2014 AIME I Problems/Problem 6
Problem
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.
Note on the previous note: we still must use the second equation since we could also use , yielding This answer however does not check out with the second equation which is why it is invalid.
Solution 3
Similar to the first two solutions, we deduce that and are of the form and , respectively, because the roots are integers and so is the -intercept of both equations. So the -intercepts should be integers also.
The first parabola gives And the second parabola gives
We know that and that . It is just a fitting coincidence that the average of and is the same as the average of and . That is .
To check, we have Those are the only two prime factors of and , respectively. So we don't need any new factorizations for those numbers.
Thus the common integer value for 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 |
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