/*
Prob#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.
Prob#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.
*/
Code:
#include<stdio.h> /* * * Prosen Ghosh * American International University - Bangladesh (AIUB) * */ int main (){ long int i=1,sum=0,testsum=0,value1=3,value2=5; for(;i < 1000;){ testsum = value1*i; if(testsum < 1000){ sum+=testsum; } testsum=value2*i; if(testsum < 1000){ if(testsum % value1 != 0){ sum += testsum; } } i++; } printf("The result is = %ld\n",sum); return 0; }