Task and Solution in Java : Project Euler #1: Multiples of 3 and 5 (HackerRank)

 

This problem is a programming version of Problem 1 from projecteuler.net

If we list all the natural numbers below  that are multiples of  or , we get  and . The sum of these multiples is .

Find the sum of all the multiples of  or  below .

Input Format

First line contains  that denotes the number of test cases. This is followed by  lines, each containing an integer, .

Constraints

Output Format

For each test case, print an integer that denotes the sum of all the multiples of  or  below .

Sample Input 0

2
10
100

Sample Output 0

23
2318

Explanation 0

For , if we list all the natural numbers below  that are multiples of  or , we get  and . The sum of these multiples is .

Similarly for , we get .

Solution 

import java.io.*;

import java.math.*;

import java.security.*;

import java.text.*;

import java.util.*;

import java.util.concurrent.*;

import java.util.function.*;

import java.util.regex.*;

import java.util.stream.*;

import static java.util.stream.Collectors.joining;

import static java.util.stream.Collectors.toList;

public class Multiples_of_3_and_5 {

     // this method gives sum of n terms in AP talking first term

    // as 'a' and common differnce as d and number of terms as n

    static long apSum(long a, long d, long n){

        long ans = a*n + n*(n-1)*d/2 ;

        return ans;

    }

    public static void main(String[] args){

        Scanner sc = new Scanner(System.in);

        int t = sc.nextInt();

        for(int i = 0; i < t; i++){

            long n = sc.nextLong();

            n = n - 1; // we need to find below n so decrement n by 1

            // number of terms which are less than n and divisible by 3 are n/3

            // similarly number of terms divisible by 5 below n are n/5

            // so we find AP Sum of 3 and 5 and remove sum for 15 as 15 comes in common for 3 and 5

            System.out.println(apSum(3,3, n/3) + apSum(5,5, n/5) - apSum(15, 15, n/15));

        }

        sc.close();

    }

}

Source : HackerRank