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

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(→‎Solution)
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== Solution ==
 
== 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:
 +
  
 
<math>50 < 49a+7b+c < 60</math>
 
<math>50 < 49a+7b+c < 60</math>
 +
  
 
<math>50 < 48a+6b < 60</math>
 
<math>50 < 48a+6b < 60</math>
 +
  
 
<math>\frac{25}{3} < 8a+b < 10</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:
 
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 < 64a+8b+c < 80</math>
 +
  
 
<math>70 < 63a+7b < 80</math>
 
<math>70 < 63a+7b < 80</math>
 +
  
 
<math>10 < 9a+b < \frac{80}{7}</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
 
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>
 
<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 = \boxed{(\textbf{C}) 3}</math>
+
 
 +
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})}.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=19|num-a=21|ab=A}}
 
{{AMC12 box|year=2011|num-b=19|num-a=21|ab=A}}

Revision as of 13:08, 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})}.

See also

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
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