A problem

Lesson 14 of 20 in Coddy's Mathematical Riddles course.

The least common multiplier can be applied to more than two numbers.

It obeys the following:

LCM(x,y,z) = LCM(LCM(x,y),z)

Challenge

Easy

Write a function calcSmallestNumberDivideableBy1toN that gets an integer, N, and calculates the smallest number that can be divided by each of the numbers from 1 to N without any remainder.

Try it yourself

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
#include "solution.h"

int main() {
    int n;
    if (scanf("%d", &n) != 1) n = 0;
    int r = calcSmallestNumberDivideableBy1toN(n);
    printf("%d\n", r);
    return 0;
}