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

(Created page with "==Problem== There are exactly <math>N</math> distinct rational numbers <math>k</math> such that <math>|k|<200</math> and <cmath>5x^2+kx+12=0</cmath> has at least one integer solu...")
 
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\textbf{(E) }78\qquad</math>
 
\textbf{(E) }78\qquad</math>
  
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== Solution ==
+
== Solution 1==
 
Factor the quadratic into
 
Factor the quadratic into
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<cmath> \left(5x - \frac{12}{n}\right)\left(x - n\right) = 0 </cmath>
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<cmath> \left(5x + \frac{12}{n}\right)\left(x + n\right) = 0 </cmath>
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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{(B) } 78}</math>.
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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>.
 +
 
 +
==Solution 2==
 +
Solve for <math>k</math> so <cmath>k=-\frac{12}{x}-5x.</cmath>  Note that <math>x</math> can be any integer in the range <math>[-39,0)\cup(0,39]</math>  so <math>k</math> is rational with <math>\lvert k\rvert<200</math>.  Hence, there are <math>39+39=\boxed{\textbf{(E) } 78}.</math>
 +
 
 +
==Solution 3==
 +
Plug in <math>k=200</math> to find the upper limit. You will find the limit to be a number from <math>0<x<-1</math> and one that is just below <math>-39.</math> All the integer values from <math>-1</math> to <math>-39</math> can be attainable through some value of <math>k</math>. Since the question asks for the absolute value of <math>k</math>, we see that the answer is <math>39\cdot2 = \boxed{\textbf{(E)  }78.}</math>
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 +
iron
 +
 
 +
==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=BoPnuYKBq30
  
 
==See Also==
 
==See Also==
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{{AMC10 box|year=2014|ab=A|num-b=18|num-a=20}}
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{{AMC12 box|year=2014|ab=A|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:33, 4 April 2024

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 $39\cdot2 = \boxed{\textbf{(E) }78.}$

iron

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=BoPnuYKBq30

See Also

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

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