Difference between revisions of "2014 AMC 12A Problems/Problem 19"
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\textbf{(E) }78\qquad</math> | \textbf{(E) }78\qquad</math> | ||
| − | == Solution == | + | == Solution 1== |
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> | ||
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<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>. | ||
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| + | ==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> | ||
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| + | ==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 | ||
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| + | ==Video Solution by SpreadTheMathLove== | ||
| + | https://www.youtube.com/watch?v=BoPnuYKBq30 | ||
==See Also== | ==See Also== | ||
{{AMC12 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}} | {{MAA Notice}} | ||
Latest revision as of 19:33, 4 April 2024
Contents
Problem
There are exactly 
distinct rational numbers 
such that 
and ![\[5x^2+kx+12=0\]](http://latex.artofproblemsolving.com/2/a/3/2a3083567ef6e4c63ddf113aad1c6efad4e79b71.png)
has at least one integer solution for 
. What is ?
Solution 1
Factor the quadratic into
![\[\left(5x + \frac{12}{n}\right)\left(x + n\right) = 0\]](http://latex.artofproblemsolving.com/f/e/5/fe5199e9f960b0c62e2ab49965e0ff96a0bd31f6.png)
where 
is our integer solution. Then,
![\[k = \frac{12}{n} + 5n,\]](http://latex.artofproblemsolving.com/3/e/b/3eb94d8ebb2ae4384f3d40a12c2ae537ea961970.png)
which takes rational values between 
and 
when 
, excluding 
. This leads to an answer of 
.
Solution 2
Solve for so ![\[k=-\frac{12}{x}-5x.\]](http://latex.artofproblemsolving.com/a/c/f/acf467a467306991cbf257dce1a9d3696123e1a7.png)
Note that can be any integer in the range ![$[-39,0)\cup(0,39]$](http://latex.artofproblemsolving.com/0/d/8/0d8a6d739d6ebf4e90398304c7d024a68032a5f7.png)
so is rational with 
. Hence, there are 
Solution 3
Plug in 
to find the upper limit. You will find the limit to be a number from 
and one that is just below 
All the integer values from 
to 
can be attainable through some value of . Since the question asks for the absolute value of , we see that the answer is 
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
