#!/usr/bin/env python
# coding: utf-8

# ### Factorials
# 
# Let's compute $n! = 1\times 2\times 3\times 4....\times n$:

# In[1]:


def fact(n):
    res = 1
    for i in range(1, n+1):
        res *= i
        
    return res


# In[2]:


fact(500)


# ### Trailing zeros
# 
# Suppose we'd like to count up the number of *trailing zeros* in $n!$ -- the zeros at the end of the number. Here is the strategy:
# 
# * Compute fact_n = n!
# * Keep dividing fact_n by 10 while it's divisible by 10
# * The number of times that res could be divided by 10 is the number of trailing zeros in n!
# 
# (Note that dividing a number by 10 is the same as removing one trailing zero.)

# In[3]:


def trailing_zeros(n):
    
    fact_n = fact(n)
    
    counter = 0
    
    while fact_n % 10 == 0:
        fact_n //= 10
        counter += 1
        
    return counter


# In[4]:


trailing_zeros(500)


# We have set up a counter variable, which is incremented every time we divide `fact_n` by 10 (i.e., remove a trailing 0.
# 
# What happens if n! has no trailing 0s? We simply never enter the `while`-loop, and return 0, which is the correct thing.