Let's compute $n! = 1\times 2\times 3\times 4....\times n$:
def fact(n):
res = 1
for i in range(1, n+1):
res *= i
return res
fact(500)
1220136825991110068701238785423046926253574342803192842192413588385845373153881997605496447502203281863013616477148203584163378722078177200480785205159329285477907571939330603772960859086270429174547882424912726344305670173270769461062802310452644218878789465754777149863494367781037644274033827365397471386477878495438489595537537990423241061271326984327745715546309977202781014561081188373709531016356324432987029563896628911658974769572087926928871281780070265174507768410719624390394322536422605234945850129918571501248706961568141625359056693423813008856249246891564126775654481886506593847951775360894005745238940335798476363944905313062323749066445048824665075946735862074637925184200459369692981022263971952597190945217823331756934581508552332820762820023402626907898342451712006207714640979456116127629145951237229913340169552363850942885592018727433795173014586357570828355780158735432768888680120399882384702151467605445407663535984174430480128938313896881639487469658817504506926365338175055478128640000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
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:
(Note that dividing a number by 10 is the same as removing one trailing zero.)
def trailing_zeros(n):
fact_n = fact(n)
counter = 0
while fact_n % 10 == 0:
fact_n //= 10
counter += 1
return counter
trailing_zeros(500)
124
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.