Problem 1
Statement
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.
Solution
The code, in C.
#include <stdio.h>
int sumMult(int a, int b){
int t=0;
int i;
for(i=0;i<1000;i++){
if(i%a==0 || i%b==0)
t+=i;
}
return t;
}
int main(void){
printf("%d\n",sumMult(3,5));
return 0;
}
Answer
233168
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