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 | + | <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>. | ||
+ | |||
+ | ==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> | ||
+ | |||
+ | iron | ||
+ | |||
+ | ==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}} |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:33, 4 April 2024
Contents
Problem
There are exactly distinct rational numbers
such that
and
has at least one integer solution for
. What is
?
Solution 1
Factor the quadratic into
where
is our integer solution. Then,
which takes rational values between
and
when
, excluding
. This leads to an answer of
.
Solution 2
Solve for so
Note that
can be any integer in the range
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.