1995 AHSME Problems/Problem 13

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Problem

The addition below is incorrect. The display can be made correct by changing one digit $d$ , wherever it occurs, to another digit $e$ . Find the sum of $d$ and $e$ .

$\begin{tabular}{ccccccc} & 7 & 4 & 2 & 5 & 8 & 6 \\+ & 8 & 2 & 9 & 4 & 3 & 0 \\\hline 1 & 2 & 1 & 2 & 0 & 1 & 6 \end{tabular}$

$\mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ \text{more than 10} }$

Solution

If we change 0 only, it would be incorrect. The same for 1, but 2 we could change to 6. If we try all the other digits, we see that it would be impossible to correct the addition. Thus $d=2$ and $e=6$ . $d+e=\boxed{8\mathrm{(C)}}$

1995 AHSME (Problems)
Preceded by Problem 12	Followed by Problem 14
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1995 AHSME Problems/Problem 13

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Solution

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