I see this problem more as a puzzle than a math problem.  It does

require some thought combined with trial-and-error:



If 2 is replaced by 6 you get:



     746586

    +869430

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

    1616016      



Also note 1212016 becomes 1616016, as required.



Thus, you have d=2, e=6  so  d+e = 8.

 

Answer by greenestamps(13198)   (Show Source): You can put this solution on YOUR website!
 


This problem can of course be solved by trial and error; but solving the problem using logical reasoning is a good mental exercise.
  7 4 2 5 8 6
+ 8 2 9 4 3 0
-------------
1 2 1 2 0 1 6


The addition is correct for the final three digits:
  5 8 6
+ 4 3 0
-------
  0 1 6


So it is unlikely that any of the digits in that part of the problem -- 0, 1, 3, 4, 5, 6, and 8 -- is the one that changes.  So very probably the digit that changes is 2, 7, or 9.


In the thousands column we have digits 2 and 9 in the addends, plus a carry, yielding digit 2 in the sum.  So the 9 can't change; however, the digit 2 can change to any other digit and the addition in the thousands column will still be correct.


In the ten thousands column, we have digits 4 and 2 in the addends yielding digit 1 in the sum.  We don't want the 1 or 4 to change, so the 2 MUST change.


There is a carry from the thousands column to the ten thousands column, so the 2 must change to a 6 to make the addition correct in the ten thousands column.


So it appears changing digit 2 to digit 6 is what we need to do.  And doing that shows that the addition is now correct.
  7 4 6 5 8 6
+ 8 6 9 4 3 0
-------------
1 6 1 6 0 1 6


ANSWER: d+e = 2+6 = 8


 

Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
 .





        First, we must assume (or the problem must say it explicitly), 

                    that the addends are written correctly.







If so, then all you need to solve this problem is





        (a)  to add the two given numbers in a correct way (it is assumed that you are able to do it),





    and





        (b)  to compare your calculated sum with the calculations presented in the post 



              and to detect the difference in some single digit.







So, I honestly do not understand, why a conscientious student comes to the forum 

with such question, where everything is as clear as at a sunny day.







 

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