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

(→‎See also)
Line 9: Line 9:
 
\textbf{(E)}\ 5 </math>
 
\textbf{(E)}\ 5 </math>
  
−
== Solution ==
+
== Solution 1 ==
 
From <math>f(1) = 0</math>, we know that <math>a+b+c = 0</math>.
 
From <math>f(1) = 0</math>, we know that <math>a+b+c = 0</math>.
  
Line 18: Line 18:
 
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 = 19305</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 = 19305</math>
  
−
Since <math>15000 < 19305 < 20000 \to 5000(3) < 19305 < 5000(4)</math>, we find that <math>k = 3 \rightarrow \boxed{(\textbf{C})}</math>.
+
Since <math>15000 < 19305 < 20000 \to 5000(3) < 19305 < 5000(4)</math>, we find that <math>k = 3 \rightarrow \boxed{\textbf{(C)}\ 3}</math>.
 +
 
 +
== Solution 2 ==
 +
<math>f(x)</math> is some non-monic quadratic with a root at <math>x=1</math>. Knowing this, we'll forget their silly <math>a</math>, <math>b</math>, and <math>c</math> and instead write it as <math>f(x)=p(x-1)(x-r)</math>.
 +
 
 +
<math>f(7)=6p(7-r)</math>, so <math>f(7)</math> is a multiple of 6. They say <math>f(7)</math> is between 50 and 60, exclusive. Notice that the only multiple of 6 in that range is 54. Thus, <math>f(7)=6p(7-r)=54</math>.
 +
 
 +
<math>f(8)=7p(8-r)</math>, so <math>f(8)</math> is a multiple of 7. They say <math>f(8)</math> is between 70 and 80, exclusive. Notice that the only multiple of 7 in that range is 77. Thus, <math>f(8)=7p(8-r)=77</math>.
 +
 
 +
Now, we solve a system of equations in two variables.
 +
 
 +
<cmath>
 +
\begin{align*}
 +
6p(7-r)&=54 \\
 +
7p(8-r)&=77 \\
 +
\\
 +
p(7-r)&=9 \\
 +
p(8-r)&=11 \\
 +
\\
 +
7p-pr&=9 \\
 +
8p-pr&=11 \\
 +
\\
 +
(8p-pr)-(7p-pr)&=11-9 \\
 +
\\
 +
p&=2 \\
 +
\\
 +
2(7-r)&=9 \\
 +
\\
 +
r&=2.5
 +
\end{align*}
 +
</cmath>
 +
 
 +
<math>f(100)=2(100-1)(100-2.5)=19305 \implies k=3 \implies \boxed{\textbf{(C)}\ 3}</math>
  
 
== 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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:58, 26 July 2017

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 1

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

Solution 2

$f(x)$ is some non-monic quadratic with a root at $x=1$. Knowing this, we'll forget their silly $a$, $b$, and $c$ and instead write it as $f(x)=p(x-1)(x-r)$.

$f(7)=6p(7-r)$, so $f(7)$ is a multiple of 6. They say $f(7)$ is between 50 and 60, exclusive. Notice that the only multiple of 6 in that range is 54. Thus, $f(7)=6p(7-r)=54$.

$f(8)=7p(8-r)$, so $f(8)$ is a multiple of 7. They say $f(8)$ is between 70 and 80, exclusive. Notice that the only multiple of 7 in that range is 77. Thus, $f(8)=7p(8-r)=77$.

Now, we solve a system of equations in two variables.

\begin{align*} 6p(7-r)&=54 \\ 7p(8-r)&=77 \\ \\ p(7-r)&=9 \\ p(8-r)&=11 \\ \\ 7p-pr&=9 \\ 8p-pr&=11 \\ \\ (8p-pr)-(7p-pr)&=11-9 \\ \\ p&=2 \\ \\ 2(7-r)&=9 \\ \\ r&=2.5 \end{align*}

$f(100)=2(100-1)(100-2.5)=19305 \implies k=3 \implies \boxed{\textbf{(C)}\ 3}$

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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