Difference between revisions of "2023 AMC 12B Problems/Problem 23"
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==Solution 4 (Easy computation)== | ==Solution 4 (Easy computation)== | ||
− | The key observation is if <math>P=2^a\cdot 3^b \cdot 5^c</math>, then given <math>b</math> and <math>c</math>, <math>a</math> can take any value from <math>0</math> to <math> | + | The key observation is if <math>P=2^a\cdot 3^b \cdot 5^c</math>, then given <math>b</math> and <math>c</math>, <math>a</math> can take any value from <math>0</math> to <math>b+2d</math> where <math>d=n-b-c</math> is the number of rolls which is neither divisible by <math>3</math> nor <math>5</math>. We are left to calculate |
<cmath> | <cmath> | ||
− | \sum_{b+c\leq n} ( | + | \sum_{b+c\leq n} (b+2d+1)=\sum_{b+c+d=n} (b+2d+1). |
</cmath> | </cmath> | ||
− | + | By symmetry, <math> \sum_{b+c+d=n} d = \sum_{b+c+d=n} c</math>. Therefore, | |
− | |||
<cmath> | <cmath> | ||
− | + | \sum_{b+c\leq n}(b+2d+1)=\sum_{b+c\leq n}(b+c+d+1)=\sum_{b+c\leq n}(n+1)=(n+1)\binom{n+2}{2}. | |
− | |||
− | |||
− | |||
− | |||
− | |||
</cmath> | </cmath> | ||
− | + | The rest is the same as above. | |
~asops | ~asops |
Revision as of 22:16, 23 November 2023
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
Video Solution 1 by OmegaLearn
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 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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