Largest power of k in n! (factorial) where k may not be prime

08 Jul, 2021
Given two numbers k and n, find the largest power of k that divides n! Constraints:  
 K > 1
Examples:  
Input : n = 7, k = 2
Output : 4
Explanation : 7! = 5040
The largest power of 2 that
divides 5040 is 24.

Input : n = 10, k = 9
Output :  2
The largest power of 9 that
divides 10! is 92.
 
We have discussed a solution in below post when k is always prime.Legendre’s formula (Given p and n, find the largest x such that p^x divides n!)Now to find the power of any non-prime number k in n!, we first find all the prime factors of the number k along with the count of number of their occurrences. Then for each prime factor, we count occurrences using Legendre’s formula which states that the largest possible power of a prime number p in n is ⌊n/p⌋ + ⌊n/(p2)⌋ + ⌊n/(p3)⌋ + ......Over all the prime factors p of K, the one with the minimum value of findPowerOfK(n, p)/count will be our answer where count is number of occurrences of p in k. 
C++
Java
Python3
C#
Javascript
// CPP program to find the largest power// of k that divides n!#include <bits/stdc++.h>using namespace std; // To find the power of a prime p in// factorial Nint findPowerOfP(int n, int p){    int count = 0;    int r=p;    while (r <= n) {         // calculating floor(n/r)        // and adding to the count        count += (n / r);         // increasing the power of p        // from 1 to 2 to 3 and so on        r = r * p;    }    return count;} // returns all the prime factors of kvector<pair<int, int> > primeFactorsofK(int k){    // vector to store all the prime factors    // along with their number of occurrence    // in factorization of k    vector<pair<int, int> > ans;     for (int i = 2; k != 1; i++) {        if (k % i == 0) {            int count = 0;            while (k % i == 0) {                k = k / i;                count++;            }             ans.push_back(make_pair(i, count));        }    }    return ans;} // Returns largest power of k that// divides n!int largestPowerOfK(int n, int k){    vector<pair<int, int> > vec;    vec = primeFactorsofK(k);    int ans = INT_MAX;    for (int i = 0; i < vec.size(); i++)         // calculating minimum power of all        // the prime factors of k        ans = min(ans, findPowerOfP(n,              vec[i].first) / vec[i].second);     return ans;} // Driver codeint main(){    cout << largestPowerOfK(7, 2) << endl;    cout << largestPowerOfK(10, 9) << endl;    return 0;}
// JAVA program to find the largest power// of k that divides n!import java.util.*; class GFG{     static class pair{    int first, second;    public pair(int first, int second)    {        this.first = first;        this.second = second;    }}// To find the power of a prime p in// factorial Nstatic int findPowerOfP(int n, int p){    int count = 0;    int r = p;    while (r <= n)    {         // calculating Math.floor(n/r)        // and adding to the count        count += (n / r);         // increasing the power of p        // from 1 to 2 to 3 and so on        r = r * p;    }    return count;} // returns all the prime factors of kstatic Vector<pair > primeFactorsofK(int k){    // vector to store all the prime factors    // along with their number of occurrence    // in factorization of k    Vector<pair> ans = new Vector<pair>();     for (int i = 2; k != 1; i++)    {        if (k % i == 0)        {            int count = 0;            while (k % i == 0)            {                k = k / i;                count++;            }             ans.add(new pair(i, count));        }    }    return ans;} // Returns largest power of k that// divides n!static int largestPowerOfK(int n, int k){    Vector<pair > vec = new Vector<pair>();    vec = primeFactorsofK(k);    int ans = Integer.MAX_VALUE;    for (int i = 0; i < vec.size(); i++)         // calculating minimum power of all        // the prime factors of k        ans = Math.min(ans, findPowerOfP(n,            vec.get(i).first) / vec.get(i).second);     return ans;} // Driver codepublic static void main(String[] args){    System.out.print(largestPowerOfK(7, 2) +"\n");    System.out.print(largestPowerOfK(10, 9) +"\n");}} // This code is contributed by 29AjayKumar
# Python3 program to find the largest power# of k that divides n!import sys # To find the power of a prime p in# factorial Ndef findPowerOfP(n, p) :     count = 0    r = p    while (r <= n) :         # calculating floor(n/r)        # and adding to the count        count += (n // r)         # increasing the power of p        # from 1 to 2 to 3 and so on        r = r * p          return count # returns all the prime factors of kdef primeFactorsofK(k) :     # vector to store all the prime factors    # along with their number of occurrence    # in factorization of k    ans = []    i = 2    while k != 1 :        if k % i == 0 :            count = 0            while k % i == 0 :                k = k // i                count += 1            ans.append([i , count])        i += 1     return ans # Returns largest power of k that# divides n!def largestPowerOfK(n, k) :     vec = primeFactorsofK(k)    ans = sys.maxsize    for i in range(len(vec)) :         # calculating minimum power of all        # the prime factors of k        ans = min(ans, findPowerOfP(n, vec[i][0]) // vec[i][1])     return ans print(largestPowerOfK(7, 2))print(largestPowerOfK(10, 9)) # This code is contributed by divyesh072019
// C# program to find the largest power// of k that divides n!using System;using System.Collections.Generic; class GFG{     class pair{    public int first, second;    public pair(int first, int second)    {        this.first = first;        this.second = second;    }} // To find the power of a prime p in// factorial Nstatic int findPowerOfP(int n, int p){    int count = 0;    int r = p;    while (r <= n)    {         // calculating Math.Floor(n/r)        // and adding to the count        count += (n / r);         // increasing the power of p        // from 1 to 2 to 3 and so on        r = r * p;    }    return count;} // returns all the prime factors of kstatic List<pair > primeFactorsofK(int k){    // vector to store all the prime factors    // along with their number of occurrence    // in factorization of k    List<pair> ans = new List<pair>();     for (int i = 2; k != 1; i++)    {        if (k % i == 0)        {            int count = 0;            while (k % i == 0)            {                k = k / i;                count++;            }             ans.Add(new pair(i, count));        }    }    return ans;} // Returns largest power of k that// divides n!static int largestPowerOfK(int n, int k){    List<pair > vec = new List<pair>();    vec = primeFactorsofK(k);    int ans = int.MaxValue;    for (int i = 0; i < vec.Count; i++)         // calculating minimum power of all        // the prime factors of k        ans = Math.Min(ans, findPowerOfP(n,            vec[i].first) / vec[i].second);     return ans;} // Driver codepublic static void Main(String[] args){    Console.Write(largestPowerOfK(7, 2) +"\n");    Console.Write(largestPowerOfK(10, 9) +"\n");}} // This code is contributed by 29AjayKumar
<script> // JavaScript program to find the largest power// of k that divides n! class pair{    constructor(first,second)    {        this.first = first;        this.second = second;    }} // To find the power of a prime p in// factorial Nfunction findPowerOfP(n,p){    let count = 0;    let r = p;    while (r <= n)    {          // calculating Math.floor(n/r)        // and adding to the count        count += Math.floor(n / r);          // increasing the power of p        // from 1 to 2 to 3 and so on        r = r * p;    }    return count;} // returns all the prime factors of kfunction primeFactorsofK(k){    // vector to store all the prime factors    // along with their number of occurrence    // in factorization of k    let ans = [];      for (let i = 2; k != 1; i++)    {        if (k % i == 0)        {            let count = 0;            while (k % i == 0)            {                k = Math.floor(k / i);                count++;            }              ans.push(new pair(i, count));        }    }    return ans;} // Returns largest power of k that// divides n!function largestPowerOfK(n,k){    let vec = [];    vec = primeFactorsofK(k);    let ans = Number.MAX_VALUE;    for (let i = 0; i < vec.length; i++)          // calculating minimum power of all        // the prime factors of k        ans = Math.min(ans, findPowerOfP(n,            vec[i].first) / vec[i].second);      return ans;} // Driver codedocument.write(largestPowerOfK(7, 2) +"<br>");document.write(largestPowerOfK(10, 9) +"<br>");  // This code is contributed by rag2127 </script>
Output:  
4
2
This article is contributed by ShivamKD. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
29AjayKumar
divyesh072019
rag2127
factorial
number-theory
prime-factor
Mathematical
number-theory
Mathematical
factorial
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.