C Program for Sum of all the Multiples of 3 or 5 below 1000 : From ProjectEuler.net
/* 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. */ #include<stdio.h> #include<conio.h> int main() { long sum=0,range=1; clrscr(); do{ if(range%3==0 || range%5==0) sum+=range; range++; }while(range<1000); printf("\nThe sum is: %ld", sum); getch(); return 0; } Answer: 233168 233168 233168 233168 Hints/Background Knowledge: Type Bytes Range --------------------------------------------------------------- short int 2 -32,768 -> +32,767 (32kb) unsigned short int 2 0 -> +65,535 (64Kb) unsigned int 4 0 -> +4,294,967,295 ( 4Gb) int 4 -2,147,483,648 -> +2,147,483,647( 2Gb) long int 4 -2,147,483,648 -> +2,147,483,647( 2Gb) signed char 1 -128 -> +127 unsi