2011 AIME I Problems/Problem 6

If the vertex is at $(\frac{1}{4}, -\frac{9}{8})$ , the equation of the parabola can be expressed in the form $y=a(x-\frac{1}{4})^2-\frac{9}{8}$ . Expanding, we find that $y=a(x^2-\frac{x}{2}+\frac{1}{16})-\frac{9}{8}$ , and $y=ax^2-\frac{ax}{2}+\frac{a}{16}-\frac{9}{8}$ . From the problem, we know that the parabola can be expressed in the form $y=ax^2+bx+c$ , where $a+b+c$ is an integer. From the above equation, we can conclude that $a=a$ , $-\frac{a}{2}=b$ , and $\frac{a}{16}-\frac{9}{8}=c$ . Adding up all of these gives us $\frac{9a-18}{16}=a+b+c$ . We know that $a+b+c$ is an integer, so 9a-18 must be divisible by 16. Let $9a=z$ . If ${z-18}\equiv {0} \pmod{16}$ , then ${z}\equiv {2} \pmod{16}$ . Therefore, if $9a=2$ , $a=\frac{2}{9}$ . Adding up gives us $2+9=\boxed{011}$

2011 AIME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2011 AIME I Problems/Problem 6

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