2011 AMC 12A Problems/Problem 20

Revision as of 12:49, 24 June 2011 by Math Kirby (talk | contribs) (Solution)

Problem

Let $f(x)=ax^2+bx+c$, where $a$, $b$, and $c$ are integers. Suppose that $f(1)=0$, $50<f(7)<60$, $70<f(8)<80$, $5000k<f(100)<5000(k+1)$ for some integer $k$. What is $k$?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Solution

From $f(1) = 0$, we know that $a+b+c = 0$.

From the first inequality, we get $50 < 49a+7b+c < 60$. Subtracting $a+b+c = 0$ from this gives us $50 < 48a+6b < 60$, and thus $\frac{25}{3} < 8a+b < 10$. Since $8a+b$ must be an integer, it follows that $8a+b = 9$.

Similarly, from the second inequality, we get $70 < 64a+8b+c < 80$. Again subtracting $a+b+c = 0$ from this gives us $70 < 63a+7b < 80$, or $10 < 9a+b < \frac{80}{7}$. It follows from this that $9a+b = 11$.

We now have a system of three equations: $a+b+c = 0$, $8a+b = 9$, and $9a+b = 11$. Solving gives us $(a, b, c) = (2, -7, 5)$ and from this we find that $f(100) = 2(100)^2-7(100)+5 = 19305$

Since $15000 < 19305 < 20000 \to 5000(3) < 19305 < 5000(4)$, we find that $k = 3 \rightarrow \boxed{(\textbf{C})}$.

See also

2011 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AMC 12 Problems and Solutions
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