# Number of trailing zeroes in base B representation of N!

Given two positive integers B and N. The task is to find the number of trailing zeroes in b-ary (base B) representation of N! (factorial of N)

**Examples:**

Input:N = 5, B = 2Output:3 5! = 120 which is represented as 1111000 in base 2.Input:N = 6, B = 9Output:1

A **naive solution** is to find the factorial of the given number and convert it into given base B. Then, count the number of trailing zeroes but that would be a costly operation. Also, it will not be easy to find the factorial of large numbers and store it in integer.

**Efficient Approach: ** Suppose, the base is 10 i.e., decimal then we’ll have to calculate the highest power of 10 that divides N! using Legendre’s formula. Thus, number B is represented as 10 when converted into base B. Let’s say base B = 13, then 13 in base 13 will be represented as 10, i.e., 13_{10} = 10_{13}. Hence, problem reduces to finding the highest power of B in N!. (Largest power of k in n!)

Below is the implementation of the above approach.

## C++

`// CPP program to find the number of trailing ` `// zeroes in base B representation of N! ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// To find the power of a prime p in ` `// factorial N ` `int` `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 k ` `vector<pair<` `int` `, ` `int` `> > primeFactorsofB(` `int` `B) ` `{ ` ` ` `// vector to store all the prime factors ` ` ` `// along with their number of occurrence ` ` ` `// in factorization of B ` ` ` `vector<pair<` `int` `, ` `int` `> > ans; ` ` ` ` ` `for` `(` `int` `i = 2; B != 1; i++) { ` ` ` `if` `(B % i == 0) { ` ` ` `int` `count = 0; ` ` ` `while` `(B % i == 0) { ` ` ` `B = B / i; ` ` ` `count++; ` ` ` `} ` ` ` ` ` `ans.push_back(make_pair(i, count)); ` ` ` `} ` ` ` `} ` ` ` `return` `ans; ` `} ` ` ` `// Returns largest power of B that ` `// divides N! ` `int` `largestPowerOfB(` `int` `N, ` `int` `B) ` `{ ` ` ` `vector<pair<` `int` `, ` `int` `> > vec; ` ` ` `vec = primeFactorsofB(B); ` ` ` `int` `ans = INT_MAX; ` ` ` `for` `(` `int` `i = 0; i < vec.size(); i++) ` ` ` ` ` `// calculating minimum power of all ` ` ` `// the prime factors of B ` ` ` `ans = min(ans, findPowerOfP(N, ` ` ` `vec[i].first) ` ` ` `/ vec[i].second); ` ` ` ` ` `return` `ans; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `cout << largestPowerOfB(5, 2) << endl; ` ` ` `cout << largestPowerOfB(6, 9) << endl; ` ` ` `return` `0; ` `} ` |

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## Java

// Java program to find the number of trailing

// zeroes in base B representation of 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 N

static int 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 k
static Vector

{

// vector to store all the prime factors

// along with their number of occurrence

// in factorization of B

Vector

for (int i = 2; B != 1; i++)

{

if (B % i == 0)

{

int count = 0;

while (B % i == 0)

{

B = B / i;

count++;

}

ans.add(new pair(i, count));

}

}

return ans;

}

// Returns largest power of B that

// divides N!

static int largestPowerOfB(int N, int B)

{

Vector

vec = primeFactorsofB(B);

int ans = Integer.MAX_VALUE;

for (int i = 0; i < vec.size(); i++)
// calculating minimum power of all
// the prime factors of B
ans = Math.min(ans, findPowerOfP(
N, vec.get(i).first) /
vec.get(i).second);
return ans;
}
// Driver code
public static void main(String[] args)
{
System.out.println(largestPowerOfB(5, 2));
System.out.println(largestPowerOfB(6, 9));
}
}
// This code is contributed by Princi Singh
[tabby title = "Python3"]

`# Python 3 program to find the number of ` `# trailing zeroes in base B representation of N! ` `import` `sys ` ` ` `# To find the power of a prime ` `# p in factorial N ` `def` `findPowerOfP(N, p): ` ` ` `count ` `=` `0` ` ` `r ` `=` `p ` ` ` `while` `(r <` `=` `N): ` ` ` ` ` `# calculating floor(n/r) ` ` ` `# and adding to the count ` ` ` `count ` `+` `=` `int` `(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 k ` `def` `primeFactorsofB(B): ` ` ` ` ` `# vector to store all the prime factors ` ` ` `# along with their number of occurrence ` ` ` `# in factorization of B' ` ` ` `ans ` `=` `[] ` ` ` `i ` `=` `2` ` ` ` ` `while` `(B!` `=` `1` `): ` ` ` `if` `(B ` `%` `i ` `=` `=` `0` `): ` ` ` `count ` `=` `0` ` ` `while` `(B ` `%` `i ` `=` `=` `0` `): ` ` ` ` ` `B ` `=` `int` `(B ` `/` `i) ` ` ` `count ` `+` `=` `1` ` ` ` ` `ans.append((i, count)) ` ` ` ` ` `i ` `+` `=` `1` ` ` ` ` `return` `ans ` ` ` `# Returns largest power of B that ` `# divides N! ` `def` `largestPowerOfB(N, B): ` ` ` `vec ` `=` `[] ` ` ` `vec ` `=` `primeFactorsofB(B) ` ` ` `ans ` `=` `sys.maxsize ` ` ` ` ` `# calculating minimum power of all ` ` ` `# the prime factors of B ` ` ` `ans ` `=` `min` `(ans, ` `int` `(findPowerOfP(N, vec[` `0` `][` `0` `]) ` `/` ` ` `vec[` `0` `][` `1` `])) ` ` ` ` ` `return` `ans ` ` ` `# Driver code ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` `print` `(largestPowerOfB(` `5` `, ` `2` `)) ` ` ` `print` `(largestPowerOfB(` `6` `, ` `9` `)) ` ` ` `# This code is contributed by ` `# Surendra_Gangwar ` |

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**Output:**

3 1

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