Given a number N, the task is to find the largest power of 2 that divides LCM of first N Natural numbers.
Examples:
Input: N = 5
Output: 2
Explanation:
LCM of {1, 2, 3, 4, 5} = 60
60 is divisible by 22Input: N = 15
Output: 3
Explanation:
LCM of {1, 2, 3…..14, 15} = 360360
360360 is divisible by 23
Naive Approach: The idea is to find the Least common multiple of first N natural numbers. Then iterate a loop from i = 1 and check if 2i Divides the LCM or not and keep the track of maximum i that divides LCM.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to find LCM of// first N natural numbersint findlcm(int n){ // Initialize result int ans = 1; // Ans contains LCM of 1, 2, 3, ..i // after i'th iteration for (int i = 1; i <= n; i++) ans = (((i * ans)) / (__gcd(i, ans))); return ans;}// Function to find the// highest power of 2// which divides LCM of// first n natural numbersint highestPower(int n){ // Find lcm of first // N natural numbers int lcm = findlcm(n); // To store the highest // required power of 2 int ans = 0; // Counting number of consecutive zeros // from the end in the given binary string for (int i = 1;; i++) { int x = pow(2, i); if (lcm % x == 0) { ans = i; } if (x > n) break; } return ans;}// Driver codeint main(){ int n = 15; cout << highestPower(n); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG{// Function to find LCM of// first N natural numbersstatic int findlcm(int n){ // Initialize result int ans = 1; // Ans contains LCM of 1, 2, 3, ..i // after i'th iteration for(int i = 1; i <= n; i++) ans = (((i * ans)) / (__gcd(i, ans))); return ans;}// Function to find the// highest power of 2// which divides LCM of// first n natural numbersstatic int highestPower(int n){ // Find lcm of first // N natural numbers int lcm = findlcm(n); // To store the highest // required power of 2 int ans = 0; // Counting number of consecutive zeros // from the end in the given binary String for(int i = 1;; i++) { int x = (int) Math.pow(2, i); if (lcm % x == 0) { ans = i; } if (x > n) break; } return ans;}static int __gcd(int a, int b) { return b == 0 ? a : __gcd(b, a % b); }// Driver codepublic static void main(String[] args){ int n = 15; System.out.print(highestPower(n));}}// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach# Function to find LCM of# first N natural numbersdef findlcm(n): # Initialize result ans = 1; # Ans contains LCM of 1, 2, 3, ..i # after i'th iteration for i in range(1, n + 1): ans = (((i * ans)) // (__gcd(i, ans))); return ans;# Function to find the highest power# of 2 which divides LCM of first n# natural numbersdef highestPower(n): # Find lcm of first # N natural numbers lcm = findlcm(n); # To store the highest # required power of 2 ans = 0; # Counting number of consecutive zeros # from the end in the given binary String for i in range(1, n): x = int(pow(2, i)); if (lcm % x == 0): ans = i; if (x > n): break; return ans;def __gcd(a, b): if (b == 0): return a; else: return __gcd(b, a % b);# Driver codeif __name__ == '__main__': n = 15; print(highestPower(n));# This code is contributed by 29AjayKumar |
C#
// C# implementation of the approachusing System;class GFG{// Function to find LCM of// first N natural numbersstatic int findlcm(int n){ // Initialize result int ans = 1; // Ans contains LCM of 1, 2, 3, ..i // after i'th iteration for(int i = 1; i <= n; i++) ans = (((i * ans)) / (__gcd(i, ans))); return ans;}// Function to find the// highest power of 2// which divides LCM of// first n natural numbersstatic int highestPower(int n){ // Find lcm of first // N natural numbers int lcm = findlcm(n); // To store the highest // required power of 2 int ans = 0; // Counting number of consecutive zeros // from the end in the given binary String for(int i = 1;; i++) { int x = (int) Math.Pow(2, i); if (lcm % x == 0) { ans = i; } if (x > n) break; } return ans;}static int __gcd(int a, int b) { return b == 0 ? a : __gcd(b, a % b); }// Driver codepublic static void Main(String[] args){ int n = 15; Console.Write(highestPower(n));}}// This code is contributed by 29AjayKumar |
Javascript
<script>// JavaScript program for the// above approach// Function to find LCM of// first N natural numbersfunction findlcm(n){ // Initialize result let ans = 1; // Ans contains LCM of 1, 2, 3, ..i // after i'th iteration for(let i = 1; i <= n; i++) ans = (((i * ans)) / (__gcd(i, ans))); return ans;} // Function to find the// highest power of 2// which divides LCM of// first n natural numbersfunction highestPower(n){ // Find lcm of first // N natural numbers let lcm = findlcm(n); // To store the highest // required power of 2 let ans = 0; // Counting number of consecutive zeros // from the end in the given binary String for(let i = 1;; i++) { let x = Math.pow(2, i); if (lcm % x == 0) { ans = i; } if (x > n) break; } return ans;} function __gcd(a, b){ return b == 0 ? a : __gcd(b, a % b); }// Driver Code let n = 15; document.write(highestPower(n));</script> |
Output
3
Time Complexity: O(N)
Auxiliary Space: O(1)
Efficient Approach: The LCM of first N natural numbers is always divisible by a power of 2 and since the LCM of first N natural numbers contains the product 2 * 4 * 8 * 16 ……N. Therefore, the largest power of 2 that divides LCM of first N Natural numbers will always be 
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to find the// highest power of 2// which divides LCM of// first n natural numbersint highestPower(int n){ return log(n) / log(2);}// Driver codeint main(){ int n = 15; cout << highestPower(n); return 0;} |
Java
// Java implementation of the approach class GFG{ // Function to find the highest // power of 2 which divides LCM of // first n natural numbers static int highestPower(int n) { return (int)(Math.log(n) / Math.log(2)); } // Driver codepublic static void main(String[] args) { int n = 15; System.out.println(highestPower(n)); } }// This code is contributed by dewantipandeydp |
Python3
# Python3 implementation of the approachimport math# Function to find the highest# power of 2 which divides LCM of# first n natural numbersdef highestPower(n): return int((math.log(n) // math.log(2)));# Driver codeif __name__ == '__main__': n = 15; print(highestPower(n));# This code is contributed by Rajput-Ji |
C#
// C# implementation of the approach using System;class GFG{ // Function to find the highest // power of 2 which divides LCM of // first n natural numbers static int highestPower(int n) { return (int)(Math.Log(n) / Math.Log(2)); } // Driver codepublic static void Main(String[] args) { int n = 15; Console.WriteLine(highestPower(n)); } }// This code is contributed by sapnasingh4991 |
Javascript
<script>// Javascript implementation of the approach// Function to find the// highest power of 2// which divides LCM of// first n natural numbersfunction highestPower(n){ return parseInt(Math.log(n) / Math.log(2));}// Driver codevar n = 15;document.write( highestPower(n));</script> |
Output
3
Time Complexity: O(1)
Auxiliary Space: O(1)
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