# 2007 AMC 8 Problems/Problem 9

## Problem

To complete the grid below, each of the digits 1 through 4 must occur once in each row and once in each column. What number will occupy the lower right-hand square? $$\begin{tabular}{|c|c|c|c|}\hline 1 & & 2 &\\ \hline 2 & 3 & &\\ \hline & &&4\\ \hline & &&\\ \hline\end{tabular}$$ $\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad\textbf{(E)}\ \text{cannot be determined}$

## Solution

The number in the first row, last column must be a $3$ due to the fact if a $3$ was in the first row, second column, there would be two threes in that column. By the same reasoning, the number in the second row, last column has to be a $1$. Therefore the number in the lower right-hand square is $\boxed{\textbf{(B)}\ 2}$.

## Solution 2

Note how the first and second row already contain a $2$. Since the third row, last column already has a $4$, the only possible place a $2$ could be in is the bottom right square. Thus our answer is $\boxed{\textbf{(B)}\ 2}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 