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

Revision as of 21:35, 11 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, 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 $f(100) = 2(100)^2-7(100)+5 = 19295$

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

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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 10: / Line 10: @@
 == Solution ==
-From <math>f(1) = 0</math>, we know that <math>a+b+c = 0</math>. From the first inequality:
+From <math>f(1) = 0</math>, we know that <math>a+b+c = 0</math>.
+From the first inequality, we get <math>50 < 49a+7b+c < 60</math>. Subtracting <math>a+b+c = 0</math> from this gives us <math>50 < 48a+6b < 60</math>, and thus <math>\frac{25}{3} < 8a+b < 10</math>. Since <math>8a+b</math> must be an integer, it follows that <math>8a+b = 9</math>.
-<math>50 < 49a+7b+c < 60</math>
+Similarly, from the second inequality, we get <math>70 < 64a+8b+c < 80</math>. Again subtracting <math>a+b+c = 0</math> from this gives us <math>70 < 63a+7b < 80</math>, or <math>10 < 9a+b < \frac{80}{7}</math>. It follows from this that <math>9a+b = 11</math>.
+We now have a system of three equations: <math>a+b+c = 0</math>, <math>8a+b = 9</math>, and <math>9a+b = 11</math>. Solving gives us <math>(a, b, c) = (2, -7, 5)</math> and from this we find that <math>f(100) = 2(100)^2-7(100)+5 = 19295</math>
-<math>50 < 48a+6b < 60</math>
+Since <math>15000 < 19295 < 20000 \to 5000(3) < 19295 < 5000(4)</math>, we find that <math>k = 3 \rightarrow \boxed{(\textbf{C})}</math>.
-<math>\frac{25}{3} < 8a+b < 10</math>
-Since <math>8a+b</math> must be an integer, it follows that <math>8a+b = 9</math>. Similarly, from the second inequality:
-<math>70 < 64a+8b+c < 80</math>
-<math>70 < 63a+7b < 80</math>
-<math>10 < 9a+b < \frac{80}{7}</math>
-And it follows that <math>9a+b = 11</math>. We now have a system of three equations. Solving it gives us <math>(a, b, c) = (2, -7, 5)</math>. From this, we find that
-<math>f(100) = 2(100)^2-7(100)+5 = 19295</math>
-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 21:35, 11 February 2011

Problem

Solution

See also