2023 AMC 12B Problems/Problem 23
Contents
Problem
When standard six-sided dice are rolled, the product of the numbers rolled can be any of possible values. What is ?
Solution 1
We start by trying to prove a function of , and then we can apply the function and equate it to to find the value of .
It is helpful to think of this problem in the format . Note that if we represent the scenario in this manner, we can think of picking a for one factor and then a for another factor to form their product - this is similar thinking to when we have the factorized form of a polynomial. Unfortunately this is not quite accurate to the problem because we can reach the same product in many ways: for example for , can be reached by picking and or and . However, this form gives us insights that will be useful later in the problem.
Note that there are only primes in the set : and . Thus if we're forming the product of possible values of a dice roll, the product has to be written in the form (the choice of variables will become clear later), for integer nonnegative values . So now the problem boils down to how many distinct triplets can be formed by taking the product of dice values.
We start our work on representing : the powers of , because it is the simplest in this scenario because there is only one factor of in the set. Because of this, having fives in our prime factorization of the product is equivalent to picking factors from the polynomial and choosing each factor to be a . Now that we've selected factors, there are factors remaining to choose our powers of and .
Suppose our prime factorization of this product contains powers of . These powers of can either come from a factor or a factor, but since both and contain only one power of , this means that a product with powers of corresponds directly to picking factors from the polynomial, each of which is either or (but this distinction doesn't matter when we consider only the powers of .
Now we can reframe the problem again. Our method will be as follows: Suppose we choose an arbitrary pair that match the requirements, corresponding to the number of 's and the number of 's our product will have. Then how many different values for the powers of are possible?
In the factors we have already chosen, we obviously can't have any factors of in the factors with . However, we can have a factor of pairing with factors of , if we choose a . The maximal possible power of in these factors is thus , which occurs when we pick every factor to be .
We now have factors remaining, and we want to allocate these to solely powers of . For each of these factors, we can choose either a or . Therefore the maximal power of achieved in these factors is when we pick for all of them, which is equivalent to .
Now if we multiply this across the total factors (or dice) we have a total of , which is the maximal power of attainable in the product for a pair . Now note that every power of below this power is attainable: we can simply just take away a power of from an existing factor by dividing by . Therefore the powers of , and thus the value ranges from to , so there are a total of distinct values for for a given pair .
Now to find the total number of distinct triplets, we must sum this across all possible s and s. Lets take note of our restrictions on : the only restriction is that , since we're picking factors from dice.
We start by calculating the first term. is constant, so we just need to find out how many pairs there are such that . Set to : can range from to , then set to : can range from to , etc. The total number of pairs is thus . Therefore the left summation evaluates to
Now we calculate . This simplifies to . Note that because is symmetric with respect to , the sum of in all of the pairs will be equal to the sum of in all of the pairs. Thus this is equal to calculating .
In the pairs, appears for ranging between and so the sum here is . Similarly appears for ranging from to , so the sum is . If we continue the pattern, the sum overall is . We can rearrange this as
We can write this in easier terms as
We multiply this by to obtain that
Thus our final answer for the number of distinct triplets is:
Now most of the work is done. We set this equal to and prime factorize. , so . Clearly cannot be anything squared and is a perfect square, so and
~KingRavi
Solution 2
The product can be written as
Therefore, we need to find the number of ordered tuples where , , , , are non-negative integers satisfying . We denote this number as .
Denote by the number of ordered tuples where with .
Thus,
Next, we compute .
Denote . Thus, for each given , the range of is from 0 to . Thus, the number of is
Therefore,
By solving , we get .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 3 (Cheese)
The product can be written as
Letting , we get , 6 possible values. But if the only restriction of the product if that , we can get possible values. We calculate the ratio
Letting , we get
, 17 possible values.
The number of possibilities in the ideal situation is , making .
Now we can predict the trend of : as increases, decreases.
Letting
, you get possible values of ideal situation=.
, the number=.
, the number=.
, the number= so 6 is not the answer.
, the number=.
, the number=,but ≈ still much smaller than 936.
, the number=,but ≈ still smaller than 936.
, the number=, ≈ is a little bigger 936, but the quotient of (possible values of real situation)/(possible values of ideal situation) is much smaller than 0.378 now, so 10 is probably not the answer,so the answer is .
Check calculation: ,the number=,≈ is much bigger than 936.
~Troublemaker
Solution 4 (Easy computation)
The key observation is if , then given and , can take any value from to where is the number of rolls which is neither divisible by nor . We are left to calculate By symmetry, . Therefore,
The rest is the same as above.
~asops
Solution 5 ("normalize" the product) by FireSirius
The difficulty is that a product can be formed by different combinations of 1,2,...,n. So the key thought of this solution is to set a "normalized" combination for each product.
, with .
For a fixed number of dice, it is easy to see that there are three basic "transformation": 1*6 = 2*3, 1*4 = 2*2, and 2*6 = 3*4, which does not change the number of dice. Other transformation like 1*6*6=3*3*4 can be viewed as a 1*6=2*3 combined with a 2*6=3*4.
So we can transform any given combination into a "normallised" combination in this way: step 1, if there are 1 and 6, then transorm them into 2*3; step 2, if there are both 1 and 4, then transform them into a 2*2; step3: if there are both 2 and 6, then transform them into 3*4. Until no more basic transformation can be made.
Now begin casework with :
case 1(): all "1"s cancel with "6"s, so the product contains only 2,3,4,5,6. And now:
If "2"s are no more than remaining "6"s: all "2"s cancel with "6"s, so the "nomarlized" product consists of only , with the total number of dice unchanged. No transformation can be made now and it is the same with n stars and 3 bars, so there are ways.
If "2"s are no less than remaining "6"s: all "6" cancels with "2", so the product consist of only . Again, there are ways.
case 2(): all "6" cancels with "1", the remaining numbers are: 1,2,3,4,5; now if "1"s are no less than "4"s, then it can be transformed into product with only ; there are, again, ways. if "1"s are no more than "4", then it is product with only 2,3,4,5, which is already counted before.
So there are three major cases: , and , each with ways. There are overlaps: , which has ways, are counted in both and , and are counted in both and .
So, the final number is: and now any possible product is counted exactly once. And then it is easily to find out .
~FireSirius
Video Solution 1 by OmegaLearn
Video Solution
https://youtu.be/lHtp83x1Hcw?si=DVuGnTC3hAcdmTsF
~MathProblemSolvingSkills.com
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2023 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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