# Difference between revisions of "1997 AHSME Problems/Problem 1"

## Problem

If $\texttt{a}$ and $\texttt{b}$ are digits for which

$\begin{array}{ccc}& 2 & a\\ \times & b & 3\\ \hline & 6 & 9\\ 9 & 2\\ \hline 9 & 8 & 9\end{array}$

then $\texttt{a+b =}$

$\mathrm{(A)\ } 3 \qquad \mathrm{(B) \ }4 \qquad \mathrm{(C) \ } 7 \qquad \mathrm{(D) \ } 9 \qquad \mathrm{(E) \ }12$

## Solution

From the units digit calculation, we see that the units digit of $a\times 3$ is $9$. Since $0 \le a \le 9$ and $a$ is an integer, the only value of $a$ that works is is $a=3$. As a double-check, that does work, since $23 \times 3 = 69$, which is the first line of the multiplication.

The second line of the multiplication can be found by doing the multiplication $23\times b = 92$. Dividing both sides by $23$ gives $b=4$.

Thus, $a + b = 3 + 4 = 7$, and the answer is $\boxed{C}$.