2007 AMC 8 Problems/Problem 9

Revision as of 22:36, 12 November 2012 by Basketball8 (talk | contribs) (Solution)

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?

AMC8 2007 9.png

$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}$ 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 $2$. $\boxed{B}$

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AJHSME/AMC 8 Problems and Solutions