2011 AMC 12A Problems/Problem 20

Revision as of 13:08, 10 February 2011 by Rpond (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:


$50 < 49a+7b+c < 60$


$50 < 48a+6b < 60$


$\frac{25}{3} < 8a+b < 10$


Since $8a+b$ must be an integer, it follows that $8a+b = 9$. Similarly, from the second inequality:


$70 < 64a+8b+c < 80$


$70 < 63a+7b < 80$


$10 < 9a+b < \frac{80}{7}$


And it follows that $9a+b = 11$. We now have a system of three equations. Solving it gives us $(a, b, c) = (2, -7, 5)$. From this, we find that


$f(100) = 2(100)^2-7(100)+5 = 19295$


And since $15000 < 19295 < 20000 \to 5000(3) < 19295 < 5000(4)$, we find that $k = 3$, which is $\boxed{(\textbf{C})}$.

See also

2011 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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