human
Trailing number of 0 s in product of two factorials | Returns number of zeros in factorial n.
Dividing x by powers of 5 and update count.
Returns count of trailing zeros in M ! x N !.
Driver program"
How can the above be solved in Python?

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What is a code to trailing number of 0 s in product of two factorials and the other steps required?
assistant
#Here is code to trailing number of 0 s in product of two factorials and the other steps required.
 def trailingZero ( x ) :
	i = 5
	count = 0
	while ( x > i ) :
		count = count + x // i
		i = i * 5
	return count
def countProductTrailing ( M , N ) :
	return trailingZero ( N ) + trailingZero ( M )
N = 67
M = 98
print ( countProductTrailing ( N , M ) )