P2.5.6de Polignac's formula
Write a Python program to determine \(f(n)\), the number of trailing zeros in \(n!\), using the special case of de Polignac's formula:
\[f(n) = \sum_{i=1} \left\lfloor \frac{n}{5^i} \right\rfloor,\]
where \(\lfloor x \rfloor\) denotes the floor of \(x\), the largest integer less than or equal to \(x\).
Show solution
Here is one solution:
# Caculate the number of trailing zeros in n! using de Polignac's formula:
# nzeros = sum_i floor(n/5^i)
import math
n = 126
nzeros = 0
term = n
while True:
term //= 5
if term == 0:
break
nzeros += term
print('{:d}! ends in {:d} zeros.'.format(n, nzeros))
print('{:d}! = {:d}'.format(n, math.factorial(n)))
For this example value of n the output is:
126! ends in 31 zeros.
126! = 2372173242880046885677147305139417080570208597380804566183737717005249769778331345722724
954407648631483944708618718727531940040183701395532517931565237692899606512332119089860313088
0000000000000000000000000000000