Difference between revisions of "2011 AMC 12A Problems/Problem 20"

Revision as of 20:52, 3 July 2013

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 $k = 3 \rightarrow \boxed{(\textbf{C})}$ .

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 == See also ==
 {{AMC12 box|year=2011|num-b=19|num-a=21|ab=A}}
+{{MAA Notice}}

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Difference between revisions of "2011 AMC 12A Problems/Problem 20"

Revision as of 20:52, 3 July 2013

Problem

Solution

See also