2014 AIME I Problems/Problem 6

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

Begin by setting $x$ to 0, then set both equations to $h^2=\frac{2013-j}{3}$ and $h^2=\frac{2014-k}{2}$ , respectively. You'll notice that because the two parabolas have to have x-intercepts, $h\le32$ .

You'll know that $h^2=\frac{2014-k}{2}$ , so you now need to find a positive integer $h$ which has [b]positive integer[/b] x-intercepts for both equations.

Notice that if $k=2014-2h^2$ is -2 times a square number, then you have found a value of $h$ for which the second equation has positive x-intercepts. We guess and check $h=36$ to obtain $k=-578=-2(17^2)$ .

Following this, we check to make sure the first equation also has positive x-intercepts (which it does), so we can conclude the answer is $\boxed{036}$ .

2014 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2014 AIME I Problems/Problem 6

Problem 6

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