# 1995 AHSME Problems/Problem 13

## 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 $0$, the units column would be incorrect.

If we change $1$, then the leading $1$ in the sum would be incorrect.

However, looking at the $2$ in the hundred-thousands column, it would be possible to change the $2$ to either a $5$ (no carry) or a $6$ (carry) to create a correct statement.

Changing the $2$ to a $5$ would give $745586 + 859430$ on top, which equals $1605016$. This does not match up to the bottom.

Changing the $2$ to a $6$ gives $746586 + 869430$ on top, which has a sum of $1616016$. This is the number on the bottom if the $2$s were changed to $6$s.

Thus $d=2$ and $e=6$. so $d+e= 8 \boxed{\mathrm{ (C)}}$

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