# your code goes here

factor2 = 0

factor3 = 0

n = 987654321

temp = n

while temp >= 2:

temp /= 2

factor2 += temp

temp = n

while temp >= 3:

temp /= 3

factor3 += temp

#print factor2, factor3

print min(factor2/2, factor3)

上面的代碼已經(jīng)很清晰了。把2和3的因子都找出。

2*2*3=12 在12進(jìn)制下就得出10，就是一個(gè)0了。 每遇到2個(gè)2和1個(gè)3就會(huì)出一個(gè)0.

n! = (n1) *  2^factor2 * 3^factor3

result = min(factor/2, factor3)