Difference between revisions of "1992 AHSME Problems/Problem 8"
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== Solution == | == Solution == | ||
− | <math>\fbox{E}</math> | + | <math>\fbox{E}</math> Let the side length of the square be <math>s</math>. If <math>s</math> is even, there are <math>s</math> black tiles on each of the two diagonals, so the total number is <math>2s</math>. If <math>s</math> is odd, we double-count the middle square which is on both diagonals, so the number is instead <math>2s-1</math>. In this case, since <math>101</math> is odd, we clearly must have the second case, so <math>2s-1=101 \implies 2s=102 \implies s=51</math>, and indeed <math>s</math> is odd as we expected. Thus the overall number of tiles is <math>51^2 = 2601</math>. |
== See also == | == See also == |
Latest revision as of 01:33, 20 February 2018
Problem
A square floor is tiled with congruent square tiles. The tiles on the two diagonals of the floor are black. The rest of the tiles are white. If there are 101 black tiles, then the total number of tiles is
Solution
Let the side length of the square be . If is even, there are black tiles on each of the two diagonals, so the total number is . If is odd, we double-count the middle square which is on both diagonals, so the number is instead . In this case, since is odd, we clearly must have the second case, so , and indeed is odd as we expected. Thus the overall number of tiles is .
See also
1992 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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All AHSME Problems and Solutions |
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