2014 AMC 12A Problems/Problem 19

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 questions asks for the absolute value of k, we see that the answer is 39*2 = 78

iron

2014 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2014 AMC 12A Problems/Problem 19

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also