title: To generate numbers that don't have num "6" in any of the two number combinations
body: <p>I have been given a problem where - whenever the user types the number '<strong>6</strong>' in the digits (it can be anywhere in the digit's place), he has to type or generate, a combination of <strong>two values that add up to the entered number</strong>.</p>
  <p><strong>Ex1</strong> : if I input number &quot;1<strong>6</strong>&quot;  ==&gt; (7,9) or (8,8) or (1,15) etc., but not (0,1<strong>6</strong>) or (<strong>6</strong>,10) etc.,</p>
  <p><strong>Ex2</strong> : if I input number &quot;<strong>6</strong>1&quot;  ==&gt; (58,3) or (59,2) or (57,4) etc., but not (0,<strong>6</strong>1) or (<strong>6</strong>0,1) etc.,</p>
  <p><strong>Ex3</strong> : if I input number &quot;1<strong>6</strong>8&quot; ==&gt; (75,93) or (80,88) or (10,158) etc., but not (1<strong>6</strong>8,0) or (<strong>6</strong>0,108) etc.,</p>
  <p>I tried to work around this problem and got :</p>
  <p>'''</p>
  <pre><code>def faulty_num():
      x=str(input())
      result=[]
  
      for i in range(1, int(x)+1):
          for j in range(1, int(x)+1):
              if (i+j == int(x) and int(x)%10==6):
                  result.append((i,j))
   
      print(*result[:(int(x)//2)])
  </code></pre>
  <p>'''</p>
  <blockquote>
  <p>when I input 1<strong>6</strong> ==&gt; (1, 15) (2, 14) (3, 13) (4, 12) (5, 11) (<strong>6</strong>, 10) (7, 9) (8, 8)</p>
  </blockquote>
  <p>Kindly forgive me, if my code is messy.</p>
  <p>In the code above, I can eliminate combinations having <strong>6</strong> by mentioning (i!=6 ,j!=6) or (i%10==6 , j%10==6) in my &quot;if condition&quot;. I won't be able to do that when, the digit's size increases. In the mentioned code block, this if condition =</p>
  <blockquote>
  <p>if (i+j == int(x) and int(x)%10==6):</p>
  </blockquote>
  <p>will only generate the required combinations, when the ones place of the digit is &quot;6&quot; and only generate combinations that have the digit <strong>6</strong>. The code only gets more convoluted, the more I think. But, I am honestly clueless at this point. Any suggestion will definitely be helpful. Thank you for your valuable time!</p>