SOLUTION: sum of two three-digit numbers is 600.both th number end with 5.if sum of the digits of one number is more than than the sum of the digits of another number by 12,which of the foll

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