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

Python

# Longer version could be if i % 3 == 0 or i % 5 == 0, which will return 1
# Negating i % 3 or i % 5 produces the same result

n = 0
for i in range(1000):
  if not i % 3 or not i % 5:
    n = n + i
print n
# 233168

Typescript

// execute with
// tsc soln.ts && node soln.js


var sum = 0;
for (var i = 0; i <= 1000; i++){
    if (i % 3 == 0 || i % 5 == 0){
        sum = sum + i;
    }
}
console.log(sum);
// 233168

Swift

// execute with
// swift soln.swift
//
// compile with
// swiftc soln.swift

var sum = 0
for index in 1..<1000{
  if (index % 3 == 0) || (index % 5 == 0) {
    sum = sum + index
  }
}
print(sum)
// 233168

Git Repository / All Euler Solutions

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