the number of trailing zeros in a factorial of a given number - Ruby

Question

Having a little trouble trying calculate the number of trailing zeros in a factorial of a given number. This is one of the challenges from Codewars- can\'t get mine to pass.

Answer 1

def zeros(n)
    zeros = 0
    zeros += n /= 5 while n >= 1
    zeros
end

Answer 2

    n=int(input())
    j=5
    c=int(0)
    while int(n/j)>0:
        c=c+int(n/j)
        j=j*5
    print(c)

Answer 3

If N is a number then number of trailing zeroes in N! is

N/5 + N/5^2 + N/5^3 ..... N/5^(m-1) WHERE (N/5^m)<1

You can learn here how this formula comes.

Answer 4

So, now that @Spunden has so artfully let the cat out of the bag, here's one way to implement it.

Code

def zeros(n)
  return 0 if n.zero?
  k = (Math.log(n)/Math.log(5)).to_i
  m = 5**k
  n*(m-1)/(4*m)
end

Examples

zeros(3)   #=>  0
zeros(5)   #=>  1
zeros(12)  #=>  2
zeros(15)  #=>  3
zeros(20)  #=>  4
zeros(25)  #=>  6
zeros(70)  #=> 16
zeros(75)  #=> 18
zeros(120) #=> 28
zeros(125) #=> 31

Explanation

Suppose n = 128.

Then each number between one and 128 (inclusive) that is divisible by 5^1=>5 provides at least one factor, and there are 128/5 => 25 such numbers. Of these, the only ones that provide more than one factor are those divisible by 5^2=>25, of which there are 128/25 => 5 (25, 50, 75, 100, 125). Of those, there is but 128/125 => 1 that provides more than two factors, and since 125/(5^4) => 0, no numbers contribute more than three divisors. Hence, the total number of five divisors is:

128/5 + 128/25 + 128/125 #=> 31

(Note that, for 125, which has three divisors of 5, one is counted in each of these three terms; for 25, 50, etc., which each have two factors of 5, one is counted in each of the first terms.)

For arbitrary n, we first compute the highest power k for which:

5**k <= n

which is:

k <= Math.log(n)/Math.log(5)

so the largest such value is:

k = (Math.log(n)/Math.log(5)).to_i

As @spundun noted, you could also calculate k by simply iterating, e.g.,

last = 1
(0..1.0/0).find { |i| (last *= 5) > n }

The total number of factors of five is therefore

(n/5) + (n/25) +...+ (n/5**k)

Defining:

r = 1/5,

this sum is seen to be:

n * s

where

s = r + r**2 +...+ r**k

The value of s is the sum of the terms of a geometric series. I forget the formula for that, but recall how it's derived:

s  = r + r**2 +...+ r**k
sr =     r**2 +...+ r**(k+1)

s-sr = r*(1-r**k)

s = r*(1-r**k)/(1-r)

I then did some rearrangement so that only only integer arithmetic would be used to calculate the result.

Answer 5

count = 0
i  =5
n = 100
k = n

while(n/i!=0):
    count+=(n/i)
    i=i*5
    n = k
print count

Answer 6

My solution

def zeros(n)
  trailing_zeros = []
  fact = (1..n).inject(:*)
  fact.to_s.split('').reverse.select {|x| break if (x.to_i != 0); trailing_zeros << x}
  return trailing_zeros.count
end