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

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right-hand square?
 
right-hand square?
  
<center>[[Image:AMC8_2007_9.png]]</center>
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<cmath> Unknown environment 'tabular' </cmath>
  
<math>\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}</math> cannot be determined
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<math>\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad\textbf{(E)}\ \text{cannot be determined}</math>
  
 
== Solution ==
 
== Solution ==
  
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. 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>2</math>. <math>\boxed{B}</math>
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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>.
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== Solution 2 ==
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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>.
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==Video Solution by SpreadTheMathLove==
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https://www.youtube.com/watch?v=omFpSGMWhFc
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==Video Solution by WhyMath==
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https://youtu.be/E8pV_WO_u9E
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==See Also==
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{{AMC8 box|year=2007|num-b=8|num-a=10}}
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{{MAA Notice}}

Latest revision as of 16:38, 28 October 2024

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

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=omFpSGMWhFc

Video Solution by WhyMath

https://youtu.be/E8pV_WO_u9E

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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