2014 AMC 12A Problems/Problem 19: Difference between revisions

Revision as of 11:00, 7 February 2014

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

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 $2 \cdot 39 = \boxed{\textbf{(E) } 78}$ .

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

@@ Line 16: / Line 16: @@
 ==See Also==
-{{AMC10 box|year=2014|ab=A|num-b=18|num-a=20}}
+{{AMC12 box|year=2014|ab=A|num-b=18|num-a=20}}
 {{MAA Notice}}

Page

Toolbox

Search

2014 AMC 12A Problems/Problem 19: Difference between revisions

Revision as of 11:00, 7 February 2014

Problem

Solution

See Also