We divide

This is my code in Python 2.7:

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a = 567948952
p = 1                       #starts with 3^1
ans=0
while 3**p<a:          # if 3^p < a then it carries on in the loop
    b = a/(3**p)       # This finds the floor of a divided by every power of 3 less than a
    ans = b+ans        # This sums all of the floors
    p+=1          # p is incremented by 1 and retested if 3^p is still less than a 
print ans     # when 3^p is more than a, the loop stops and prints the sum of the floors

This gives the output:

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>>>283974468

PHP solution -

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$count = 0;
for ($i = 1; $i < 19; $i++)
    $count += floor(567948952 / pow(3, $i));
echo $count;

Yes I'm learning PHP :)

This is exactly similar to counting trailing zeroes . Here is a general formula to calculate largest power of $x$ in $n!$ :

$\huge{\sum _{ i=1 }^{ n }{ \left\lfloor \frac { n }{ { x }^{ i } } \right\rfloor }}$

Note: you don't actually need to calculate upto $n$

Largest power $\alpha$ of a prime $p$ in $n!$ is calculated by $\alpha = \dfrac{n-s_p(n)}{p-1}$ where $s_p(n)$ is the sum of the digits of $n$ when written in base $p$ . This can easily be proved by using the formula for the sum of a finite geometric series.