Difference between revisions of "2014 AIME I Problems/Problem 6"
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== Solution == | == Solution == | ||
− | + | Begin by setting <math>x</math> to 0, then set both equations to <math>h^2=\frac{2013-j}{3}</math> and <math>h^2=\frac{2014-k}{2}</math>, respectively. You'll notice that because the two parabolas have to have x-intercepts, <math>h\le32</math>. | |
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+ | You'll know that <math>h^2=\frac{2014-k}{2}</math>, so you now need to find a positive integer <math>h</math> which has [b]positive integer[/b] x-intercepts for both equations. | ||
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+ | Notice that if <math>k=2014-2h^2</math> is -2 times a square number, then you have found a value of <math>h</math> for which the second equation has positive x-intercepts. We guess and check <math>h=36</math> to obtain <math>k=-578=-2(17^2)</math>. | ||
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+ | 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 <math>\boxed{036}</math>. | ||
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== 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 12:44, 15 March 2014
Problem 6
The graphs and have y-intercepts of and , respectively, and each graph has two positive integer x-intercepts. Find .
Solution
Begin by setting to 0, then set both equations to and , respectively. You'll notice that because the two parabolas have to have x-intercepts, .
You'll know that , so you now need to find a positive integer which has [b]positive integer[/b] x-intercepts for both equations.
Notice that if is -2 times a square number, then you have found a value of for which the second equation has positive x-intercepts. We guess and check to obtain .
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 .
See also
2014 AIME I (Problems • Answer Key • Resources) | ||
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.