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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 <iostream> #include <time.h> using namespace std; int main() { int n=0; for(int i=1;i<1000;i++){ if(i%3==0||i%5==0){ n=n+i; } } cout<<n<<endl; }