2011 AMC 12A Problems/Problem 20

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 (Long but easy system of equations)

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

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

(Ex. $X=2$ provides the value $52$ )

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

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 $b=-8a+9$

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

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

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

$f(100)=10000a+100b+c$

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

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2011 AMC 12A Problems/Problem 20

Contents

Problem

Solution 1

Solution 2

Solution 3 (Long but easy system of equations)

See also