2019 AIME II Problems/Problem 4
Contents
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 , where
and
are relatively prime positive integers. Find
.
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 outcomes).
Case 1 (easy): Four 5's are rolled. This has probability 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 , let
equal the number of outcomes after rolling the die
times, with the property that the product is a square. Thus,
as 1 and 4 are the only possibilities.
To find given
(where
), we observe that if the first
rolls multiply to a perfect square, then the last roll must be 1 or 4. This gives
outcomes. Otherwise, the first
rolls do not multiply to a perfect square (
outcomes). In this case, we claim that the last roll is uniquely determined (either 2, 3, or 6). If the product of the first
rolls is
where
and
are not both even, then we observe that if
and
are both odd, then the last roll must be 6; if only
is odd, the last roll must be 2, and if only
is odd, the last roll must be 3. Thus, we have
outcomes in this case, and
.
Computing ,
,
gives
,
, and
. Thus for Case 3, there are 157 outcomes. For case 2, we multiply
by
to distribute the two 5's among four rolls. Thus the probability is
-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:
Probability of rolling 3 duds:
Probability of rolling 2 duds:
Probability of rolling 1 dud:
Probability of rolling 0 duds:
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: (as long as our two non-duds are the same, our product will be square)
Probability of getting a square product given 1 dud: =
(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
ways to roll 3 non-duds).
Probability of getting a square product given 0 duds: =
(We can have any two non-duds twice. For example, 2,2,5,5. There are
ways of choosing which two non-duds to use and
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.
=
-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 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
ways to choose the two 1/4's, 4 ways to choose the remaining pair of numbers, and
ways to arrange, so there are
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
ways to order, meaning there are
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
to choose the numbers and
ways to arrange them, so
. If all four numbers are the same there are
combinations, so there are
combinations for this case.
Hence there are combinations where the product of the dice is a perfect square, and there are
total combinations, so the desired probability is
, yielding an answer of
.
-Stormersyle
Solution 4 (Casework)
Another way to solve this problem is to do casework on all the perfect squares from to
, and how many ways they can be ordered
-
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way.
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or
-
ways.
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-
ways.
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,
, or
-
ways.
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-
ways.
-
,
,
,
-
ways.
- Since there is a prime greater than 6 in its prime factorization there are
ways.
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or
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ways.
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-
way.
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or
-
ways.
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ways for the same reason as
.
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,
,
, or
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ways.
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ways.
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ways.
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ways.
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way.
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ways.
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ways.
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ways.
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ways.
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ways.
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ways.
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ways.
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ways.
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way.
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ways.
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ways.
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ways.
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ways.
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ways.
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ways.
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ways.
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ways.
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ways.
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ways.
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-
way.
There are ways that the dice can land. Summing up the ways, it is easy to see that there are
ways.
This results in a probability of
-superninja2000
Solution 5 (Recursion)
We can do recursion on the number of rolls to find the number of ways we can get rolls to multiply to a square.
After rolls, let us say that the product is
.
Define the following:
the number of ways to have a product after
rolls where
is odd, and
,
are even
the number of ways to have a product after
rolls where
is odd, and
,
are even
the number of ways to have a product after
rolls where
is odd, and
,
are even
the number of ways to have a product after
rolls where
is even, and
,
are odd
the number of ways to have a product after
rolls where
is even, and
,
are odd
the number of ways to have a product after
rolls where
is even, and
,
are odd
the number of ways to have a product after
rolls where
and
are all odd
the number of ways to have a product after
rolls where
and
are all even (square!)
We have the following equations after considering the possible values of the nth roll:
We have the following values after considering the possible values of the 1st roll:
After applying recursion twice, we get:
Finally, we have ,
meaning our answer is
.
~Mathkiddie
Solution 6
Consider all the distinct "fundamental" groups of integers from to
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,
is a fundamental group, while
is not, because it can be broken up into
and
.
and
are already perfect squares, so they each form a "fundamental" group and cannot belong in any other group. Pairs of the other
numbers (
,
, etc. ) form "fundamental" groups as well. The last "fundamental" group is
. It can be easily seen that no more groups exist.
Thus, we have the "fundamental" groups ,
,
,
,
,
, and
.
We now consider the ways to use these groups to form a sequence of numbers whose product is a perfect square. To form a set, we can simply select zero to two groups of size
or
and fill in any remaining spots with
s and
s. We can do this in one of
ways: Using only
s and
s, using one group of size
, using one group of size
, using two different groups of size
, and using the same group of size
twice.
If we only use s and
s, each of the
slots can be filled with one of the
numbers, so there are
possibilities.
If we use one group of size , there are
options for the group to use,
ways to place the two numbers (since they are identical), and
ways to fill in the remaining slots with
s and
s, so there are
possibilities.
If we use one group of size , there is only
option for the group to use,
ways to place the three numbers (since they are distinct), and
ways to fill in the remaining slot, so there are
possibilities.
If we use two different groups of size , there are
options for the groups to use and
ways to place the four numbers (since there are
groups of identical numbers, and one group's placement uniquely determines the other's), so there are
possibilities.
If we use the same group of size twice, there are
options for the group to use and
way to place the four numbers (since they are all identical), so there are
possibilities.
This gives us a total of possibilities, and since there are
total sequences that can be rolled, the probability is equal to
, so the answer is
. ~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
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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