Difference between revisions of "2019 AMC 12B Problems/Problem 9"

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==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
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-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 (ProblemsAnswer KeyResources)
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

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