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

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== Problem ==
 
== Problem ==
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== Solution ==
+
Let <math>f(x)=ax^2+bx+c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers. Suppose that <math>f(1)=0</math>, <math>50<f(7)<60</math>, <math>70<f(8)<80</math>, <math>5000k<f(100)<5000(k+1)</math> for some integer <math>k</math>. What is <math>k</math>?
−
== See also ==
+
 
−
{{AMC12 box|year=2011|num-b=19|num-a=21|ab=A}}
+
<math>
 +
\textbf{(A)}\ 1 \qquad
 +
\textbf{(B)}\ 2 \qquad
 +
\textbf{(C)}\ 3 \qquad
 +
\textbf{(D)}\ 4 \qquad
 +
\textbf{(E)}\ 5 </math>
 +
 
 +
== Solution 1 ==
 +
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>.
 +
 
 +
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 = 19305</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>
 +
 
 +
 
 +
 
 +
== Solution 3 (Essentially the same thing as Solution 1) ==
 +
 
 +
So we know that <math>a,b,c</math> are integers so we can use this to our advantage
 +
 
 +
<math>\quad</math>
 +
 
 +
Using <math>f(1)=0</math>, we get the equation <math>a+b+c=0</math> and <math>f(7)=49a+7b+c=5X</math> where <math>X</math> is a decimal digit placeholder. (Ex. <math>X=2</math> provides the value <math>52</math>)
 +
 
 +
<math>\quad</math>
 +
 
 +
Solving for <math>b</math> using the system of equations, we get <math>48a+6b=5X</math> <math>\implies</math> <math>b=-8a+ \frac{5X}{6}</math>
 +
 
 +
<math>\quad</math>
 +
 
 +
Since we know that <math>a</math> and <math>b</math> are both integers, we know that <math>\frac{5X}{6}</math> <math>\in</math> <math>\mathbb{Z}</math> <math>\implies</math> <math>X=4</math> and by extension <math>b=-8a+9</math>
 +
 
 +
<math>\quad</math>
 +
 
 +
Attempting to solve for <math>b</math> again using the system <math>f(8)=64a+8b+c=7Y</math> (<math>Y</math> is another decimal digit placeholder),<math>f(1)=a+b+c=0</math> gives us <math>b=-9a+ \frac{7Y}{7}</math> <math>\implies</math> <math>Y=7</math> <math>\implies</math> <math>b=-9a+11</math>
 +
 
 +
<math>\quad</math>
 +
 
 +
This leads to <math>-8a+9=-9a+11</math> <math>\implies</math> <math>a=2</math> <math>\implies</math> <math>b=-7</math>
 +
 
 +
<math>\quad</math>
 +
 
 +
Plugging in the values of <math>a</math> and <math>b</math> into <math>f(1)=a+b+c=0</math>, we get <math>c=5</math>
 +
 
 +
<math>\quad</math>
 +
 
 +
Substituting the values of <math>a,b,c</math> into <math>f(100)=10000a+100b+c</math>, we get <math>f(100)=19305</math> and <math>5000k<19305<5000(k+1)</math> <math>\implies</math> <math>k=3</math> <math>\implies</math>
 +
<math>\boxed{\textbf{(C)}\ 3}</math>
 +
 
 +
<math>\quad</math>
 +
 
 +
<math>\bf{Note}</math>: We can say that <math>f(7)=5X</math> and <math>f(8)=7Y</math> because we are given that <math>50<f(7)<60</math> and <math>70<f(8)<80</math>
 +
 
 +
 
 +
== See also == {{AMC12 box|year=2011|num-b=19|num-a=21|ab=A}}

Latest revision as of 16:36, 9 January 2024

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*}

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


Solution 3 (Essentially the same thing as Solution 1)

So we know that $a,b,c$ are integers so we can use this to our advantage

$\quad$

Using $f(1)=0$, we get the equation $a+b+c=0$ and $f(7)=49a+7b+c=5X$ where $X$ is a decimal digit placeholder. (Ex. $X=2$ provides the value $52$)

$\quad$

Solving for $b$ using the system of equations, we get $48a+6b=5X$ $\implies$ $b=-8a+ \frac{5X}{6}$

$\quad$

Since we know that $a$ and $b$ are both integers, we know that $\frac{5X}{6}$ $\in$ $\mathbb{Z}$ $\implies$ $X=4$ and by extension $b=-8a+9$

$\quad$

Attempting to solve for $b$ again using the system $f(8)=64a+8b+c=7Y$ ($Y$ is another decimal digit placeholder),$f(1)=a+b+c=0$ gives us $b=-9a+ \frac{7Y}{7}$ $\implies$ $Y=7$ $\implies$ $b=-9a+11$

$\quad$

This leads to $-8a+9=-9a+11$ $\implies$ $a=2$ $\implies$ $b=-7$

$\quad$

Plugging in the values of $a$ and $b$ into $f(1)=a+b+c=0$, we get $c=5$

$\quad$

Substituting the values of $a,b,c$ into $f(100)=10000a+100b+c$, we get $f(100)=19305$ and $5000k<19305<5000(k+1)$ $\implies$ $k=3$ $\implies$ $\boxed{\textbf{(C)}\ 3}$

$\quad$

$\bf{Note}$: We can say that $f(7)=5X$ and $f(8)=7Y$ because we are given that $50<f(7)<60$ and $70<f(8)<80$


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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