Difference between revisions of "2014 AIME I Problems/Problem 6"

Line 4: Line 4:
  
 
== Solution ==
 
== Solution ==
 
+
We use the fact that 2013 and 2014 are y-intercepts by plugging them into the equations to get <math>2013=3h^2+j</math> and <math>2014=2h^2+k.</math>
 
== See also ==
 
== See also ==
 
{{AIME box|year=2014|n=I|num-b=5|num-a=7}}
 
{{AIME box|year=2014|n=I|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:16, 14 March 2014

Problem 6

The graphs $y=3(x-h)^2+j$ and $y=2(x-h)^2+k$ have y-intercepts of $2013$ and $2014$, respectively, and each graph has two positive integer x-intercepts. Find $h$.

Solution

We use the fact that 2013 and 2014 are y-intercepts by plugging them into the equations to get $2013=3h^2+j$ and $2014=2h^2+k.$

See also

2014 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png