Difference between revisions of "2007 AMC 8 Problems/Problem 9"
m (Reverted edits by Raina0708 (talk) to last revision by Nathan wailes) (Tag: Rollback) |
|||
| Line 16: | Line 16: | ||
{{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
Contents
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?
![\[\begin{tabular}{|c|c|c|c|}\hline 1 & & 2 &\\ \hline 2 & 3 & &\\ \hline & &&4\\ \hline & &&\\ \hline\end{tabular}\]](http://latex.artofproblemsolving.com/7/9/0/7906b7cd161b30f23457e7d33aa6f86c753371f3.png)
Solution
The number in the first row, last column must be a 
due to the fact if a 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 
. Therefore the number in the lower right-hand square is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
Solution 2
Note how the first and second row already contain a 
. Since the third row, last column already has a 
, the only possible place a could be in is the bottom right square. Thus our answer is .
