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 22:20, 24 May 2014
Contents
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. 
