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.
public class Multiple {
/**
* @param args
*/
public static void main(String[] args) {
int sum = 0;
for(int i=3;i<1000;i++){
// 避免3和5的公倍数重复相加求和
if(i%3==0 || i%5==0){
sum += i;
}
}
System.out.println(sum);
}
}
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3和5的倍数问题: :