2014 AMC 12A Problems/Problem 19

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Problem

There are exactly $N$ distinct rational numbers $k$ such that $|k|<200$ and \[5x^2+kx+12=0\] has at least one integer solution for $x$. What is $N$?

$\textbf{(A) }6\qquad \textbf{(B) }12\qquad \textbf{(C) }24\qquad \textbf{(D) }48\qquad \textbf{(E) }78\qquad$

Solution 1

Factor the quadratic into \[\left(5x + \frac{12}{n}\right)\left(x + n\right) = 0\] where $-n$ is our integer solution. Then, \[k = \frac{12}{n} + 5n,\] which takes rational values between $-200$ and $200$ when $|n| \leq 39$, excluding $n = 0$. This leads to an answer of $2 \cdot 39 = \boxed{\textbf{(E) } 78}$.

Solution 2

Solve for $k$ so \[k=-\frac{12}{x}-5x.\] Note that $x$ can be any integer in the range $[-39,0)\cup(0,39]$ so $k$ is rational with $\lvert k\rvert<200$. Hence, there are $39+39=\boxed{\textbf{(E) } 78}.$

Solution 3

Plug in $k=200$ to find the upper limit. You will find the limit to be a number from $0<x<-1$ and one that is just below $-39.$ All the integer values from $-1$ to $-39$ can be attainable through some value of $k$. Since the question asks for the absolute value of $k$, we see that the answer is $39\cdot2 = \boxed{\textbf{(C)  }78.}$

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See Also

2014 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AMC 12 Problems and Solutions

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