Problem 1
Problem
Find the sum of all the numbers evenly divisible by 3 or 5 below 1000.
General solution
Given the small search space, the simplest solution is to iterate over the numbers from 1 to 999 and add them to a running total if they are evenly divisible by 3 or 5.
Solution: Go
package main
import (
"fmt"
)
func main() {
var sum int = 0
for num := 1; num < 1000; num++ {
if num % 3 == 0 || num % 5 == 0 {
sum += num
}
}
fmt.Println(sum)
}
Solution: PHP
<?php
declare(strict_types=1);
error_reporting(E_ALL);
$sum = 0;
for ($num = 1; $num < 1000; $num++)
{
if ($num % 3 === 0 || $num % 5 === 0)
{
$sum += $num;
}
}
print("$sum\n");
Solution: C
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int sum = 0;
for (int i = 1; i < 1000; i++)
{
if (i % 3 == 0 || i % 5 == 0)
{
sum += i;
}
}
printf("Sum of multiples of 3 or 5 below 1000: %d\n", sum);
return EXIT_SUCCESS;
}