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