Simplify this python code
<p>I've written a program to check if my thought about solution on paper is right (and it is).</p>

<p>The task: how many zeros is in the back of multiplication of all numbers from 10 to 200.</p>

<p>It is 48 and it is a simple to calculate manually.</p>

<p>I never write on python seriously and this is what I get:</p>

<pre><code>mul = 1
for i in range(10, 200 + 1):
    mul *= i

string = str(mul)
string = string[::-1]
count = 0;
for c in str(string):
    if c == '0':
        count += 1
    else:
        break

print count
print mul
</code></pre>

<p>I bet it is possible to write the same more elegant in such language like a python.</p>

<p>ps: yes, it is a homework, but not mine - i just helped a guy ;-)</p>
