Power of a prime number r in n! in C++

In this problem, we are given two integer n and r. Our task is to find the power of the given prime number r in the factorial of the number n.

Let’s take an example to understand the problem

Input − n = 6 r = 2

Output − 4

Explanation −

Factorial n, 6! = 6*5*4*3*2*1 = 720
720 = 24 * 32 * 5, power of 2 is 4

To solve this problem, a simple solution would be directly finding the factorial and then finding the power of the prime number. But this is not the best solution.

Another efficient solution is using a formula,

Power of ‘r’ in n! = floor(n/r) + floor(n/r2) + floor(n/r3) + ...

Example

Program to show the implementation of our solution,

 Live Demo

#include <iostream>
using namespace std;
int primePower(int n, int r) {
   int count = 0;
   for (int i = r; (n / i) >= 1; i = i * r)
      count = count+n/i;
   return count;
}
int main() {
   int n = 6, r = 2;
   cout<<"Power of prime number "<<r<<"in factorial "<<n<<" is : "<<primePower(n, r);
   return 0;
}

Output

Power of prime number 2in factorial 6 is : 4