title: Count to 80, skipping every 5th number in a range of 100 numbers
body: <p>I'm working on a quiz with 80 numbered questions. After every 4 questions there will be an unnumbered 5th bonusquestion. These 100 questions have index-numbers (1-100) in a database table. Maybe it'd have been more logical to use a separate table for the bonusquestions, but the client wants the questions in one table in that order.</p>
  
  <p>I don't want to display the bonusquestion index-numbers in the interface of the quiz, therefore i need to skip every 5th index-number from the range of 100 questions.</p>
  
  <p>I did manage to make a simple calculation which works well when i check it in a loop, but somehow i feel it's a rather dirty solution (ceil). Is there any way of writing this code in a neater or more logical way?</p>
  
  <p><strong>ActionScript:</strong></p>
  
  <pre><code>for (var i:Number = 1; i &lt; 101; i++) {
      var displayIndex:Number = Math.ceil((i/5) * 4);
      trace("i:" + i + " / " + displayIndex);
  }
  </code></pre>
  
  <p><strong>PHP:</strong></p>
  
  <pre><code>for ($i = 1; $i &lt; 101; $i++) {
      echo "i: " . $i . " / displayIndex: " . (ceil(($i/5) * 4) . "&lt;br&gt;");
  }
  </code></pre>
  
  <p><strong>Edit:</strong>
  Let me try to visualize this</p>
  
  <pre><code>index-number =  1 -&gt; display-number =  1
  index-number =  2 -&gt; display-number =  2
  index-number =  3 -&gt; display-number =  3
  index-number =  4 -&gt; display-number =  4
  index-number =  5 -&gt; display-number =  none
  index-number =  6 -&gt; display-number =  5
  index-number =  7 -&gt; display-number =  6
  index-number =  8 -&gt; display-number =  7
  index-number =  9 -&gt; display-number =  8
  index-number = 10 -&gt; display-number =  none
  etc...
  </code></pre>