Difference between revisions of "2019 AIME II Problems/Problem 4"

Revision as of 19:56, 23 March 2019

Problem 4

A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution

Notice that, other than the number 5, the remaining numbers 1, 2, 3, 4, 6 are only divisible by 2 and/or 3. We can do some cases on the number of 5's rolled (note that there are $6^4 = 1296$ outcomes).

Case 1 (easy): Four 5's are rolled. This has probability $\frac{1}{6^4}$ of occurring.

Case 2: Two 5's are rolled.

Case 3: No 5's are rolled.

To find the number of outcomes for the latter two cases, we will use recursion. Consider a 5-sided die with faces numbered 1, 2, 3, 4, 6. For $n \ge 1$ , let $a_n$ equal the number of outcomes after rolling the die $n$ times, with the property that the product is a square. Thus, $a_1 = 2$ as 1 and 4 are the only possibilities.

To find $a_{n+1}$ given $a_n$ (where $n \ge 1$ ), we observe that if the first $n$ rolls multiply to a perfect square, then the last roll must be 1 or 4. This gives $2a_n$ outcomes. Otherwise, the first $n$ rolls do not multiply to a perfect square ( $5^n - a_n$ outcomes). In this case, we claim that the last roll is uniquely determined (either 2, 3, or 6). If the product of the first $n$ rolls is $2^x 3^y$ where $x$ and $y$ are not both even, then we observe that if $x$ and $y$ are both odd, then the last roll must be 6; if only $x$ is odd, the last roll must be 2, and if only $y$ is odd, the last roll must be 3. Thus, we have $5^n - a_n$ outcomes in this case, and $a_{n+1} = 2a_n + (5^n - a_n) = 5^n + a_n$ .

Computing $a_2$ , $a_3$ , $a_4$ gives $a_2 = 7$ , $a_3 = 32$ , and $a_4 = 157$ . Thus for Case 3, there are 157 outcomes. For case 2, we multiply by $\binom{4}{2} = 6$ to distribute the two 5's among four rolls. Thus the probability is

$\frac{1 + 6 \cdot 7 + 157}{6^4} = \frac{200}{6^4} = \frac{25}{162} \implies m+n = \boxed{187}$

-scrabbler94

Solution 2

We can solve this without finding the amount of cases. Notice when we roll a 1 or a 4, it does not affect whether or not the product is a square number. We have a 1/3 chance of rolling either a 1 or 4. We have a 2/3 chance of rolling a 2,3,5 or 6. Lets call rolling 1 or 4 rolling a dud.

Probability of rolling 4 duds: $(\frac{1}{3})^4$

Probability of rolling 3 duds: $4 * (\frac{1}{3})^3 * \frac{2}{3}$

Probability of rolling 2 duds: $6 * (\frac{1}{3})^2 * (\frac{2}{3})^2$

Probability of rolling 1 dud: $4 * \frac{1}{3} * (\frac{2}{3})^3$

Probability of rolling 0 duds: $(\frac{2}{3})^4$

Now we will find the probability of a square product given we have rolled each amount of duds

Probability of getting a square product given 4 duds: 1

Probability of getting a square product given 3 duds: 0 (you will have 1 non-dud and that's never going to be square)

Probability of getting a square product given 2 duds: $\frac{1}{4}$ (as long as our two non-duds are the same, our product will be square)

Probability of getting a square product given 1 duds: $\frac{3!}{4^3}$ = $\frac{3}{32}$ (the only way to have a square product is rolling a 2,3 and 6. There are 3! ways of doing that and a total of $4^3$ ways to roll 3 non-duds).

Probability of getting a square product given 0 duds: $\frac{40}{4^4}$ = $\frac{5}{32}$ (We can have any two non-duds twice. For example, 2,2,5,5. There are $\binom{4}{2} = 6$ ways of choosing which two non-duds to use and $\binom{4}{2} = 6$ ways of choosing how to arrange those 4 numbers. That gives us 6*6=36 combinations. We can also have 2,2,2,2 or 3,3,3,3 or 5,5,5,5 or 6,6,6,6. This gives us a total of 40 combinations).

We multiply each probability of rolling k duds with the probability of getting a square product given k duds and then sum all the values.

$(\frac{1}{3})^4 * 1 + 4 * (\frac{1}{3})^3 * \frac{2}{3} * 0 + 6 * (\frac{1}{3})^2 * (\frac{2}{3})^2 * \frac{1}{4} + 4 * \frac{1}{3} * (\frac{2}{3})^3 * \frac{3}{32} + (\frac{2}{3})^4 * \frac{5}{32} = \frac{25}{162}.$

$25+162$ = $\boxed{187}$

-dnaidu (silverlizard)

Solution 3

Note that rolling a 1/4 will not affect whether or not the product is a perfect square. This means that in order for the product to be a perfect square, all non 1/4 numbers rolled must come in pairs, with the only exception being the triplet 2,3, 6. Now we can do casework:

If there are four 1/4's, then there are $2^4=16$ combinations. If there are three 1/4's, then there are 0 combinations, because the fourth number isn't a square. If there are two 1/4's, there are $2^2=4$ ways to choose the two 1/4's, 4 ways to choose the remaining pair of numbers, and $\frac{4!}{2!2!}=6$ ways to arrange, so there are $4\cdot 4\cdot 6=96$ combinations for this case. If there is one 1/4, then there are 2 ways to choose whether it is a 1 or 4, and the remaining three numbers must be 2, 3, and 6, so there are $4!$ ways to order, meaning there are $2\cdot 4!=48$ combinations for this case. Our final case is if there are no 1/4's, in which case we must have two pairs. If the two pairs are of different numbers, then there $\binom{4}{2}$ to choose the numbers and $\frac{4!}{2!2!}=6$ ways to arrange them, so $6\cdot 6=36$ . If all four numbers are the same there are $4$ combinations, so there are $4+36=40$ combinations for this case.

Hence there are $16+0+96+48+40=200$ combinations where the product of the dice is a perfect square, and there are $6^4=1296$ total combinations, so the desired probability is $\frac{200}{1296}=\frac{25}{162}$ , yielding an answer os $25+162=\boxed{187}$ .

-Stormersyle

@@ Line 54: / Line 54: @@
 -dnaidu (silverlizard)
+==Solution 3==
+Note that rolling a 1/4 will not affect whether or not the product is a perfect square. This means that in order for the product to be a perfect square, all non 1/4 numbers rolled must come in pairs, with the only exception being the triplet 2,3, 6. Now we can do casework:
+If there are four 1/4's, then there are <math>2^4=16</math> combinations.
+If there are three 1/4's, then there are 0 combinations, because the fourth number isn't a square.
+If there are two 1/4's, there are <math>2^2=4</math> ways to choose the two 1/4's, 4 ways to choose the remaining pair of numbers, and <math>\frac{4!}{2!2!}=6</math> ways to arrange, so there are <math>4\cdot 4\cdot 6=96</math> combinations for this case.
+If there is one 1/4, then there are 2 ways to choose whether it is a 1 or 4, and the remaining three numbers must be 2, 3, and 6, so there are <math>4!</math> ways to order, meaning there are <math>2\cdot 4!=48</math> combinations for this case.
+Our final case is if there are no 1/4's, in which case we must have two pairs. If the two pairs are of different numbers, then there <math>\binom{4}{2}</math> to choose the numbers and <math>\frac{4!}{2!2!}=6</math> ways to arrange them, so <math>6\cdot 6=36</math>. If all four numbers are the same there are <math>4</math> combinations, so there are <math>4+36=40</math> combinations for this case.
+Hence there are <math>16+0+96+48+40=200</math> combinations where the product of the dice is a perfect square, and there are <math>6^4=1296</math> total combinations, so the desired probability is <math>\frac{200}{1296}=\frac{25}{162}</math>, yielding an answer os <math>25+162=\boxed{187}</math>.
+-Stormersyle
 ==See Also==
 {{AIME box|year=2019|n=II|num-b=3|num-a=5}}
 {{MAA Notice}}

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