# 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 byProblem 5 Followed byProblem 7 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions