Difference between revisions of "2015 AMC 12A Problems/Problem 24"
(4 intermediate revisions by 3 users not shown) | |||
Line 5: | Line 5: | ||
is a real number? | is a real number? | ||
− | <math> \textbf{(A)}\ \frac{3}{ | + | <math> \textbf{(A)}\ \frac{3}{50} \qquad\textbf{(B)}\ \frac{4}{25} \qquad\textbf{(C)}\ \frac{41}{200} \qquad\textbf{(D)}\ \frac{6}{25} \qquad\textbf{(E)}\ \frac{13}{50}</math> |
==Solution== | ==Solution== | ||
Line 16: | Line 16: | ||
<math>\text{Case~1:}</math> Either <math>\cos(a\pi)</math> or <math>\sin(b\pi)</math> is <math>0</math>. | <math>\text{Case~1:}</math> Either <math>\cos(a\pi)</math> or <math>\sin(b\pi)</math> is <math>0</math>. | ||
− | The two <math>\text{a's}</math> satisfying this are <math>\tfrac{1}{2}</math> and <math>\tfrac{3}{2}</math>, and the two <math>\text{b's}</math> satisfying this are <math>0</math> and <math>1</math>. Because <math>a</math> and <math>b</math> can both be expressed as fractions with a denominator less than or equal to <math>5</math>, | + | The two <math>\text{a's}</math> satisfying this are <math>\tfrac{1}{2}</math> and <math>\tfrac{3}{2}</math>, and the two <math>\text{b's}</math> satisfying this are <math>0</math> and <math>1</math>. Because <math>a</math> and <math>b</math> can both be expressed as fractions with a denominator less than or equal to <math>5</math>, there are a total of <math>20</math> possible values for <math>a</math> and <math>b</math>: |
<cmath>0, 1, \frac{1}{2}, \frac{3}{2}, \frac{1}{3},</cmath> | <cmath>0, 1, \frac{1}{2}, \frac{3}{2}, \frac{1}{3},</cmath> | ||
Line 43: | Line 43: | ||
Thus, our final answer is <math>\frac{(20 + 76)}{400} = \frac{6}{25}</math>, which is <math>\boxed{\text{(D)}}</math>. | Thus, our final answer is <math>\frac{(20 + 76)}{400} = \frac{6}{25}</math>, which is <math>\boxed{\text{(D)}}</math>. | ||
+ | |||
+ | === Video Solution by Richard Rusczyk === | ||
+ | |||
+ | https://artofproblemsolving.com/videos/amc/2015amc12a/401 | ||
+ | |||
+ | ~ dolphin7 | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2015|ab=A|num-b=23|num-a=25}} | {{AMC12 box|year=2015|ab=A|num-b=23|num-a=25}} |
Revision as of 14:02, 15 May 2020
Problem
Rational numbers and are chosen at random among all rational numbers in the interval that can be written as fractions where and are integers with . What is the probability that is a real number?
Solution
Let and . Consider the binomial expansion of the expression:
We notice that the only terms with are the second and the fourth terms. Thus for the expression to be a real number, either or must be , or the second term and the fourth term cancel each other out (because in the fourth term, you have ).
Either or is .
The two satisfying this are and , and the two satisfying this are and . Because and can both be expressed as fractions with a denominator less than or equal to , there are a total of possible values for and :
Calculating the total number of sets of results in sets. Calculating the total number of invalid sets (sets where doesn't equal or and doesn't equal or ), resulting in .
Thus the number of valid sets is .
: The two terms cancel.
We then have:
So:
which means for a given value of or , there are valid values(one in each quadrant).
When either or are equal to , however, there are only two corresponding values. We don't count the sets where either or equals , for we would get repeated sets. We also exclude values where the denominator is an odd number, for we cannot find any corresponding values(for example, if is , then must be , which we don't have). Thus the total number of sets for this case is .
Thus, our final answer is , which is .
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2015amc12a/401
~ dolphin7
See Also
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 |