Difference between revisions of "2017 AMC 8 Problems/Problem 11"
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==Solution== | ==Solution== | ||
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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| + | ==Video Solution== | ||
| + | Associated video: https://youtu.be/QCWOZwYVJMg | ||
==See Also:== | ==See Also:== | ||
Revision as of 10:17, 30 December 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 
.
Video Solution
Associated video: https://youtu.be/QCWOZwYVJMg
See Also:
| 2017 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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. 
