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. 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> |
Revision as of 16:16, 15 February 2010
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 
. 
