Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2016 AIME I Problems/Problem 2"

m (→‎Solution)
(→‎Solution)
Line 3: Line 3:
 
==Solution==
 
==Solution==
 
It is easier to think of the dice as <math>21</math> sided dice with <math>6</math> sixes, <math>5</math> fives, etc.  Then there are <math>21^2=441</math> possible rolls. There are <math>2\cdot(1\cdot 6+2\cdot 5+3\cdot 4)=56</math> rolls that will result in a seven.  The odds are therefore <math>\frac{56}{441}=\frac{8}{63}</math>.  The answer is <math>8+63=\boxed{071}</math>
 
It is easier to think of the dice as <math>21</math> sided dice with <math>6</math> sixes, <math>5</math> fives, etc.  Then there are <math>21^2=441</math> possible rolls. There are <math>2\cdot(1\cdot 6+2\cdot 5+3\cdot 4)=56</math> rolls that will result in a seven.  The odds are therefore <math>\frac{56}{441}=\frac{8}{63}</math>.  The answer is <math>8+63=\boxed{071}</math>
 +
 +
See also [[2006 AMC 12B Problems/Problem 17]]
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2016|n=I|num-b=1|num-a=3}}
 
{{AIME box|year=2016|n=I|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:09, 9 June 2016

Problem 2

Two dice appear to be normal dice with their faces numbered from $1$ to $6$, but each die is weighted so that the probability of rolling the number $k$ is directly proportional to $k$. The probability of rolling a $7$ with this pair of dice is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

It is easier to think of the dice as $21$ sided dice with $6$ sixes, $5$ fives, etc. Then there are $21^2=441$ possible rolls. There are $2\cdot(1\cdot 6+2\cdot 5+3\cdot 4)=56$ rolls that will result in a seven. The odds are therefore $\frac{56}{441}=\frac{8}{63}$. The answer is $8+63=\boxed{071}$

See also 2006 AMC 12B Problems/Problem 17

See also

2016 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_I_Problems/Problem_2&oldid=78882"