title: Check if a number can be formed by sum of a number and its reverse
body: <p>I want to check if a given number can be formed by another number say <code>b</code> and <code>reverse(b)</code>. For example <code>12 == 6+6</code>, <code>22 == 11 + 11</code> and <code>121 == 29+92</code>. One thing I have figured out is if the number is multiple of 11 or it is an even number less than 20, then it can be formed. I tried to implement this below:</p>
  <pre><code>num = 121
  if num%11==0:
      print('Yes')
  else:
      if num%2==0 and num&lt;20:
          print('Yes')
      else:
          for j in range(11,(int(num)//2)+1):
              if j+int(str(j)[::-1])==num:
                  print('Yes')
                  break
           
  </code></pre>
  <p>However, if the condition goes into the <code>for</code> loop, it gives <code>TLE</code>. Can any other conditions be given?</p>
  <p><strong>Update</strong>: If the reversed number has trailing zeroes, it should be removed and then added. For example: <code>101 == 100+1</code>. I am looking for an optimized form of my code. Or I think I am missing some conditions which can take O(1) time, similar to the condition <code>if num%11==0: print('Yes')</code></p>