Factorial Trailing Zeroes in C++

Here we will see how to calculate the number of trailing 0s for the result of factorial of any number. So if the n = 5, then 5! = 120. There is only one trailing 0. For 20! it will be 4 zeros as 20! = 2432902008176640000.

The easiest approach is just calculating the factorial and count the 0s. But this approach fails for a large value of n. So we will follow another approach. The trailing zeros will be there if the prime factors are 2 and 5. If we count the 2s and 5s we can get the result. For that, we will follow this rule.

Trailing 0s = Count of 5s in prime factors of factorial(n)

so Trailing 0s = $$\lvert\frac{n}{5}\rvert+\lvert\frac{n}{25}\rvert+\lvert\frac{n}{125}\rvert+...$$

To solve this, we have to follow these steps −

• set count = 0
• for i = 5, (n/i) > 1, update i = i * 5, do
  • count = count + (n /i)
• return count

Example (C++)

 Live Demo

#include <iostream>
#include <cmath>
#define MAX 20
using namespace std;
int countTrailingZeros(int n) {
   int count = 0;
   for (int i = 5; n / i >= 1; i *= 5)
      count += n / i;
   return count;
}
main() {
   int n = 20;
   cout << "Number of trailing zeros: " << countTrailingZeros(n);
}

Input

Number of trailing zeroes: 20

Output

Number of trailing zeros: 4