Multiples of 3 or 5

/**
 * Checks whether `y` is a multiple of `x`.
 *
 * @param {number} x
 * @param {number} y
 * @returns {boolean}
 */
function isMultOf(x, y) {
  return y % x === 0;
}

function mult3or5(n) {
  if (n < 0) return 0;

  let total = 0;

  for (let i = 1; i < n; ++i) {
    if (isMultOf(3, i) && isMultOf(5, i))
      total += i;
    else if (isMultOf(3, i))
      total += i;
    else if (isMultOf(5, i))
      total += i;
  }

  return total;
}

Solution 2

When they say “If the number is a multiple of both 3 and 5, only count it once”, it means we can really simplify the conditions to this:

if (isMultOf(3, i) || isMultOf(5, i))
  total += i;

Therefore, here’s the improved version:

C

Solution 1

Same as JS solution 2.

#include <stdio.h>

/**
 * Checks whether `y` is a multiple of `x`.
 */
int is_mult_of(int x, int y) {
  return y % x == 0;
}

int sum_mults_3_or_5(int n) {
  int total = 0;

  for (int i = 1; i < n; ++i)
    if (is_mult_of(5, i) || is_mult_of(3, i))
      total += i;

  return total;
}

int main(void) {

  printf("%d\n", sum_mults_3_or_5(10));

  return 0;
}

Haskell

Solution 1

Main idea is is to use filter and list comprehension.

multsOf3or5 :: [Int] -> [Int]
multsOf3or5 = filter (\x -> rem x 3 == 0 || rem x 5 == 0)

mults :: Int -> Int
mults n = sum . multsOf3or5 $ [1 .. n - 1]


main :: IO ()
main = do
  print $ show $ mults 10

Python

Solution 1

Using range() and an is_mult_of() tiny helper function.

def is_mult_of(x, y):
    """Checks if `x` is a multiple of `y`.

    ASSUME: `y` is not 0 (zero).

    Parameters
    ----------
    x : int
    y : int

    Returns
    -------
    int
    """

    return x % y == 0

def sum_mults(num):
    """Sum all ints below `num` which are multiples of 3 or 5.

    Parameters
    ----------
    num : int

    Returns
    -------
    int
    """

    if num < 0:
        return 0

    total = 0;

    for n in range(1, num):
        if is_mult_of(n, 3) or is_mult_of(n, 5):
            total = total + n;

    return total