1
def prime?(num)
  return false if num == 1
  (2..num/2).each do |x|
    if num%x==0
      return false
    end
  end
  true
end
Sergio Tulentsev
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John Morris
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    For what it's worth, you can shorten this brute force attempt a bit by going up to the *square root* of `num`, not the half. – vcsjones Jun 09 '17 at 19:58

3 Answers3

2

Because you never enter your each. If the value is 2, then (2..num/2) becomes (2..2/2), which is (2..1). Ruby's range operators don't go backwards, so there is nothing to "each" over.

vcsjones
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0

I believe that since (num/2) is 1 in that case, the loop never runs, so it falls through to the 'true' at the bottom.

PMar
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0

I think your method is working:

puts prime?(2)
#=> true

As someone mentioned in the comments, you can also save some steps by only going through the sqrt of the number:

(2..Math.sqrt(num)).each do |x|
Kathryn
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