(written question) The sums of the numbers divisible by 3 and 5

Title: Given a number n, the number of all the numbers that can be divisible by 3 or 5 is not more than N. For example: n = 9, answer 3 + 6 + 5 + 9 = 23. Ideas:

What are the numbers that can be divisible by 3 or 5?

Number divisible by 3:3,6,9 .... [N/3]*3

Number divisible by 5:5,10,15 ... [N/5]*5

Number of repetitions (the number divisible by 3 and 5, which is divisible by 15): 15,30 ... [n/15]*15

So the answer to the question is obvious:

Sum of the sum of the numbers divisible by 3 or 5 and = the sum of the numbers divisible by 5 and the number of the numbers divided by 15-the sum of the numbers divisible by the sum

Because the sum of the sequence is arithmetic progression, the use of arithmetic progression summation formula can be easily solved.

• X is the first item, Y is the number of items, and D is the tolerance

• (x + x + d * (y–1)) * Y/2, note y = 0 also applicable

Code:

#include <iostream>using namespace Std;int sumofarithmeticseries (int x,int c,int d) {    return (x+x+ (c-1) *d) *c/ 2;} int main () {    int sum_3=0,sum_5=0,sum_15=0;    int n=9;    int sum=0;    Sum_3=sumofarithmeticseries (3,n/3,3);    Sum_5=sumofarithmeticseries (5,n/5,5);    Sum_15=sumofarithmeticseries (15,n/15,15);    sum=sum_3+sum_5-sum_15;    cout<<sum<<endl;    return 0;}

　　

(written question) The sums of the numbers divisible by 3 and 5