Difference between revisions of "2015 AMC 12A Problems/Problem 18"
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− | By the quadratic formula, | + | By the quadratic formula, the roots <math>r</math> can be represented by |
− | <cmath>r=\frac{a\pm\sqrt{a^2-8a}}{2}</cmath> | + | <cmath>r=\frac{a\pm\sqrt{a^2-8a}}{2}</cmath> |
− | + | For <math>r\in\mathbb{Z}</math>, <math>a\in\mathbb{Z}</math> (or <math>\frac{\sqrt{a^2-8a}}{2}</math> and \frac{a}{2} will have different mantissae). So we observe the discriminant <math>\sqrt{a^2-8a}=\sqrt{a(a-8)}</math>. We then have two cases. | |
Revision as of 22:26, 6 September 2019
Problem
The zeros of the function are integers. What is the sum of the possible values of
?
Solution 1
The problem asks us to find the sum of every integer value of such that the roots of
are both integers.
The quadratic formula gives the roots of the quadratic equation:
As long as the numerator is an even integer, the roots are both integers. But first of all, the radical term in the numerator needs to be an integer; that is, the discriminant equals
, for some nonnegative integer
.
From this last equation, we are given a hint of the Pythagorean theorem. Thus, must be a Pythagorean triple unless
.
In the case , the equation simplifies to
. From this equation, we have
. For both
and
,
yields two integers, so these values satisfy the constraints from the original problem statement. (Note: the two zero roots count as "two integers.")
If is a positive integer, then only one Pythagorean triple could match the triple
because the only Pythagorean triple with a
as one of the values is the classic
triple. Here,
and
. Hence,
. Again,
yields two integers for both
and
, so these two values also satisfy the original constraints.
There are a total of four possible values for :
and
. Hence, the sum of all of the possible values of
is
.
Solution 2 (Quick and Dirty)
By the quadratic formula, the roots can be represented by
For
,
(or
and \frac{a}{2} will have different mantissae). So we observe the discriminant
. We then have two cases.
Positive
and
, since any
yields imaginary roots. Testing positive
values, it quickly becomes clear that
. After
and
, the difference between the closest nonzero factor pairs of perfect squares exceeds
. In the interval
,
. Checking
yields an integer.
Negative
We can instead test with , where
. If
, we have our original discriminant. For the same reasons,
.
(0 also works but does not affect the answer).
Solution 3
Let and
be the roots of
By Vieta's Formulas, and
Substituting gets us
Using Simon's Favorite Factoring Trick:
This means that the values for are
giving us
values of
and
. Adding these up gets
.
Solution 4
The quadratic formula gives . For
to be an integer, it is necessary (and sufficient!) that
to be a perfect square. So we have
; this is a quadratic in itself and the quadratic formula gives
We want to be a perfect square. From smartly trying small values of
, we find
as solutions, which correspond to
. These are the only ones; if we want to make sure then we must hand check up to
. Indeed, for
we have that the differences between consecutive squares are greater than
so we can't have
be a perfect square. So summing our values for
we find 16 (C) as the answer.
Additional note: You can use the quadratic and plug in squares for a (since for b^2 to be an integer a would have to be some square), and eventually you can notice a limit to get the answer~
See Also
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |