Difference between revisions of "2023 AMC 12B Problems/Problem 23"

Latest revision as of 03:01, 7 October 2024

Problem

When $n$ standard six-sided dice are rolled, the product of the numbers rolled can be any of $936$ possible values. What is $n$ ?

$\textbf{(A)}~11\qquad\textbf{(B)}~6\qquad\textbf{(C)}~8\qquad\textbf{(D)}~10\qquad\textbf{(E)}~9$

Solution 1

We start by trying to prove a function of $n$ , and then we can apply the function and equate it to $936$ to find the value of $n$ .

It is helpful to think of this problem in the format $(1+2+3+4+5+6) \cdot (1+2+3+4+5+6) \dots$ . Note that if we represent the scenario in this manner, we can think of picking a $1$ for one factor and then a $5$ 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 $n=2$ , $4$ can be reached by picking $1$ and $4$ or $2$ and $2$ . However, this form gives us insights that will be useful later in the problem.

Note that there are only $3$ primes in the set $\{1,2,3,4,5,6\}$ : $2,3,$ and $5$ . Thus if we're forming the product of possible values of a dice roll, the product has to be written in the form $2^h \cdot 3^i \cdot 5^j$ (the choice of variables will become clear later), for integer nonnegative values $h,i,j$ . So now the problem boils down to how many distinct triplets $(h,i,j)$ can be formed by taking the product of $n$ dice values.

We start our work on representing $j$ : the powers of $5$ , because it is the simplest in this scenario because there is only one factor of $5$ in the set. Because of this, having $j$ fives in our prime factorization of the product is equivalent to picking $j$ factors from the polynomial $(1+\dots + 6) \cdots$ and choosing each factor to be a $5$ . Now that we've selected $j$ factors, there are $n-j$ factors remaining to choose our powers of $3$ and $2$ .

Suppose our prime factorization of this product contains $i$ powers of $3$ . These powers of $3$ can either come from a $3$ factor or a $6$ factor, but since both $3$ and $6$ contain only one power of $3$ , this means that a product with $i$ powers of $3$ corresponds directly to picking $i$ factors from the polynomial, each of which is either $3$ or $6$ (but this distinction doesn't matter when we consider only the powers of $3$ .

Now we can reframe the problem again. Our method will be as follows: Suppose we choose an arbitrary pair $(i,j)$ that match the requirements, corresponding to the number of $3$ 's and the number of $5$ 's our product will have. Then how many different $h$ values for the powers of $2$ are possible?

In the $i+j$ factors we have already chosen, we obviously can't have any factors of $2$ in the $j$ factors with $5$ . However, we can have a factor of $2$ pairing with factors of $3$ , if we choose a $6$ . The maximal possible power of $2$ in these $i$ factors is thus $2^i$ , which occurs when we pick every factor to be $6$ .

We now have $n-i-j$ factors remaining, and we want to allocate these to solely powers of $2$ . For each of these factors, we can choose either a $1,2,$ or $4$ . Therefore the maximal power of $2$ achieved in these factors is when we pick $4$ for all of them, which is equivalent to $2^{2\cdot (n-i-j)}$ .

Now if we multiply this across the total $n$ factors (or $n$ dice) we have a total of $2^{2n-2i-2j} \cdot 2^i = 2^{2n-i-2j}$ , which is the maximal power of $2$ attainable in the product for a pair $(i,j)$ . Now note that every power of $2$ below this power is attainable: we can simply just take away a power of $2$ from an existing factor by dividing by $2$ . Therefore the powers of $2$ , and thus the $h$ value ranges from $h=0$ to $h=2n-i-2j$ , so there are a total of $2n+1-i-2j$ distinct values for $h$ for a given pair $(i,j)$ .

Now to find the total number of distinct triplets, we must sum this across all possible $i$ s and $j$ s. Lets take note of our restrictions on $i,j$ : the only restriction is that $i+j \leq n$ , since we're picking factors from $n$ dice.

$\sum_{i+j\leq n}^{} 2n+1-i-2j = \sum_{i+j \leq n}^{} 2n+1 - \sum_{i+j \leq n}^{} i+2j$

We start by calculating the first term. $2n+1$ is constant, so we just need to find out how many pairs there are such that $i+j \leq n$ . Set $i$ to $0$ : $j$ can range from $0$ to $n$ , then set $i$ to $1$ : $j$ can range from $0$ to $n-1$ , etc. The total number of pairs is thus $n+1+n+n-1+\dots+1 = \frac{(n+1)(n+2)}{2}$ . Therefore the left summation evaluates to $\frac{(2n+1)(n+1)(n+2)}{2}$

Now we calculate $\sum_{i+j \leq n}^{} i+2j$ . This simplifies to $\sum_{i+j \leq n}^{} i + 2 \cdot \sum_{i+j \leq n}^{} j$ . Note that because $i+j = n$ is symmetric with respect to $i,j$ , the sum of $i$ in all of the pairs will be equal to the sum of $j$ in all of the pairs. Thus this is equal to calculating $3 \cdot \sum_{i+j \leq n}^{} i$ .

In the pairs, $i=1$ appears for $j$ ranging between $0$ and $n-1$ so the sum here is $1 \cdot (n)$ . Similarly $i=2$ appears for $j$ ranging from $0$ to $n-2$ , so the sum is $2 \cdot (n-1)$ . If we continue the pattern, the sum overall is $(n)+2 \cdot (n-1) + 3 \cdot (n-2) + \dots + (n) \cdot 1$ . We can rearrange this as $((n)+(n-1)+ \dots + 1) + ((n-1)+(n-2)+ \dots + 1)+ ((n-2)+(n-3)+ \dots + 1) + \dots + 1)$

$= \frac{(n)(n+1)}{2} + \frac{(n-1)(n)}{2}+ \dots + 1$

We can write this in easier terms as $\sum_{k=0}^{n} \frac{(k)(k+1)}{2} = \frac{1}{2} \cdot \sum_{k=0}^{n} k^2+k$ $=\frac{1}{2} \cdot( \sum_{k=0}^{n} k^2 + \sum_{k=0}^{n} k)$ $= \frac{1}{2} \cdot ( \frac{(n)(n+1)(2n+1)}{6} + \frac{(n)(n+1)}{2})$ $= \frac{1}{2} \cdot ( \frac{(n)(n+1)(2n+1)}{6} + \frac{3n(n+1)}{6}) = \frac{1}{2} \cdot \frac{n(n+1)(2n+4)}{6}$ $= \frac{n(n+1)(n+2)}{6}$

We multiply this by $3$ to obtain that $\sum_{i+j \leq n}^{} i+2j = \frac{n(n+1)(n+2)}{2}$

Thus our final answer for the number of distinct triplets $(h,i,j)$ is: $\sum_{i+j\leq n}^{} 2n+1-i-2j = \frac{(2n+1)(n+1)(n+2)}{2} - \frac{n(n+1)(n+2)}{2}$ $= \frac{(n+1)(n+2)}{2} \cdot (2n+1-n) = \frac{(n+1)(n+2)}{2} \cdot (n+1)$ $= \frac{(n+1)^2(n+2)}{2}$

Now most of the work is done. We set this equal to $936$ and prime factorize. $936 = 12 \cdot 78 = 2^3 \cdot 3^2 \cdot 13$ , so $(n+1)^2(n+2) = 936 \cdot 2 = 2^4 \cdot 3^2 \cdot 13$ . Clearly $13$ cannot be anything squared and $2^4 \cdot 3^2$ is a perfect square, so $n+2 = 13$ and $n = 11 = \boxed{A}$

~KingRavi

Solution 2

The product can be written as $\begin{align*} 2^a 3^b 4^c 5^d 6^e & = 2^{a + 2c + e} 3^{b + e} 5^d . \end{align*}$

Therefore, we need to find the number of ordered tuples $\left( a + 2c + e, b+e, d \right)$ where $a$ , $b$ , $c$ , $d$ , $e$ are non-negative integers satisfying $a+b+c+d+e \leq n$ . We denote this number as $f(n)$ .

Denote by $g \left( k \right)$ the number of ordered tuples $\left( a + 2c + e, b+e \right)$ where $\left( a, b, c, e \right) \in \Delta_k$ with $\Delta_k \triangleq \left\{ (a,b,c,e) \in \Bbb Z_+^4: a+b+c+e \leq k \right\}$ .

Thus, $\begin{align*} f \left( n \right) & = \sum_{d = 0}^n g \left( n - d \right) \\ & = \sum_{k = 0}^n g \left( k \right) . \end{align*}$

Next, we compute $g \left( k \right)$ .

Denote $i = b + e$ . Thus, for each given $i$ , the range of $a + 2c + e$ is from 0 to $2 k - i$ . Thus, the number of $\left( a + 2c + e, b + e \right)$ is $\begin{align*} g \left( k \right) & = \sum_{i=0}^k \left( 2 k - i + 1 \right) \\ & = \frac{1}{2} \left( k + 1 \right) \left( 3 k + 2 \right) . \end{align*}$

Therefore, $\begin{align*} f \left( n \right) & = \sum_{k = 0}^n g \left( k \right) \\ & = \sum_{k=0}^n \frac{1}{2} \left( k + 1 \right) \left( 3 k + 2 \right) \\ & = \frac{3}{2} \sum_{k=0}^n \left( k + 1 \right)^2 - \frac{1}{2} \sum_{k=0}^n \left( k + 1 \right) \\ & = \frac{3}{2} \cdot \frac{1}{6} \left( n+1 \right) \left( n+2 \right) \left( 2n + 3 \right) - \frac{1}{2} \cdot \frac{1}{2} \left( n + 1 \right) \left( n + 2 \right) \\ & = \frac{1}{2} \left( n + 1 \right)^2 \left( n + 2 \right) . \end{align*}$

By solving $f \left( n \right) = 936$ , we get $n = \boxed{\textbf{(A) 11}}$ .

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 3 (Cheese)

The product can be written as $\begin{align*} 2^x 3^y 5^z \end{align*}$

Letting $n=1$ , we get $(x,y,z)=(0,0,0),(0,0,1),(0,1,0),(1,0,0),(1,1,0),(2,0,0)$ , 6 possible values. But if the only restriction of the product if that $2x\le n,y\le n,z\le n$ , we can get $(2+1)(1+1)(1+1)=12$ possible values. We calculate the ratio $r = \frac{\text{possible values of real situation}}{\text{possible values of ideal situation}} = \frac{6}{12}=0.5.$

Letting $n=2$ , we get $(x,y,z)=(0,0,0),(0,0,1),(0,0,2),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1),(1,2,0),$
$(2,0,0),(2,0,1),(2,1,0),(2,2,0),(3,0,0),(3,1,0),(4,0,0)$ , 17 possible values. The number of possibilities in the ideal situation is $5*3*3=45$ , making $r = 17/45 \approx 0.378$ .

Now we can predict the trend of $r$ : as $n$ increases, $r$ decreases. Letting $n=3$ , you get possible values of ideal situation= $7*4*4=112$ . $n=4$ , the number= $9*5*5=225$ .
$n=5$ , the number= $11*6*6=396$ .
$n=6$ , the number= $13*7*7=637,637<936$ so 6 is not the answer.
$n=7$ , the number= $15*8*8=960$ .
$n=8$ , the number= $17*9*9=1377$ ,but $1377*0.378$ ≈ $521$ still much smaller than 936.
$n=9$ , the number= $19*10*10=1900$ ,but $1900*0.378$ ≈ $718$ still smaller than 936.
$n=10$ , the number= $21*11*11=2541$ , $2541*0.378$ ≈ $960$ 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 $\boxed{\textbf{(A) 11}}$ .

Check calculation: $n=11$ ,the number= $23*12*12=3312$ , $3312*0.378$ ≈ $1252$ is much bigger than 936.

~Troublemaker

Solution 4 (Easy computation)

The key observation is if $P=2^a\cdot 3^b \cdot 5^c$ , then given $b$ and $c$ , $a$ can take any value from $0$ to $b+2d$ where $d=n-b-c$ is the number of rolls which is neither divisible by $3$ nor $5$ . We are left to calculate $\sum_{b+c\leq n} (b+2d+1)=\sum_{b+c+d=n} (b+2d+1).$ By symmetry, $\sum_{b+c+d=n} d = \sum_{b+c+d=n} c$ . Therefore, $\sum_{b+c\leq n}(b+2d+1)=\sum_{b+c\leq n}(b+c+d+1)=\sum_{b+c\leq n}(n+1)=(n+1)\binom{n+2}{2}.$

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.

$P=1^{k_1}\cdot 2^{k_2} \cdot 3^{k_3}\cdot 4^{k_4}\cdot 5^{k_5} \cdot 6^{k_6}$ , with $k_1+k_2+...+k_6=n$ .

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 $k_1$ :

case 1( $k_1 <= k_6$ ): 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 $"3456"$ , 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 $C^3_{n+3}$ ways.

If "2"s are no less than remaining "6"s: all "6" cancels with "2", so the product consist of only $"2345"$ . Again, there are $C^3_{n+3}$ ways.

case 2( $k_1>=k_6$ ): 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 $"1235"$ ; there are, again, $C^3_{n+3}$ 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: $"2345"$ , $"3456"$ and $"1235"$ , each with $C^3_{n+3}$ ways. There are overlaps: $3^{k_3} 4^{k_4} 5^{k_5}$ , which has $C^2_{n+2}$ ways, are counted in both $"2345"$ and $"3456"$ , and $2^{k_2} 3^{k_3} 5^{k_5}$ are counted in both $"2345"$ and $"1235"$ .

So, the final number is: $3C^3_{n+3}-2C^2_{n+2}=\frac{(n+1)^2 (n+2)}{2},$ and now any possible product is counted exactly once. And then it is easily to find out $n = \boxed{\textbf{(A) 11}}$ .

~FireSirius

Video Solution 1 by OmegaLearn

https://youtu.be/FZG1j95owTo

Video Solution

https://youtu.be/lHtp83x1Hcw?si=DVuGnTC3hAcdmTsF

~MathProblemSolvingSkills.com

Video Solution

https://youtu.be/BWb1dS4Jba0

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

@@ Line 185: / Line 185: @@
 C^3_{n+3}-2C^2_{n+2}=\frac{(n+1)^2 (n+2)}{2},
 </cmath>
-and now any possible product is counted exactly once. And then it is easily to find out <math>n=11</math>.
+and now any possible product is counted exactly once. And then it is easily to find out <math>n = \boxed{\textbf{(A) 11}}</math>.
 ~FireSirius

2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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Difference between revisions of "2023 AMC 12B Problems/Problem 23"

Latest revision as of 03:01, 7 October 2024

Contents

Problem

Solution 1

Solution 2

Solution 3 (Cheese)

Solution 4 (Easy computation)

Solution 5 ("normalize" the product) by FireSirius

Video Solution 1 by OmegaLearn

Video Solution

Video Solution

See Also