15. BOTH+ DRESS SMART (Use the numbers: 2, 0, 4, 1, 5, 7, 9, 6, 8, and 3) Each letter stands for 1-digit...

This type of problem is considered a logic
puzzle.

1. It is easiest to figure out if anything
   is an obvious zero first. To do this you would have any letter plus another letter
   equaling the first letter.  In this example that does not happen.  Neither B, D, H, nor
   S can be zero.


2. B + R + any carryover from O + E must be
   greater than or equal to ten, because D and S in the 10,000 place are different.  S must
   be one greater than D.  Since D cannot be one from step 1, then S cannot be zero, one,
   or two.


3. At this point you can begin plugging numbers in
   and see if they work or when you would get a conflict. Since there are two Rs and three
   Ss, these are where you will find the most conflicts. 
   


4. The most logical place to start is to substitute 1 for H
   and test numbers 3 - 9 for S. None of these worked. Then test 2 for H and 3-9 for S.
   


5. When you use 2 for H and 8 for S, there are no conflicts
   with the S & R while solving the problem. H + S = 2 + 8 = 10, therefore T = 0,
   carry the 1 and add T + S = 1 + 0 + 8 = 9, therefore R = 9.
   


6. From step 2, you know that D must be one less than S, so
   D must be 7.


7. At this point you can substitute the numbers
   for letters you have solved: __ __ 0 2 + 7 9 __ 8 8 = 8 __ __ 9
   0  


8. The remaining numbers that fill in the blanks are 1,
   3, 4, 5, & 6. There was nothing carried forward from the tens colums, so you
   need to figure out from the remaining numbers what two can add to a third (for hundreds
   place) and also what number added to a nine will equal the last digit with a ten carried
   forward (for the thousands place).


9. If you add 6 + 9 you
   get 15, this would use the 6 and 5 and have the ten carryforward to the 10,000 column.
   Therefore B = 6 and M = 5.


10. You now know: 6 _ 0 2 + 7 9 _
    8 8 = 8 5 _ 9 0


11. For the last three blanks we have 1, 3,
    & 4. The A must be 4 because 1 + 3 = 4. The O and E must be 1 & 3 in
    some order, since there are no other clues, these two can actually fit in either
    way.


12.  B(6)O(1)T(0)H(2) + D(7)R(9)E(3)S(8)S(8) =
    S(8)M(5)A(4)R(9)T(0)