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}} | ||
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Revision as of 17:38, 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?
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
.
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
.
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.