Program to find number m such that it has n number of 0s at end in Python

Suppose we have a number n. We have to find smallest number m, such that factorial of m has at least n number of 0s.

So, if the input is like n = 2, then the output will be 10 because 10! = 3628800 and 9! = 362880, minimum number with 2 zeros is 10.

To solve this, we will follow these steps −

• Define a function count_fives() . This will take n
• cnt := 0
• while n > 0, do
  • n := floor of (n / 5)
  • cnt := cnt + n
• return cnt
• From the main method, do the following −
• left := 1
• right := 5^24
• while right - left > 5, do
  • mid := floor of ((right + left) / 10) * 5
  • fives := count_fives(mid)
  • if fives is same as n, then
    • right := mid
    • left := right - 5
    • come out from the loop
  • otherwise when fives < n, then
    • left := mid
  • otherwise,
    • right := mid
• return right

Example

Let us see the following implementation to get better understanding −

def count_fives(n):
   cnt = 0
   while n > 0:
      n = n // 5
      cnt += n
   return cnt

def solve(n):
   left = 1
   right = 5**24
   while right - left > 5:
      mid = int((right + left) / 10) * 5
      fives = count_fives(mid)
      if fives == n:
         right = mid
         left = right - 5
         break
      elif fives < n:
         left = mid
      else:
         right = mid
   return right

n = 2
print(solve(n))

Input

2

Output

10