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>2</math>. <math>\boxed{B}</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>2</math>. <math>\boxed{B}</math> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2007|num-b=8|num-a=10}} | ||
Revision as of 23:36, 12 November 2012
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?
cannot be determined
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 | ||
