PROJECT EULER SOLUTION # 1

Solution to problem number 1 of Project Euler.

Question # 1

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 1000.

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#include<stdio.h>

#include<conio.h>

#include<time.h>

int main()

{

                long sum=0;

                int i;

                for(i=0;i<1000;i++)

                                if(i%3==0 || i%5==0)

                                {

                                                sum+=i;

                                                printf("%d\t",i);

                                }

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

                printf("EXECUTION TIME = %f",clock()/(float)CLK_TCK);

                system("pause");

}

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