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

Latest revision as of 02:24, 14 August 2024

Problem

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 1

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 $a_2$ 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. Let's call rolling 1 or 4 rolling a dud (a perfect square).

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

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

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

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

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

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

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

-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 of $25+162=\boxed{187}$ .

-Stormersyle

Solution 4 (Casework)

Another way to solve this problem is to do casework on all the perfect squares from $1^2$ to $36^2$ , and how many ways they can be ordered $1^2$ - $1,1,1,1$ - $1$ way. $2^2$ - $4,1,1,1$ or $2,2,1,1$ - $\binom{4}{2}+4=10$ ways. $3^2$ - $3,3,1,1$ - $\binom{4}{2}=6$ ways. $4^2$ - $4,4,1,1$ , $2,2,2,2$ , or $2,2,4,1$ - $\binom{4}{2}+1+12=19$ ways. $5^2$ - $5,5,1,1$ - $\binom{4}{2}=6$ ways. $6^2$ - $6,6,1,1$ , $1,2,3,6$ , $2,3,2,3$ , $3,3,4,1$ - $2*\binom{4}{2}+4!+12=48$ ways. $7^2$ - Since there is a prime greater than 6 in its prime factorization there are $0$ ways. $8^2$ - $4,4,4,1$ or $2,4,2,4$ - $\binom{4}{2}+4=10$ ways. $9^2$ - $3,3,3,3$ - $1$ way. $10^2$ - $2,2,5,5$ or $1,4,5,5$ - $6+12=18$ ways. $11^2$ - $0$ ways for the same reason as $7^2$ . $12^2$ - $6,6,2,2$ , $4,4,3,3$ , $2,3,4,6$ , or $1,4,6,6$ - $2*\binom{4}{2}+4!+12=48$ ways. $13^2$ - $0$ ways. $14^2$ - $0$ ways. $15^2$ - $3,3,5,5$ - $\binom{4}{2}=6$ ways. $16^2$ - $4,4,4,4$ - $1$ way. $17^2$ - $0$ ways. $18^2$ - $3,3,6,6$ - $\binom{4}{2}=6$ ways. $19^2$ - $0$ ways. $20^2$ - $4,4,5,5$ - $\binom{4}{2}=6$ ways. $21^2$ - $0$ ways. $22^2$ - $0$ ways. $23^2$ - $0$ ways. $24^2$ - $4,4,6,6$ - $\binom{4}{2}=6$ ways. $25^2$ - $5,5,5,5$ - $1$ way. $26^2$ - $0$ ways. $27^2$ - $0$ ways. $28^2$ - $0$ ways. $29^2$ - $0$ ways. $30^2$ - $5,5,6,6$ - $\binom{4}{2}$ ways. $31^2$ - $0$ ways. $32^2$ - $0$ ways. $33^2$ - $0$ ways. $34^2$ - $0$ ways. $35^2$ - $0$ ways. $36^2$ - $6,6,6,6$ - $1$ way.

There are $6^4=1296$ ways that the dice can land. Summing up the ways, it is easy to see that there are $200$ ways. This results in a probability of $\frac{200}{1296}=\frac{25}{162}\implies\boxed{187}$ -superninja2000

Solution 5 (Recursion)

We can do recursion on the number of rolls to find the number of ways we can get $4$ rolls to multiply to a square.

After $n$ rolls, let us say that the product is $p = 2^a3^b5^c$ .

Define the following:

$A_{n} =$ the number of ways to have a product after $n$ rolls where $a$ is odd, and $b$ , $c$ are even

$B_{n} =$ the number of ways to have a product after $n$ rolls where $b$ is odd, and $a$ , $c$ are even

$C_{n} =$ the number of ways to have a product after $n$ rolls where $c$ is odd, and $a$ , $b$ are even

$D_{n} =$ the number of ways to have a product after $n$ rolls where $c$ is even, and $a$ , $b$ are odd

$E_{n} =$ the number of ways to have a product after $n$ rolls where $b$ is even, and $a$ , $c$ are odd

$F_{n} =$ the number of ways to have a product after $n$ rolls where $a$ is even, and $b$ , $c$ are odd

$G_{n} =$ the number of ways to have a product after $n$ rolls where $a, b,$ and $c$ are all odd

$S_{n} =$ the number of ways to have a product after $n$ rolls where $a, b,$ and $c$ are all even (square!)

We have the following equations after considering the possible values of the nth roll:

$A_{n} = S_{n-1}+B_{n-1}+D_{n-1}+E_{n-1}+2A_{n-1}$

$B_{n} = A_{n-1}+D_{n-1}+F_{n-1}+S_{n-1}+2B_{n-1}$

$C_{n} = S_{n-1}+E_{n-1}+F_{n-1}+G_{n-1}+2C_{n-1}$

$D_{n} = S_{n-1}+A_{n-1}+B_{n-1}+G_{n-1}+2D_{n-1}$

$E_{n} = A_{n-1}+C_{n-1}+F_{n-1}+G_{n-1}+2E_{n-1}$

$F_{n} = B_{n-1}+E_{n-1}+C_{n-1}+G_{n-1}+2F_{n-1}$

$G_{n} = C_{n-1}+D_{n-1}+F_{n-1}+E_{n-1}+2G_{n-1}$

$S_{n} = A_{n-1}+C_{n-1}+B_{n-1}+D_{n-1}+2S_{n-1}$

We have the following values after considering the possible values of the 1st roll:

$A_1 = B_1 = C_1 = D_1 = 1; E_1 = F_1 = G_1 = 0; S_1 = 2$

After applying recursion twice, we get:

$A_2 = B_2 = D_2 = 6, C_2 = 4, E_2 = F_2 = G_2 = 2, S_2 = 8$

$A_3 = B_3 = D_3 = 34, C_3 = 22, E_3 = F_3 = G_3 = 18, S_3 = 38$

Finally, we have $S_4 = 200$ , $\frac{m}{n} = \frac{200}{1296} = \frac{25}{162}$ meaning our answer is $\boxed{187}$ .

Solution 6

Consider all the distinct "fundamental" groups of integers from $1$ to $6$ whose product is a perfect square. A "fundamental" group is one that cannot be broken into two smaller groups that each have a perfect square product. For example, $\{2,2\}$ is a fundamental group, while $\{3,3,4\}$ is not, because it can be broken up into $\{3,3\}$ and $\{4\}$ .

$1$ and $4$ are already perfect squares, so they each form a "fundamental" group and cannot belong in any other group. Pairs of the other $4$ numbers ( $\{2,2\}$ , $\{3,3\}$ , etc. ) form "fundamental" groups as well. The last "fundamental" group is $\{2,3,6\}$ . It can be easily seen that no more groups exist.

Thus, we have the "fundamental" groups $\{1\}$ , $\{4\}$ , $\{2,2\}$ , $\{3,3\}$ , $\{5,5\}$ , $\{6,6\}$ , and $\{2,3,6\}$ .

We now consider the ways to use these groups to form a sequence of $4$ numbers whose product is a perfect square. To form a set, we can simply select zero to two groups of size $2$ or $3$ and fill in any remaining spots with $1$ s and $4$ s. We can do this in one of $5$ ways: Using only $1$ s and $4$ s, using one group of size $2$ , using one group of size $3$ , using two different groups of size $2$ , and using the same group of size $2$ twice.

If we only use $1$ s and $4$ s, each of the $4$ slots can be filled with one of the $2$ numbers, so there are $2^4=16$ possibilities.

If we use one group of size $2$ , there are $4$ options for the group to use, $\binom{4}{2}$ ways to place the two numbers (since they are identical), and $2^2$ ways to fill in the remaining slots with $1$ s and $4$ s, so there are $4\cdot\binom{4}{2}\cdot2^2=96$ possibilities.

If we use one group of size $3$ , there is only $1$ option for the group to use, $4\cdot3\cdot2$ ways to place the three numbers (since they are distinct), and $2$ ways to fill in the remaining slot, so there are $4\cdot3\cdot2\cdot2=48$ possibilities.

If we use two different groups of size $2$ , there are $\binom{4}{2}$ options for the groups to use and $\binom{4}{2}$ ways to place the four numbers (since there are $2$ groups of identical numbers, and one group's placement uniquely determines the other's), so there are $\binom{4}{2}\cdot\binom{4}{2}=36$ possibilities.

If we use the same group of size $2$ twice, there are $4$ options for the group to use and $1$ way to place the four numbers (since they are all identical), so there are $4=4$ possibilities.

This gives us a total of $16+96+48+36+4=200$ possibilities, and since there are $6^4=1296$ total sequences that can be rolled, the probability is equal to $\frac{200}{1296}=\frac{25}{162}$ , so the answer is $25+162=\boxed{187}$ . ~emerald_block

Solution 7(Generating functions)

Let's look at the prime factorization of some of these rolls:

 01 => 2^0*3^0*5^0
 02 => 2^1*3^0*5^0
 03 => 2^0*3^1*5^0
 04 => 2^2*3^0*5^0
 05 => 2^0*3^0*5^1
 06 => 2^1*3^1*5^0

Now, using multi-variable generating functions, we get:

 f(x,y,z)=1+x+y+x^2+z+xy
            | |
           \| |/
            'v'
 f(x,y,z)=1+x+y+z+x^2+xy
   =(for our purposes)
       =2+x+y+z+xy

Let's square that!

 4+2x+2y+2z+2xy+2x+x^2+xy+xz+x^2y+2y+xy+y^2+yz+xy^2+2z+xz+yz+z^2+xyz+2xy+x^2y+xy^2+xyz+x^2y^2
 Combining like terms . . .
 4+4x+4y+4z+x^2+y^2+z^2+6xy+2xz+2yz+2xyz+2x^2y+2xy^2+x^2y^2

Since we only want the parity of each of the exponents, we can collapse it again.

 8+6x+6y+4z+6xy+2xz+2yz+2xyz

Last simplification: factor out a factor of two and save it for later.

 4+3x+3y+2z+3xy+xz+yz+xyz

Let's take a better approach this time.

 ___ term :4*4+3*3+3*3+2*2+3*3+1*1+1*1+1*1=16+09+09+04+09+01+01+01=50
 __x term :4*3+3*4+3*3+2*1+3*3+1*2+1*1+1*1=12+12+09+02+09+02+01+01=48
 __y term :4*3+3*3+3*4+2*1+3*3+1*1+1*2+1*1=12+09+12+02+09+01+02+01=48
 __z term :4*2+3*1+3*1+2*4+3*1+1*3+1*3+1*3=08+03+03+08+03+03+03+03=34
 _xy term :4*3+3*3+3*3+2*1+3*4+1*1+1*1+1*2=12+09+09+02+12+01+01+02=48
 _xz term :4*1+3*2+3*1+2*3+3*1+1*4+1*3+1*3=04+06+03+06+03+04+03+03=32
 _yz term :4*1+3*1+3*2+2*3+3*1+1*3+1*4+1*3=04+03+06+06+03+03+04+03=32
 xyz term :4*1+3*1+3*1+2*3+3*2+1*3+1*3+1+4=04+03+03+06+06+03+03+04=32
 
 Result: 50+48x+48y+34z+48xy+32xz+32yz+32xyz

I know we could use vectors and dot products to make it look neater but come on. It already looks neat enough. Also, we didn't need the other parts, but it just looks nicer. Now let's stick back the two that turned into a four.

 200+192x+192y+136z+192xy+128xz+128yz+128xyz

We seek the constant term which is 200. 200/1296=100/648=50/324=25/162, 25+162=187

~ Afly (talk)

Solution 8

There are a total of $6^4=1296$ possible die rolls.

We use casework:

Case 1: All 4 numbers are the same. There are obviously $6$ ways.

Case 2: Sets of 2 different numbers.

A set of two different numbers is basically $(x,x,y,y)$ . There are a total of $\frac{4!}{2!\cdot 2!}=6$ ways to arrange the numbers.

By listing these cases, we quickly see a pattern:

$(1,1,2,2)$

$(1,1,3,3)$

$...$

$(1,1,6,6)$

$(2,2,3,3)$

$...$

$(2,2,6,6)$

$...$

$(5,5,6,6)$

There are a total of $5+4+3+2+1=15$ cases. Multiplying this by $6$ yields $15\cdot 6=90$ ways.

Case 3: Sets of numbers in the form of $(x,x,1,4)$

A special case must be made for the number $4$ because $4$ itself is a perfect square.

$(1,1,1,4)$ $4$

$(2,2,1,4)$ $12$

$(3,3,1,4)$ $12$

$(4,4,1,4)$ $4$

$(5,5,1,4)$ $12$

$(6,6,1,4)$ $12$

Summing these up yields a total of $4+12+12+4+12+12=56$ ways.

Case 4: Sets with all 4 numbers different

Note that the sets

$(1,2,3,6)$

$(2,3,4,6)$

Multiply to perfect square. The total of these cases are $24+24=48$ .

Adding all these cases together yields $6+90+56+48=200$ ways that the product of the values of the die can be a perfect square.

Therefore the probability is

$\frac{200}{1296}=\frac{25}{162}$

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

-elchen3_f

Solution 9 (Clean Generating Functions and Roots of Unity Filter)

Recall that a number is a perfect square iff every prime in its prime factorization has even multiplicity. We can use multi-variable generating functions to track the contribution of each prime (namely 2, 3, 5) in one roll. Let $f(a,b,c)$ be the generating function such that the coefficient of each term $a^xb^yc^z$ is the number of ways to get a product of $2^x3^y5^z$ when rolling 4 dice. For each roll, we can either roll a 1 (basically contribute nothing), roll a 2 (contribute a factor of 2), roll a 3 (contribute a factor of 3), roll a 4 (contribute two factors of 2), roll a 5 (contribute a factor of 5), or roll a 6 (contribute a factor of 2 and a factor of 3), so the generating function for one roll is $1+a+b+a^2+c+ab$ . Hence, $f(a,b,c)= (1+a+b+a^2+c+ab)^4$ . We seek the sum of coefficients of the $a^{2p}b^{2q}c^{2r}$ terms $(p, q, r, \in \mathbb{Z}*)$ . $\newline$

To filter out the desired coefficients, apply Roots of Unity filter 3 times. Filter out the coefficients with even $a$ multiplicity using $\frac{f(1, b, c)+f(-1, b, c)}{2}=\frac{(3+2b+c)^4+(c+1)^4}{2}=g(b,c).$ Similarly, now filter out the coefficients with even $b$ multiplicity with $\frac{g(1, c)+g(-1, c)}{2}=\frac{(c+5)^4+3(c+1)^4}{4}=h(c).$ Apply this one more time to get $\frac{h(1)+h(-1)}{2}=\frac{6^4+3(2)^4+4^4}{8}.$ The requested probability is $\frac{6^4+3(2)^4+4^4}{8\cdot 6^4}=\frac{25}{162} \implies \boxed{187}.$

~bomberdoodles

@@ Line 162: / Line 162: @@
 Finally, we have <math>S_4 = 200</math>, <math>\frac{m}{n} = \frac{200}{1296} = \frac{25}{162}</math>meaning our answer is <math>\boxed{187}</math>.
-~Mathkiddie
 ==Solution 6==
@@ Line 245: / Line 243: @@
 ==Solution 8==
-There are a total of <math>2^6=1296</math> possible die rolls.
+There are a total of <math>6^4=1296</math> possible die rolls.
@@ Line 332: / Line 330: @@
 -elchen3_f
+==Solution 9 (Clean Generating Functions and Roots of Unity Filter)==
+Recall that a number is a perfect square iff every prime in its prime factorization has even multiplicity. We can use multi-variable generating functions to track the contribution of each prime (namely 2, 3, 5) in one roll. Let <math>f(a,b,c)</math> be the generating function such that the coefficient of each term <math>a^xb^yc^z</math> is the number of ways to get a product of <math>2^x3^y5^z</math> when rolling 4 dice.  For each roll, we can either roll a 1 (basically contribute nothing), roll a 2 (contribute a factor of 2), roll a 3 (contribute a factor of 3), roll a 4 (contribute two factors of 2), roll a 5 (contribute a factor of 5), or roll a 6 (contribute a factor of 2 and a factor of 3), so the generating function for one roll is <math>1+a+b+a^2+c+ab</math>. Hence, <math>f(a,b,c)= (1+a+b+a^2+c+ab)^4</math>. We seek the sum of coefficients of the <math>a^{2p}b^{2q}c^{2r}</math> terms <math>(p, q, r, \in \mathbb{Z}*)</math>. <math>\newline</math>
+To filter out the desired coefficients, apply Roots of Unity filter 3 times. Filter out the coefficients with even <math>a</math> multiplicity using <cmath>\frac{f(1, b, c)+f(-1, b, c)}{2}=\frac{(3+2b+c)^4+(c+1)^4}{2}=g(b,c).</cmath> Similarly, now filter out the coefficients with even <math>b</math> multiplicity with <cmath>\frac{g(1, c)+g(-1, c)}{2}=\frac{(c+5)^4+3(c+1)^4}{4}=h(c).</cmath> Apply this one more time to get <cmath>\frac{h(1)+h(-1)}{2}=\frac{6^4+3(2)^4+4^4}{8}.</cmath> The requested probability is <math>\frac{6^4+3(2)^4+4^4}{8\cdot 6^4}=\frac{25}{162} \implies \boxed{187}.</math>
+~bomberdoodles
 ==See Also==

2019 AIME II (Problems • Answer Key • Resources)
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Difference between revisions of "2019 AIME II Problems/Problem 4"

Latest revision as of 02:24, 14 August 2024

Contents

Problem

Solution 1

Solution 2

Solution 3

Solution 4 (Casework)

Solution 5 (Recursion)

Solution 6

Solution 7(Generating functions)

Solution 8

Solution 9 (Clean Generating Functions and Roots of Unity Filter)

See Also