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

Revision as of 14:08, 10 February 2011

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 $\boxed{(\textbf{C})}$ .

2011 AMC 12A (Problems • Answer Key • Resources)
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

@@ Line 40: / Line 40: @@
-And since <math>15000 < 19295 < 20000 \to 5000(3) < 19295 < 5000(4)</math>, we find that <math>k = 3</math>, which is \boxed{(\textbf{C})}.
+And since <math>15000 < 19295 < 20000 \to 5000(3) < 19295 < 5000(4)</math>, we find that <math>k = 3</math>, which is <math>\boxed{(\textbf{C})}</math>.
 == See also ==
 {{AMC12 box|year=2011|num-b=19|num-a=21|ab=A}}

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

Revision as of 14:08, 10 February 2011

Problem

Solution

See also