# 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, $70, $5000k 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 and $70