Difference between revisions of "2014 AMC 12A Problems/Problem 19"
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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{( | + | 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>. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=A|num-b=18|num-a=20}} | {{AMC10 box|year=2014|ab=A|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:58, 7 February 2014
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
Factor the quadratic into
![\[\left(5x - \frac{12}{n}\right)\left(x - n\right) = 0\]](http://latex.artofproblemsolving.com/f/0/7/f073e27e2cb908a971e8e0c38248d18f709bdbc2.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 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
