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 "2002 Indonesia MO Problems/Problem 2"

m
m (→‎Solution)
 
Line 14: Line 14:
 
** For the case of no <math>1</math>, the two ways that work are <math>2,2,2,3,6</math> and <math>4,2,2,3,3</math>, for a total of <math>\tfrac{5!}{3!} + \tfrac{5!}{2!2!} = 20+30 = 50</math> possibilities.
 
** For the case of no <math>1</math>, the two ways that work are <math>2,2,2,3,6</math> and <math>4,2,2,3,3</math>, for a total of <math>\tfrac{5!}{3!} + \tfrac{5!}{2!2!} = 20+30 = 50</math> possibilities.
 
** For the case of one <math>1</math>, the three ways that work are <math>1,4,4,3,3</math> and <math>1,4,2,6,3</math> and <math>1,2,2,6,6</math>, for a total of <math>\tfrac{5!}{2!2!} + 5! + \tfrac{5!}{2!2!} = 30+120+30 = 180</math> possibilities.
 
** For the case of one <math>1</math>, the three ways that work are <math>1,4,4,3,3</math> and <math>1,4,2,6,3</math> and <math>1,2,2,6,6</math>, for a total of <math>\tfrac{5!}{2!2!} + 5! + \tfrac{5!}{2!2!} = 30+120+30 = 180</math> possibilities.
−
** For the case of two <math>1</math>, the only way that works is <math>1,1,4,4,6</math>, for a total of <math>\tfrac{5!}{2!2!} = 30</math> possibilities.
+
** For the case of two <math>1</math>, the only way that works is <math>1,1,4,6,6</math>, for a total of <math>\tfrac{5!}{2!2!} = 30</math> possibilities.
  
 
Tallying up the results yields <math>240</math> ways to get <math>180</math> and <math>260</math> ways to get <math>144</math>, so the bigger probability is the product being <math>\boxed{144}</math>.
 
Tallying up the results yields <math>240</math> ways to get <math>180</math> and <math>260</math> ways to get <math>144</math>, so the bigger probability is the product being <math>\boxed{144}</math>.

Latest revision as of 23:38, 30 January 2024

Problem

Five regular dices are thrown, one at each time, then the product of the $5$ numbers shown are calculated. Which probability is bigger; the product is $180$ or the product is $144$?

Solution

Let $a$ be the roll of the first dice, $b$ be the roll of the second dice, $c$ be the roll of the third dice, $d$ be the roll of the fourth dice, and $e$ be the roll of the fifth dice. To calculate which probability is bigger, find the number of ways to roll dice that result in the two wanted values. Note that the prime factorization of $180$ is $2^2 \cdot 3^2 \cdot 5$ and the prime factorization of $144$ is $2^4 \cdot 3^2$.

  • If the product of the five dices is $180$, then $abcde = 180$, where $1 \le a,b,c,d,e \le 6$. To find the number of ways, create casework based on the number of ones.
    • For the case of no $1$, the only way that works is $2,2,3,3,5$, for a total of $\tfrac{5!}{2!2!} = 30$ possibilities.
    • For the case of one $1$, the two ways that work are $1,4,3,3,5$ and $1,6,2,3,5$, for a total of $\tfrac{5!}{2!} + 5! = 60+120 = 180$ possibilities.
    • For the case of two $1$, the only way that works is $1,1,6,6,5$, for a total of $\tfrac{5!}{2!2!} = 30$ possibilities.
  • If the product of the five dices is $144$, then $abcde = 144$, where $1 \le a,b,c,d,e \le 6$. To find the number of ways, create casework based on the number of ones.
    • For the case of no $1$, the two ways that work are $2,2,2,3,6$ and $4,2,2,3,3$, for a total of $\tfrac{5!}{3!} + \tfrac{5!}{2!2!} = 20+30 = 50$ possibilities.
    • For the case of one $1$, the three ways that work are $1,4,4,3,3$ and $1,4,2,6,3$ and $1,2,2,6,6$, for a total of $\tfrac{5!}{2!2!} + 5! + \tfrac{5!}{2!2!} = 30+120+30 = 180$ possibilities.
    • For the case of two $1$, the only way that works is $1,1,4,6,6$, for a total of $\tfrac{5!}{2!2!} = 30$ possibilities.

Tallying up the results yields $240$ ways to get $180$ and $260$ ways to get $144$, so the bigger probability is the product being $\boxed{144}$.

See Also

2002 Indonesia MO (Problems)
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 • 7 Followed by
Problem 3
All Indonesia MO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_Indonesia_MO_Problems/Problem_2&oldid=213877"