# Solving the Fizzbuzz Puzzle with Ruby

So back in February Jeff Atwood over at codinghorror.com was talking about a puzzle to give prospective new-hires when interviewing them entitled “Fizzbuzz”. You can read more about it here.

So with a little thought I decided to solve the fizzbuzz puzzle using Rudy and an Array.

``````count = 0
100.times do
count += 1
fb = []
fb << "Fizz" if (count % 3) == 0
fb << "Buzz" if (count % 5) == 0
fb << count if (count % 3) != 0 and (count % 5) != 0
puts (fb.join "")
end``````

For those of you that are curious as to why I used a local variable of count rather then the index identifier, is because the article by Jeff Atwood requests a loop from 1-100 and Ruby starts looping at 0.

## 2 thoughts on “Solving the Fizzbuzz Puzzle with Ruby”

1. S2K says:

Why not 1.upto(100)???

2. Tim says:

Well, the post was from a year and a half ago, so at the time I wasn’t really familiar with upto, but you’re right… and this point if I was to write this it would look something like:

1.upto(100) do |x|
fb = []
fb << “Fizz” if (x % 3) == 0
fb << “Buzz” if (x % 5) == 0
fb << x if (x % 3) != 0 and (x % 5) != 0
puts (fb.join “”)
end