Difference between revisions of "1995 AHSME Problems/Problem 13"

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{{AHSME box|year=1995|num-b=12|num-a=14}}
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[[Category:Introductory Algebra Problems]]

Revision as of 08:50, 17 August 2008

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 $e=6$. $d+e=\boxed{8\mathrm{(C)}}$

See also

1995 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AHSME Problems and Solutions