Difference between revisions of "2007 AMC 8 Problems/Problem 9"

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== Solution ==
 
== 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 B:2
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The number in the first row, last column must be a <math>3</math> due to the fact if a <math>3</math> 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 <math>1</math>. Therefore the number in the lower right-hand square is <math>\boxed{\textbf{(B)}\ 2}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2007|num-b=8|num-a=10}}
 
{{AMC8 box|year=2007|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:22, 16 August 2021

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 $\boxed{\textbf{(B)}\ 2}$.

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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