2014 AIME I Problems/Problem 6
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.
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
.
Solution 4 (unnecessarily long)
First, we expand both equations to get and
. The
-intercept for the first equation can be expressed as
. From this, the x-intercepts for the first equation can be written as
. Since the
-intercepts must be integers,
must also be an integer. From solution 1, we know
must be greater than or equal to 32. Solving for
, we find that
.
We can test this using the second equation, whose -intercepts are
. Substituting 36 in for
, we get
, which satisfies the requirement that all x-intercepts must be (positive) integers.
Thus, .
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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