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}}</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>. |
==See Also== | ==See Also== | ||
Revision as of 15:06, 22 November 2017
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 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 | ||
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