Problem of the Day
A new programming or logic puzzle every Mon-Fri

Today's challenge is a new take on a classic. Without using the modulus operator or converting the number to binary to check the least significant bit, can you write a function to determine if an integer is odd or not?

Comments:

• Anonymous - 10 years, 1 month ago

  def isOdd(n):
      return n != int(n/2)*2
  

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• Akshay Bist - 10 years, 1 month ago

  Prime or odd? Which is it?

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• Max Burstein - 10 years, 1 month ago

  My bad on that. Meant to say odd

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• Driphter - 10 years, 1 month ago

  Clojure!

  (defn odd? [x]
    (let [half (float (/ x 2))]
      (not= half (Math/floor half))))
  
  (defn prime? [x]
    (let [y (float (/ x 2))
          y* (Math/floor y)]
      (if (= y y*)
        false
        (loop [s (vec (range 3 y*))]
          (if (empty? s)
            true
            (let [y (float (/ x (first s)))
                  y* (Math/floor y)]
              (if (= y y*)
                false
                (recur (remove #(= % y*) (next s))))))))))
  

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• phansen73 - 10 years, 1 month ago

  C++

  int isOdd (int x) {
    if (x == x/2*2) {
      return 0;
    } 
    return 1;
  }
  
  

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• asheehan - 10 years, 1 month ago

  felt like playing with floor... in Ruby

  def is_odd(x)
      y = ((x - 1) / 2).floor
      x = (x / 2).floor
      return (x == y)
  end
  
  def is_whole(x)
      return (x == x.floor and x == x.ceil)
  end
  
  def is_prime(x)
      if is_odd(x)
          n = (x / 2).floor
          if !is_odd(n)
              n -= 1
          end
          while n > 1 do
              if is_whole(x.to_f / n)
                  return false
              end
              n -= 2
          end
          return true
      end
      return false
  end
  
  p is_prime(37)
  

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• Jojo Masala - 9 years, 10 months ago

  Java:

  private boolean nIsOdd(double n) { boolean isOdd = false; n /= 2.0; double roundDown = (int) n; double roundUp = roundDown + 1; if((n > roundDown) && (n < roundUp)) { isOdd = true; } return isOdd;

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