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Difference between revisions of "2014 AMC 12A Problems/Problem 19"

(→‎Solution)
(→‎Solution)
Line 10: Line 10:
 
== Solution ==
 
== Solution ==
 
Factor the quadratic into
 
Factor the quadratic into
−
<cmath> \left(5x - \frac{12}{n}\right)\left(x - n\right) = 0 </cmath>
+
<cmath> \left(5x + \frac{12}{n}\right)\left(x + n\right) = 0 </cmath>
−
where <math>n</math> is our integer solution. Then,  
+
where <math>-n</math> is our integer solution. Then,  
 
<cmath> k = \frac{12}{n} + 5n, </cmath>
 
<cmath> k = \frac{12}{n} + 5n, </cmath>
 
which takes rational values between <math>-200</math> and <math>200</math> when <math>|n| \leq 39</math>, excluding <math>n = 0</math>. This leads to an answer of <math>2 \cdot 39 = \boxed{\textbf{(E) } 78}</math>.
 
which takes rational values between <math>-200</math> and <math>200</math> when <math>|n| \leq 39</math>, excluding <math>n = 0</math>. This leads to an answer of <math>2 \cdot 39 = \boxed{\textbf{(E) } 78}</math>.

Revision as of 10:58, 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 $2 \cdot 39 = \boxed{\textbf{(E) } 78}$.

See Also

2014 AMC 10A (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 10 Problems and Solutions

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