Difference between revisions of "2002 Indonesia MO Problems/Problem 2"

(Solution to Problem 2)
 
m
 
Line 19: Line 19:
  
 
==See Also==
 
==See Also==
{{Indonesia MO 7p box
+
{{Indonesia MO box
 
|year=2002
 
|year=2002
 
|num-b=1
 
|num-b=1
 
|num-a=3
 
|num-a=3
 +
|eight=
 
}}
 
}}
  
 
[[Category:Intermediate Probability Problems]]
 
[[Category:Intermediate Probability Problems]]

Latest revision as of 00:09, 4 August 2018

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 $1,1,4,4,6$, 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
Invalid username
Login to AoPS