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Difference between revisions of "2007 AMC 8 Problems/Problem 9"
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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>.
 
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>.
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== Solution 2 ==
 +
Note how the first and second row already contain a <math>2</math>. Since the third row, last column already has a <math>4</math>, the only possible place a <math>2</math> could be in is the bottom right square. Thus our answer 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}}
−
 
−
== Solution 2 ==
 
−
Note how the first and second row already contain a <math>2</math>. Since the third row, last column already has a <math>4</math>, the only possible place a <math>2</math> could be in is the bottom right square. Thus our answer is <math>\boxed{\textbf{(B)}\ 2}</math>.
 

Revision as of 16:38, 1 March 2022

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

See Also

2007 AMC 8 (Problems • Answer Key • Resources)
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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