!) x with each-left (\:) item in y
      ~        / not
    |/         / min-over to flatten into single list
   &           / indices where true
 +/            / sum up

edited Nov 11, 2018 at 22:16

answered Nov 11, 2018 at 22:06

mkst

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user96495user96495 · Accepted Answer · 2020-07-30 06:30:27Z

Io, 63 bytes

Port of NikoNyrh's answer.

method(N,A,A map(i,Range 0toBy(N-1,i)asList)flatten unique sum)

Try it online!

Explanation

method(N,A,                                                   ) // Take two inputs, the number and the array
           A map(i,                        )                    // For each item in the array A:
                   Range 0toBy(N-1,i)                           //     Generate a range from 0 to N-1, with the step of i
                                     asList                     //     Convert it to a list
                                            flatten             // Flatten the lists
                                                    unique      // Uniquify them
                                                           sum  // Sum the resulting list

ZippyMagician · Accepted Answer · 2020-08-12 14:36:22Z

Arn, 15 bytes

pIÔX╗▀‹·¬○¯vËa?

Explained

Unpacked: +\$v{$:{!v%}?1}1->

  \ Fold with...
+ ...addition
    $ Filter with...
      v{ ...block with key of v
        $: Any operation
          { Block with key of _
            ! Boolean not
              v% v modulo _ (implied)
          }
          ?1 Each element on second line of _ (implied)
        }
        1-> Range, [1, _)

When passing an array in string or integer context, Arn will automatically take the 0th element. When passing a string/integer in array context, Arn will split on spaces if there are any or every character otherwise.

STDIN is initialized to the _ variable, split on newlines if there are any or just a string if one line.

Originally I had +\$v{$:{0=v%}?1}1->, which was 1 byte longer.

user · Accepted Answer · 2020-12-03 19:22:56Z

1

Scala, 45 41 bytes

s=>0 to _-1 filter(x=>s.exists(x%_<1))sum

Try it in Scastie

Takes as its input a List[Int] s and an Int, creates a range from 0 to that number, filters every number in that range that's divisible by an element in s, and sums it.

Requires language:postfixOps.

edited Dec 3, 2020 at 19:22

answered Aug 3, 2020 at 21:20

user

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tybocopperkettle · Accepted Answer · 2022-10-10 02:05:24Z

1

Vyxal `s`, 5 bytes

ɽ'⁰Ḋa

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Explanation:

ɽ      # Range 1 to n-1
 '     # Keep those that are
   Ḋ   #   Divisible by
    a  #   Any of
  ⁰    #   The second input
       # Sum with the s flag

answered Oct 10, 2022 at 2:05

tybocopperkettle

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The Thonnu · Accepted Answer · 2023-07-08 18:38:34Z

Thunno 2 `S`, 5 bytes

Læ¹ḊS

Try it online!

Thunno 2 `S`, 6 bytes

LsȷḊʂV

Try it online!

Explanations

Læ¹ḊS   # Implicit input           ->  [2,3,5], 50
L       # Lowered range            ->  [2,3,5], [0..49]
 æ      # Filtered by:             ->     0       1    ...   48      49
  ¹Ḋ    #  Divisible by 2nd input  ->  [1,1,1] [0,0,0]     [1,1,0] [0,0,0]
    S   #  Sum (any are true)      ->     3       0    ...    2       0
        # (End filter)             ->  [0,2,3,4,5,...,42,44,45,46,48]
        # (S flag) Sum the list    ->  857
        # Implicit output

LsȷḊʂV  # Implicit input
        #   [2,3,5], 50
L       # Take the lowered range of the first input
        #   [2,3,5], [0..49]
 s      # Swap the two values on the stack
        #   [0..49], [2,3,5]
  ȷḊ    # Outer product over divisibility check
        #   [[0%2==0, 0%3==0, 0%5==0], ..., [49%2==0, 49%3==0, 49%5==0]]
        #   [[1,1,1], [0,0,0], [1,0,0], ..., [0,0,0], [1,1,0], [0,0,0]]
    ʂ   # Sum each inner list (check if any are true)
        #   [3,0,1,1,1,...,2,1,0,2,0]
     V  # Get the indices of truthy (non-zero) items
        #   [0,2,3,4,5,...,42,44,45,46,48]
        # (S flag) Sum the resulting list
        #   857
        # Implicit output

thejonymyster · Accepted Answer · 2023-08-25 03:20:46Z

Headass, 74 bytes

{N)OUO}:+E.UO^UO[{UON)}:](D)RP:N+E.U+O-[UOU{R)U-ORO^:R-OUO;UN)}:D)R:R];ONE

Try It Online! - Input is taken as a newline delimited list of numbers, with the first one being N, and the rest being the values of A

tattoo on my forehead that says I LOVE EASY CHALLENGES

Explanation

PREAMBLE: this program counts up from 0 to N, with this increasing value being
referred to as the "counter". along the way it reads members of A, which
will be given "countdown timers", which start at 0 and are reset to the
associated A member value minus one, so that every A[i]th cycle, they reach
0 and trigger the adding of the counter to the accumulator.
The accumulator, of course, accumulatizes.

{N)OUO}:+E.  code block 0 (initialization)
{N)   }:     for each input
   OUO         save 0, save input
               this inits the counter, stores N,
               creates a list of countdown timers starting at 0
               and implicitly, an accumulator starting at 0
               all on the queue. the queue now looks like this:
               0 N 0 A1 0 A2 0 A3... 0
        +E   go to code block 1
          .  end code block

UO^UO[{UON)}:](D)RP:N+E.  code block 1 (checking if we're done)
UO^                       set r1 to the counter value
   UO[                    set r2 to N
      {UON)}:             put the rest of the values back on the queue,
                          noting that the accumulator is the last value
             ](D)         if r1 == r2 (the counter has reached N)
                 RP         print the last value added to the queue
                            (which is the accumulator)
                            (and halt)
                   :      else
                    N+E     go to code block 2
                       .  end code block

U+O-[UOU{R)U-ORO^:R-OUO;UN)}:D)R:R];ONE  code block 2
U+O                                      put the counter + 1 onto the queue
U+ -[                                    set r2 to the counter value
     UO                                  put N back on the queue
        {                N)}:            for each member of A
       U R)             U                  if the countdown timer has reached 0
           U-O                               reset the countdown timer
              RO                             put the A member back on the queue
                ^                            raise a flag in r1
                 :                         else
                  R-O                        decrement the countdown timer
                     UO                      put the A member back on the queue
                       ;                   endif
                             D)          if the flag in r1 is down
                               R:  ;O      put the accumulator back on the queue
                                :        else
                                 R];O      increase the accumulator by the
                                           value of the counter
                                     NE  return to code block 1

There's a bunch of implicit re-adding of values to the queue that I don't mention, basically any time I say that I've modified some value, that means I'm saving the modified version to the queue.

Also, registers r0-r3 are reset between code blocks, so you can basically think about them as local variables in each code block. Register r0 is constantly in use but not mentioned due to the language structure, and register r3 is in use whenever ( or ) show up, but it's much easier to abstract these to simple comparisons, since that's all they're used for in this program.

I'm still always learning the best way to explain this language, trying to balance between overexplaining and overabstracting, if anyone has any questions of course feel free to ask, you may have a golf on your hands :P

Ashlin Harris · Accepted Answer · 2023-08-28 13:10:51Z

1

Julia 1.0, 31 bytes

N^A=sum(x->any(@.x%A<1)x,1:N-1)

Try it online!

The function sum can accept a subfunction to be called on each element before summation. I use this to multiply each element x in the range \$[1, N-1]\$ by the boolean value any(@.x%A<1), essentially filtering the range. Julia supports logical indexing for arrays, but this generally requires defining the array first, which often isn't ideal for code golf.

The expression @. vectorizes the functions that follow, but in this case, it doesn't save any bytes over the alternative, x.%A.<1.

-1 byte thanks to MarcMush: use the result of any as a literal coefficient

edited Aug 28, 2023 at 13:10

answered Aug 24, 2023 at 15:53

Ashlin Harris

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1

\$\begingroup\$ any(...)x should work for -1 byte \$\endgroup\$
– MarcMush
Commented Aug 27, 2023 at 19:18

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int 21h Glory to Ukraine · Accepted Answer · 2024-09-13 15:30:01Z

Husk, 14 bytes

ΣuΣzom*⁰Ṫ¦⁰tŀ²

Try it online!

I must admit that the handling of two lists (a number range and the list of divisors) was quite difficult and, subsequently, this code might look quite long.

This program takes a number and a list of divisors as an input and outputs the sum of all multiples the input list.

Explanation:

            ŀ²  # Make a range 0 to input number minus 1
           t    # Drop 0
        Ṫ¦⁰     # Make a table of divisors:
                # if the value is divisible, put quotient, otherwise put 0
   z            # Use zip to get the dividends back
    om*⁰        # By multiplying with the list of divisors (zeroes stay zeroes)
  Σ             # Simplify the list
 u              # Deduplicate 
Σ               # Calculate the total sum

Community · Accepted Answer · 2020-06-17 09:04:33Z

0

Processing, 88 bytes

int q(int a,int[]b){int s=0,i=0;for(;++i<a;)for(int j:b)if(i%j<1){s+=i;break;}return s;}

Uses the simple for-loop approach, sums all the multiples up and returns it. Input is the format int, int[]array.

edited Jun 17, 2020 at 9:04

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1

answered Dec 19, 2016 at 8:52

user41805

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\$\begingroup\$ Isn't it simpler to just name it java? \$\endgroup\$
– Razetime
Commented Aug 12, 2020 at 14:39

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miles · Accepted Answer · 2016-12-19 10:38:42Z

0

J, 18 bytes

[:+/@I.0=*/@(|/i.)

Try it online!

Explanation

[:+/@I.0=*/@(|/i.)  Input: A (LHS), N (RHS)
               i.   Make the range [0, 1, ..., n-1]
             |/     Form the modulus table between each value in A and the range N
         */@(    )  Reduce columns by multiplication
       0=           Test if each value if equal to 0, 1 if true else 0
[:   I.             Find the indicies of 1s
  +/@               Reduce by addition and return

answered Dec 19, 2016 at 10:38

miles

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Xanderhall · Accepted Answer · 2016-12-19 14:01:09Z

PHP, 80 bytes

for($c=1;++$i<$argv[1];$c++?:$t+=$i)for($j=1;$v=$argv[++$j];$i%$v?:$c=0);echo$t;

Run from command line with -r. First argument is the limit N, any subsequent arguments are our positive integers A.

Our outer loop goes through the range 1..N, while the inner loop goes through each value of A and checks to see if $i is a multiple of it. If it is, we set a flag that tells our outer loop to add $i to our total ($t). We use a flag so that we don't add $i more than once. We then output our total at the end.

NikoNyrh · Accepted Answer · 2016-12-19 23:13:49Z

0

Clojure, 44 bytes

#(apply +(set(mapcat(partial range 0 %)%2)))

Instead of generating the whole range form 0 to N it is generating values on that range with step length of A_0, A_1, etc. and uses set to get unique values.

Calling convention:

(def f #(apply +(set(mapcat(partial range 0 %)%2))))
(f 50 [2 3 5])

Steps are more clear when ->> macro is used:

(defn f [N A] (->> A (mapcat #(range 0 N %)) set (apply +)))

answered Dec 19, 2016 at 23:13

NikoNyrh

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NikoNyrh · Accepted Answer · 2016-12-19 23:24:16Z

0

Python 3 & 2.7, 55 bytes

Based on the same set idea as my Clojure version.

lambda N,A:sum(set(i for a in A for i in range(0,N,a)))

answered Dec 19, 2016 at 23:24

NikoNyrh

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Neil · Accepted Answer · 2016-12-20 01:39:16Z

Batch, 110 bytes

@set t=-%1
@for /l %%i in (1,1,%1)do @set p=1&(for %%e in (%*)do @set/ap*=%%e%%%%i)&set/at+=%%i*!p
@echo %t%

Takes input as command-line arguments; the first argument is n, the rest are the set of numbers. Works by computing the product of the remainders of all numbers up to and including n with each of the arguments including n itself and summing those numbers with a zero product; this causes n to always be included in the total so we offset that by starting the total at -n.

Community · Accepted Answer · 2020-06-17 09:04:33Z

C#, 92 bytes

n=>{int i=0,j,s=0;for(;i<n[0];i++)for(j=1;j<n.Length;)if(i%n[j++]<1){s+=i;break;}return s;};

Anonymous function which receives an array containing the upper limit and all required divisors.

This method does not use a set (actually HashSet in C#) since it was way too verbose. Not that this code is too succinct...

Full program and test cases:

using System;

public class Program
{
    public static void Main()
    {
        Func<int[], int> f =
        n =>
        {
            int i = 0, j, s = 0;
            for ( ; i < n[0] ; i++)
                for (j = 1 ; j < n.Length ; )
                    if (i % n[j++] < 1)
                    {
                        s += i;
                        break;
                    }
            return s;
        };

        // test cases:
        Console.WriteLine(f(new int[]{50, 2})); // 600
        Console.WriteLine(f(new int[]{10, 3, 5}));  // 23
        Console.WriteLine(f(new int[]{28, 4, 2}));  // 182
        Console.WriteLine(f(new int[]{19, 7, 5}));  // 51
        Console.WriteLine(f(new int[]{50, 2, 3, 5}));   // 857
    }
}

James Brown · Accepted Answer · 2016-12-21 14:53:19Z

0

awk, 64 bytes

{for(;i++<NF;j=0)while((c=$i*++j)&&c<$NF)a[c];for(i in a)$0+=i}1

Write the code to file program.awk and execute:

$ echo run 2 3 5 50 | awk -f this.awk 
857

run is a cludge to allow using record $0 for summing and shorter printing, emphasis on the latter.

Try it online

answered Dec 21, 2016 at 14:53

James Brown

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James Brown · Accepted Answer · 2016-12-21 15:15:54Z

0

R, 39 bytes

z=2:N-1;z%*%!!rowSums(!outer(z,X,"%%"))

N is the number, X is a vector of divisors

edited Dec 21, 2016 at 15:15

James Brown

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answered Dec 19, 2016 at 20:21

skan

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rnso · Accepted Answer · 2016-12-21 16:11:21Z

0

Racket 94 bytes

(apply +(remove-duplicates(for*/list((i(range 1(car l)))(j(cdr l))#:when(= 0(modulo i j)))i)))

Ungolfed:

(define (f l)
  (apply +
         (remove-duplicates
          (for*/list
              ((i (range 1 (first l)))
               (j (rest l))
               #:when (= 0 (modulo i j)))
            i))))

Testing:

(f '(10 3 5))
(f '(50 2))
(f '(28 4 2))
(f '(19 7 5))
(f '(50 2 3 5))

Output:

edited Dec 21, 2016 at 16:11

answered Dec 21, 2016 at 15:55

rnso

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Xcali · Accepted Answer · 2019-06-27 17:22:37Z

0

Perl 5 `-pa`, 44 bytes

map{$"="*$_%";$\+=!eval"$_%@F"&&$_}1..<>-1}{

Try it online!

Takes the list of A on the first line of input, space separated. Takes N on the second line.

answered Jun 27, 2019 at 17:22

Xcali

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Community · Accepted Answer · 2020-06-17 09:04:33Z

0

APL(NARS), 17 chars, 34 bytes

{+/∪⍺∼⍨∊⍵×⍳¨⌊⍺÷⍵}

test:

  f←{+/∪⍺∼⍨∊⍵×⍳¨⌊⍺÷⍵}
  50 f 2
600
  10 f 3 5
23
  28 f 4 2
182
  19 f 7 5
51
  50 f 2 3 5
857

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 5, 2018 at 20:04

user58988

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Find the sum of all numbers below n that are a multiple of some set of numbers

Challenge:

Test cases:

57 Answers 57

Common Lisp, 77

Ungolfed

PowerShell, 57 bytes

PHP, 78 76 74 bytes

Scala, 47 bytes

Java 8, 75 bytes

C11, 177 bytes

K (oK), 15 14 bytes

Io, 63 bytes

Explanation

Arn, 15 bytes

Explained

Scala, 45 41 bytes

Vyxal s, 5 bytes

Thunno 2 S, 5 bytes

Thunno 2 S, 6 bytes

Explanations

Headass, 74 bytes

Explanation

Julia 1.0, 31 bytes

Husk, 14 bytes

Explanation:

Processing, 88 bytes

J, 18 bytes

Explanation

PHP, 80 bytes

Batch, 110 bytes

C#, 92 bytes

awk, 64 bytes

R, 39 bytes

Racket 94 bytes

Perl 5 -pa, 44 bytes

APL(NARS), 17 chars, 34 bytes

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