Given a number N. Given two numbers X and Y, the task is to find the sum of all those numbers from 1 to N that are divisible by X or by Y.
Examples:
Input : N = 20
Output : 98Input : N = 14
Output : 45
Approach: To solve the problem, follow the below steps:
->Find the sum of numbers that are divisible by X upto N. Denote it by S1.
->Find the sum of numbers that are divisible by Y upto N. Denote it by S2.
->Find the sum of numbers that are divisible by both X and Y upto N. Denote it by S3.
->The final answer will be S1 + S2 – S3.
In order to find the sum, we can use the general formula of A.P. which is:
Sn = (n/2) * {2*a + (n-1)*d}
For S1: The total numbers that will be divisible by X upto N will be N/X and the sum will be:
Hence, S1 = ((N/X)/2) * (2 * X + (N/X - 1) * X)
For S2: The total numbers that will be divisible by Y upto N will be N/Y and the sum will be:
Hence, S2 = ((N/Y)/2) * (2 * Y + (N/Y - 1) * Y)
For S3: The total numbers that will be divisible by both X and Y upto N will be N/lcm(X, Y) and the sum will be:
Hence, S2 = ((N/lcm(X, Y))/2) * ((2 * lcm(X,Y)) + (N/lcm(X,Y) - 1) * lcm(X,Y))/2
Therefore, the result will be:
S = S1 + S2 - S3
Below is the implementation of the above approach:
C++
// C++ program to find sum of numbers from// 1 to N which are divisible by X or Y#include <bits/stdc++.h>using namespace std;int gcd(int a, int b){ if (a == 0) return b; return gcd(b % a, a);} // Function to return LCM of two numbersint lcm(int a, int b){ return (a / gcd(a, b)) * b;}// Function to calculate the sum// of numbers divisible by X or Yint sum(int N, int X, int Y){ int S1, S2, S3; S1 = ((N / X)) * (2 * X + (N / X - 1) * X) / 2; S2 = ((N / Y)) * (2 * Y + (N / Y - 1) * Y) / 2; S3 = ((N / lcm(X,Y))) * (2 * (lcm(X,Y)) + (N / (lcm(X,Y)) - 1) * (lcm(X,Y)))/ 2; return S1 + S2 - S3;}// Driver codeint main(){ int N = 24; int X = 4, Y = 6; cout << sum(N, X, Y); return 0;} |
Java
// Java program to find sum of numbers from// 1 to N which are divisible by X or Ypublic class GFG{ static int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a); } // method to return LCM of two numbers static int lcm(int a, int b) { return (a / gcd(a, b)) * b; } // Function to calculate the sum // of numbers divisible by X or Y static int sum(int N, int X, int Y) { int S1, S2, S3; S1 = ((N / X)) * (2 * X + (N / X - 1) * X) / 2; S2 = ((N / Y)) * (2 * Y + (N / Y - 1) * Y) / 2; S3 = ((N / ( lcm(X, Y)))) * (2 * ( lcm(X, Y)) + (N / ( lcm(X, Y)) - 1) * ( lcm(X, Y)))/ 2; return S1 + S2 - S3; } // Driver code public static void main(String []args) { int N = 14; int X = 3, Y = 5; System.out.println(sum(N, X, Y)); } // This code is contributed by Ryuga} |
Python3
# Python 3 program to find sum of numbers from# 1 to N which are divisible by X or Yfrom math import ceil, floordef gcd(a,b): if a == 0: return b return gcd(b % a, a) def lcm(a,b): return (a / gcd(a,b))* b # Function to calculate the sum# of numbers divisible by X or Ydef sum(N, X, Y): S1 = floor(floor(N / X) * floor(2 * X + floor(N / X - 1) * X) / 2) S2 = floor(floor(N / Y)) * floor(2 * Y + floor(N / Y - 1) * Y) / 2 S3 = floor(floor(N / (lcm(X,Y)))) * floor (2 * (lcm(X,Y)) + floor(N / (lcm(X,Y)) - 1) * (lcm(X,Y)))/ 2 return S1 + S2 - S3# Driver codeif __name__ == '__main__': N = 14 X = 3 Y = 5 print(int(sum(N, X, Y)))# This code is contributed by # Surendra_Gangwar |
C#
// C# program to find sum of numbers from// 1 to N which are divisible by X or Y using System;public class GFG{ static int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a); } // method to return // LCM of two numbers static int lcm(int a, int b) { return (a / gcd(a, b)) * b; } // Function to calculate the sum // of numbers divisible by X or Y static int sum(int N, int X, int Y) { int S1, S2, S3; S1 = ((N / X)) * (2 * X + (N / X - 1) * X) / 2; S2 = ((N / Y)) * (2 * Y + (N / Y - 1) * Y) / 2; S3 = ((N / (lcm(X, Y)))) * (2 * (lcm(X, Y)) + (N / (lcm(X, Y)) - 1) * (lcm(X, Y)))/ 2; return S1 + S2 - S3; } // Driver code public static void Main() { int N = 14; int X = 3, Y = 5; Console.Write(sum(N, X, Y)); } } |
PHP
<?php function gcd( $a, $b){ if ($a == 0) return $b; return gcd($b % $a, $a);} // Function to return LCM// of two numbersfunction lcm( $a, $b){ return ($a / gcd($a, $b)) * $b;} // PHP program to find sum of numbers from// 1 to N which are divisible by X or Y// Function to calculate the sum// of numbers divisible by X or Yfunction sum($N, $X, $Y){ $S1; $S2; $S3; $S1 = floor(((int)$N / $X)) * (2 * $X + (int)((int)$N / $X - 1) * $X) / 2; $S2 = floor(((int)$N / $Y)) * (2 * $Y + (int)((int)$N / $Y - 1) * $Y) / 2; $S3 = floor(((int)$N / (lcm( $X, $Y)))) * (2 * (lcm( $X, $Y)) + ((int)$N / (lcm( $X, $Y)) - 1) * (int)(lcm( $X, $Y)))/ 2; return ceil($S1 + ($S2 - $S3));}// Driver code $N = 14; $X = 3; $Y = 5; echo sum($N, $X, $Y);#This code is contributed by ajit.?> |
Javascript
<script>// javascript program to find sum of numbers from// 1 to N which are divisible by X or Yfunction gcd(a, b){if (b == 0) return a;return gcd(b, a % b);} // Function to return LCM of two numbersfunction lcm(a, b){ return (a / gcd(a, b)) * b;}// Function to calculate the sum// of numbers divisible by X or Yfunction sum(N , X , Y){ var S1, S2, S3; S1 = (parseInt(N / X)) * (2 * X + parseInt(N / X - 1) * X) / 2; S2 = (parseInt(N / Y)) * (2 * Y + parseInt(N / Y - 1) * Y) / 2; S3 = (parseInt(N / (lcm(X, Y)))) * (2 * (lcm(X, Y)) + parseInt(N / (lcm(X, Y)) - 1) * (lcm(X, Y)))/ 2; return S1 + S2 - S3;}// Driver codevar N = 14;var X = 3, Y = 5;document.write(sum(N, X, Y));// This code is contributed by Princi Singh </script> |
C
// C program to find sum of numbers from// 1 to N which are divisible by X or Y#include <stdio.h>int gcd(int a, int b){ if (a == 0) return b; return gcd(b % a, a);} // Function to return LCM of two numbersint lcm(int a, int b){ return (a / gcd(a, b)) * b;}// Function to calculate the sum// of numbers divisible by X or Yint sum(int N, int X, int Y){ int S1, S2, S3; S1 = ((N / X)) * (2 * X + (N / X - 1) * X) / 2; S2 = ((N / Y)) * (2 * Y + (N / Y - 1) * Y) / 2; S3 = ((N / (lcm(X, Y)))) * (2 * (lcm(X, Y)) + (N / (lcm(X, Y)) - 1) * (lcm(X, Y)))/ 2; return S1 + S2 - S3;}// Driver codeint main(){ int N = 14; int X = 3, Y = 5; printf("%d ",sum(N, X, Y)); return 0;} |
Output
108
Time Complexity: O(log(min(X,Y)), for calculating gcd
Auxiliary Space: O(log(min(X,Y))
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