Given a number N. and the task is to find the sum of all multiples of 2 and 5 below N ( N may be up to 10^10).
Examples:
Input : N = 10 Output : 25 Explanation : 2 + 4 + 6 + 8 + 5
Input : N = 20 Output : 110
Approach :
We know that multiples of 2 form an AP as:
2, 4, 6, 8, 10, 12, 14….(1)
Similarly, multiples of 5 form an AP as:
5, 10, 15……(2)
Now, Sum(1) + Sum(2) = 2, 4, 5, 6, 8, 10, 10, 12, 14, 15….
Here, 10 is repeated. In fact, all of the multiples of 10 or 2*5 are repeated because it is counted twice, once in the series of 2 and again in the series of 5. Hence we’ll subtract the sum of the series of 10 from Sum(1) + Sum(2).
The formula for the sum of an A.P is :
n * ( a + l ) / 2
Whereis the number of terms,
is the starting term, and
is the last term.
is the number of terms, 
is the starting term, and 
is the last term. So, the final answer is:
S2 + S5 – S10
Below is the implementation of the above approach:
C++
// CPP program to find the sum of all// multiples of 2 and 5 below N#include <bits/stdc++.h>using namespace std;// Function to find sum of AP serieslong long sumAP(long long n, long long d){ // Number of terms n /= d; return (n) * (1 + n) * d / 2;}// Function to find the sum of all// multiples of 2 and 5 below Nlong long sumMultiples(long long n){ // Since, we need the sum of // multiples less than N n--; return sumAP(n, 2) + sumAP(n, 5) - sumAP(n, 10);}// Driver codeint main(){ long long n = 20; cout << sumMultiples(n); return 0;} |
C
// C program to find the sum of all// multiples of 2 and 5 below N#include <stdio.h>// Function to find sum of AP serieslong long sumAP(long long n, long long d){ // Number of terms n /= d; return (n) * (1 + n) * d / 2;}// Function to find the sum of all// multiples of 2 and 5 below Nlong long sumMultiples(long long n){ // Since, we need the sum of // multiples less than N n--; return sumAP(n, 2) + sumAP(n, 5) - sumAP(n, 10);}// Driver codeint main(){ long long n = 20; printf("%lld",sumMultiples(n)); return 0;}// This code is contributed by kothavvsaakash. |
Java
// Java program to find the sum of all// multiples of 2 and 5 below Nclass GFG{// Function to find sum of AP seriesstatic long sumAP(long n, long d){ // Number of terms n /= d; return (n) * (1 + n) * d / 2;}// Function to find the sum of all// multiples of 2 and 5 below Nstatic long sumMultiples(long n){ // Since, we need the sum of // multiples less than N n--; return sumAP(n, 2) + sumAP(n, 5) - sumAP(n, 10);}// Driver codepublic static void main(String[] args){ long n = 20; System.out.println(sumMultiples(n));}}// This code is contributed by mits |
Python3
# Python3 program to find the sum of # all multiples of 2 and 5 below N # Function to find sum of AP series def sumAP(n, d): # Number of terms n = int(n / d); return (n) * (1 + n) * (d / 2); # Function to find the sum of all # multiples of 2 and 5 below N def sumMultiples(n): # Since, we need the sum of # multiples less than N n -= 1; return (int(sumAP(n, 2) + sumAP(n, 5) - sumAP(n, 10))); # Driver code n = 20; print(sumMultiples(n)); # This code is contributed by mits |
C#
// C# program to find the sum of all // multiples of 2 and 5 below N using System;public class GFG{ // Function to find sum of AP series static long sumAP(long n, long d) { // Number of terms n /= d; return (n) * (1 + n) * d / 2; } // Function to find the sum of all // multiples of 2 and 5 below N static long sumMultiples(long n) { // Since, we need the sum of // multiples less than N n--; return sumAP(n, 2) + sumAP(n, 5) - sumAP(n, 10); } // Driver code static public void Main (){ long n = 20; Console.WriteLine(sumMultiples(n)); } } |
PHP
<?php// PHP program to find the sum of all // multiples of 2 and 5 below N // Function to find sum of AP series function sumAP($n, $d) { // Number of terms $n = (int)($n /$d); return ($n) * ((1 + $n) * (int)$d / 2); } // Function to find the sum of all // multiples of 2 and 5 below N function sumMultiples($n) { // Since, we need the sum of // multiples less than N $n--; return sumAP($n, 2) + sumAP($n, 5) - sumAP($n, 10); } // Driver code $n = 20; echo sumMultiples($n); // This code is contributed // by Sach_Code?> |
Javascript
<script>// Java script program to find the sum of all // multiples of 2 and 5 below N // Function to find sum of AP series function sumAP(n, d) { // Number of terms n = parseInt(n / d); return (n) * ((1 + n) * parseInt(d) / 2); } // Function to find the sum of all // multiples of 2 and 5 below N function sumMultiples(n) { // Since, we need the sum of // multiples less than N n--; return sumAP(n, 2) + sumAP(n, 5) - sumAP(n, 10); } // Driver code n = 20; document.write( sumMultiples(n)); // This code is contributed by bobby </script> |
Output:
110
Time complexity: O(1), since there is no loop or recursion.
Auxiliary Space: O(1), since no extra space has been taken.
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