Addition Problem with Missing Digits

Question

In the addition problem shown each $\ast$ denotes a missing digit and the $\ast$'s are not necessarily identical. What final four digit sum will result from the proper restoration of the missing digits?

   82*
   1*9
+ 1*64
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  1**9

Answer 1

What you have is a sum of the form $$ S = [8(100)+2(10)+a]+[1(100)+b(10)+9] + [1(1000)+c(100)+6(10)+4], $$ where all of this equals $$ 1(1000)+d(100)+e(10)+9. $$ Now consider that \begin{align} S &= (800+20+100+9+1000+60+4)+(100c+10b+a)\\[0.5em] &= 1993 + (100c+10b+a). \end{align} Thus, we have that $$ 1993 + (100c+10b+a) = 1(1000)+d(100)+e(10)+9. $$ We can represent this more sensibly as $$ 1(1000)+(9+c)(100)+(9+b)(10)+(3+a)=1(1000)+d(100)+e(10)+9. $$ Look at how the parts of the equation relate. You can see that we must have the following:

• $9+c = d$
• $9+b = e$
• $3+a = 9$