Difference between revisions of "2017 AMC 8 Problems/Problem 11"
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Since the number of tiles lying on both diagonals is <math>37</math>, counting one tile twice, there are <math>37=2x-1\implies x=19</math> tiles on each side. Hence, our answer is <math>19^2=361=\boxed{\textbf{(C)}\ 361}</math>. | Since the number of tiles lying on both diagonals is <math>37</math>, counting one tile twice, there are <math>37=2x-1\implies x=19</math> tiles on each side. Hence, our answer is <math>19^2=361=\boxed{\textbf{(C)}\ 361}</math>. | ||
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| + | ==See Also:== | ||
| + | {{AMC8 box|year=2017|num-b=9|num-a=11}} | ||
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| + | {{MAA Notice}} | ||
Revision as of 19:49, 8 November 2020
Problem 11
A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?
Solution
Since the number of tiles lying on both diagonals is 
, counting one tile twice, there are 
tiles on each side. Hence, our answer is 
.
See Also:
| 2017 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
