Question 769354
<pre>
Suppose the larger number is AB5 and the smaller number is CD5

If we add them we must get 600:

   AB5
  +CD5
   ---
   600
    
Since the rightmost column 5+5 adds to 10 there will be 1 to carry to
the 10's column:

    1 
   AB5
  +CD5
   ---
   600

So the second column must add to give 10 to get a 0 at the bottom, so
B+D must equal 9, 

 and since 1+B+D=10 there will be a 1 to carry to
the 100's column:

   11 
   AB5
  +CD5
   ---
   600

So 1+A+C = 6 so A+C must = 5.

Since A is larger than C we only have two possibilities for A and C,
A=3 and C=2 or A=4 and C=1.  We try the first way:

   11        
   3B5      
  +2D5       
   ---      
   600      

The sum of the digits of 3B5 is B+8
The sum of the digits of 2D5 is D+7

Since the sum of the digits of one number is more than than the sum of the
digits of another number by 12, either B+8 = D+7+12 or D+7 = B+8+12

If B+8 = D+7+12, then
   B+8 = D+19
   B-D = 11

But no two digits can differ by 11, so that's out

If D+7 = B+8+12, then
   B+7 = D+20
   B-D = 13 

But no two digits can differ by 13, so that's out, also. So

   11        
   3B5      
  +2D5       
   ---      
   600

 is ruled out.

-----------------------

So the addition has to be such that A=4 and C=1:

   11 
   4B5
  +1D5 
   ---      
   600     

The sum of the digits of 4B5 is B+9
The sum of the digits of 1D5 is D+6

So either B+9 = D+6+12 or D+6 = B+9+12

If B+9 = D+6+12, then
   B+9 = D+18
   B-D = 9

The only way that two digits can differ by 9 is for B=9 and D=0,
So that is a possibility.  Let's see if the other is ruled out:

If D+6 = B+9+12, then
   D+6 = B+21
   D-B = 15 

But no two digits can differ by 15, so that's out.

Therefore B=9 and D=0, so the addition is

    
   11 
   495
  +105 
   ---      
   600 

So the smaller number is 105.

Edwin</pre>