If we take 47, reverse and add, 47 + 74 = 121, which is palindromic.
  
  Not all numbers produce palindromes so quickly. For example,

349 + 943 = 1292,
1292 + 2921 = 4213
4213 + 3124 = 7337

  
  That is, 349 took three iterations to arrive at a palindrome.
  
  Although no one has proved it yet, it is thought that some numbers,
  like 196, never produce a palindrome. A number that never forms a
  palindrome through the reverse and add process is called a Lychrel
  number. Due to the theoretical nature of these numbers, and for the
  purpose of this problem, we shall assume that a number is Lychrel
  until proven otherwise. In addition you are given that for every
  number below ten-thousand, it will either (i) become a palindrome in
  less than fifty iterations, or, (ii) no one, with all the computing
  power that exists, has managed so far to map it to a palindrome. In
  fact, 10677 is the first number to be shown to require over fifty
  iterations before producing a palindrome: 4668731596684224866951378664
  (53 iterations, 28-digits).
  
  Surprisingly, there are palindromic numbers that are themselves
  Lychrel numbers; the first example is 4994.
  
  How many Lychrel numbers are there below ten-thousand?


This is my solution:

from datetime import datetime

candidates = []

def reverse(number):
    ''' Returns the reverse order of an integer,
        for example: reverse(123) returns the integer 321 '''
    number = str(number)
    return int(number[::-1])

def is_lychrel(number):
    ''' Returns True if 'number' is a lychrel candidate,
        but works only with numbers from 1 to 10000 '''
    iterated = number + reverse(number)
    for i in range(25):
        rev_iterated = reverse(iterated)
        if iterated != rev_iterated:
            iterated = iterated + rev_iterated
        else:
            return False
    return True

if __name__ == '__main__':

    start_time = datetime.now()

    for number in range(10001):
        if is_lychrel(number):
            candidates.append(number)

    time = datetime.now() - start_time # To get the time of the calculation

    print('In {} were found {} lychrel candidates'.format(time, len(candidates)))
    print(' '.join(map(str, candidates)), 'are candidates.')
    suml = input('Press ENTER to close ')


On my PC (with Intel Atom 2GHz), I got it with .4 seconds.  Could you offer suggestions on making it run faster?


Question: Customising SharePoint 2007 Survey Pages with pagebreaks
Answer: You could achieve it by generating introductory content on the client side as demonstrated below.   

Steps

1) Add CEWP into Survey Edit Form page

2) Insert the following content into CEWP 

&lt;h2 id="questionSummary"&gt;Please specify your development skills​​​&lt;/h2&gt; 
&lt;script type="text/javascript"&gt;
function getParameterByName(name) {
    name = name.replace(/[\[]/, "\\[").replace(/[\]]/, "\\]");
    var regex = new RegExp("[\\?&amp;]" + name + "=([^&amp;#]*)"),
        results = regex.exec(location.search);
    return results == null ? "" : decodeURIComponent(results[1].replace(/\+/g, " "));
}


(function(){
   var questionSummary = $("#questionSummary");
   var fieldName = getParameterByName('FirstField');
   switch(fieldName){
       case "Skills":
          questionSummary.append("Please specify your development skills");
       break;
   }    


})();
&lt;/script&gt;​


Key points:


during page navigation in Survey form, query string parameter
FirstField is appended 
function getParameterByName is used to retrieve query string
parameter, in our case FirstField


Results




  Note: The specified example is for SharePoint 2013 but the same
  approach could be applied in SharePoint 2007