2019 AMC 12B Problems/Problem 9: Difference between revisions
| Line 3: | Line 3: | ||
==Solution== | ==Solution== | ||
The lower bound for x would be x=4, where the sides of the triangle would be (2,1,3). The upper bound for x would be x=4, where the sides of the triangle would be (6,3,3). The number of integers strictly between 4 and 64 is 64 - 4 - 1 = 59 | The lower bound for x would be x=4, where the sides of the triangle would be (2,1,3). The upper bound for x would be x=4, where the sides of the triangle would be (6,3,3). The number of integers strictly between 4 and 64 is 64 - 4 - 1 = 59 | ||
-DrJoyo | |||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2019|ab=B|num-b=8|num-a=10}} | {{AMC12 box|year=2019|ab=B|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 13:24, 14 February 2019
Problem
Solution
The lower bound for x would be x=4, where the sides of the triangle would be (2,1,3). The upper bound for x would be x=4, where the sides of the triangle would be (6,3,3). The number of integers strictly between 4 and 64 is 64 - 4 - 1 = 59
-DrJoyo
See Also
| 2019 AMC 12B (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 AMC 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
