Difference between revisions of "2015 AMC 12A Problems/Problem 18"
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By the quadratic formula, the roots <math>r</math> can be represented by | 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> |
Latest revision as of 20:16, 4 December 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
By the quadratic formula, the roots can be represented by For , , since and will have different mantissas (mantissae?).
Now observe the discriminant and have two cases.
Positive
and , since gives imaginary roots. Testing positive values, quickly see that . After and , the difference between the closest nonzero factor pairs of perfect squares exceeds . For , . Checking both yields an integer.
Negative
We can instead test with . If , we have our original expression. Thus, for the same reasons, . (0 does not affect the answer).
(Solution by BJHHar)
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 |