/*

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 N.

'''

'''

This solution is based on the pattern that the successive numbers in the series follow: 0+3,+2,+1,+3,+1,+2,+3.

*/

#include <stdio.h>

int main() {

int n = 0;

int sum = 0;

scanf("%d", &n);

int terms = (n - 1) / 3;

sum += ((terms)*(6 + (terms - 1) * 3)) / 2; //sum of an A.P.

terms = (n - 1) / 5;

sum += ((terms)*(10 + (terms - 1) * 5)) / 2;

terms = (n - 1) / 15;

sum -= ((terms)*(30 + (terms - 1) * 15)) / 2;

printf("%d\n", sum);

}