Difference between revisions of "2014 AIME I Problems/Problem 6"
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== Solution 1 == | == Solution 1 == | ||
− | 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. | + | 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. Notice that because the two parabolas have to have positive x-intercepts, <math>h\ge32</math>. |
− | + | We see that <math>h^2=\frac{2014-k}{2}</math>, so we now need to find a positive integer <math>h</math> which has positive integer x-intercepts for both equations. | |
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>. | 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>. | ||
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>. | 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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== Solution 2 == | == Solution 2 == |
Revision as of 21:20, 24 May 2014
Contents
[hide]Problem 6
The graphs and have y-intercepts of and , respectively, and each graph has two positive integer x-intercepts. Find .
Solution 1
Begin by setting to 0, then set both equations to and , respectively. Notice that because the two parabolas have to have positive x-intercepts, .
We see that , so we now need to find a positive integer which has positive integer 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 .
Solution 2
Let and for the first equation, resulting in . Substituting back in to the original equation, we get .
Now we set equal to zero, since there are two distinct positive integer roots. Rearranging, we get , which simplifies to . Applying difference of squares, we get .
Now, we know that and are both integers, so we can use the fact that , and set and (note that letting gets the same result). Therefore, .
Note that we did not use the second equation since we took advantage of the fact that AIME answers must be integers. However, one can enter into the second equation to verify the validity of the answer.
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.