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Difference between revisions of "2023 AMC 12B Problems/Problem 23"
(→‎Solution 5 ("normalize" the product) by FireSirius)
 
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==Solution 1 (In Progress)==
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==Solution 1==
  
 
We start by trying to prove a function of <math>n</math>, and then we can apply the function and equate it to <math>936</math> to find the value of <math>n</math>.
 
We start by trying to prove a function of <math>n</math>, and then we can apply the function and equate it to <math>936</math> to find the value of <math>n</math>.
Line 27: Line 27:
 
Now to find the total number of distinct triplets, we must sum this across all possible <math>i</math>s and <math>j</math>s. Lets take note of our restrictions on <math>i,j</math>: the only restriction is that <math>i+j \leq n</math>, since we're picking factors from <math>n</math> dice.
 
Now to find the total number of distinct triplets, we must sum this across all possible <math>i</math>s and <math>j</math>s. Lets take note of our restrictions on <math>i,j</math>: the only restriction is that <math>i+j \leq n</math>, since we're picking factors from <math>n</math> dice.
  
 +
<cmath> \sum_{i+j\leq n}^{} 2n+1-i-2j = \sum_{i+j \leq n}^{} 2n+1 - \sum_{i+j \leq n}^{} i+2j</cmath>
  
 +
We start by calculating the first term. <math>2n+1</math> is constant, so we just need to find out how many pairs there are such that <math>i+j \leq n</math>. Set <math>i</math> to <math>0</math>: <math>j</math> can range from <math>0</math> to <math>n</math>, then set <math>i</math> to <math>1</math>: <math>j</math> can range from <math>0</math> to <math>n-1</math>, etc. The total number of pairs is thus <math>n+1+n+n-1+\dots+1 = \frac{(n+1)(n+2)}{2}</math>. Therefore the left summation evaluates to <cmath>\frac{(2n+1)(n+1)(n+2)}{2}</cmath>
  
 +
Now we calculate <math>\sum_{i+j \leq n}^{} i+2j</math>. This simplifies to <math>\sum_{i+j \leq n}^{} i + 2 \cdot \sum_{i+j \leq n}^{} j</math>. Note that because <math>i+j = n</math> is symmetric with respect to <math>i,j</math>, the sum of <math>i</math> in all of the pairs will be equal to the sum of <math>j</math> in all of the pairs. Thus this is equal to calculating <math>3 \cdot \sum_{i+j \leq n}^{} i</math>.
  
−
~KingRavi
+
In the pairs, <math>i=1</math> appears for <math>j</math> ranging between <math>0</math> and <math>n-1</math> so the sum here is <math>1 \cdot (n)</math>. Similarly <math>i=2</math> appears for <math>j</math> ranging from <math>0</math> to <math>n-2</math>, so the sum is <math>2 \cdot (n-1)</math>. If we continue the pattern, the sum overall is <math>(n)+2 \cdot (n-1) + 3 \cdot (n-2) + \dots + (n) \cdot 1</math>. We can rearrange this as <math>((n)+(n-1)+ \dots + 1) + ((n-1)+(n-2)+ \dots + 1)+ ((n-2)+(n-3)+ \dots + 1) + \dots + 1)</math>
 +
 
 +
<cmath> = \frac{(n)(n+1)}{2} + \frac{(n-1)(n)}{2}+ \dots + 1</cmath>
 +
 
 +
We can write this in easier terms as <math>\sum_{k=0}^{n} \frac{(k)(k+1)}{2} = \frac{1}{2} \cdot \sum_{k=0}^{n} k^2+k</math>
 +
<cmath>=\frac{1}{2} \cdot( \sum_{k=0}^{n} k^2 + \sum_{k=0}^{n} k)</cmath>
 +
<cmath>= \frac{1}{2} \cdot ( \frac{(n)(n+1)(2n+1)}{6} + \frac{(n)(n+1)}{2})</cmath>
 +
<cmath>= \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}</cmath>
 +
<cmath> = \frac{n(n+1)(n+2)}{6}</cmath>
 +
 
 +
We multiply this by <math>3</math> to obtain that <cmath>\sum_{i+j \leq n}^{} i+2j = \frac{n(n+1)(n+2)}{2}</cmath>
 +
 
 +
Thus our final answer for the number of distinct triplets <math>(h,i,j)</math> is:
 +
<cmath>\sum_{i+j\leq n}^{} 2n+1-i-2j = \frac{(2n+1)(n+1)(n+2)}{2} - \frac{n(n+1)(n+2)}{2}</cmath>
 +
<cmath> = \frac{(n+1)(n+2)}{2} \cdot (2n+1-n) = \frac{(n+1)(n+2)}{2} \cdot (n+1)</cmath>
 +
<cmath> = \frac{(n+1)^2(n+2)}{2}</cmath>
 +
 
 +
Now most of the work is done. We set this equal to <math>936</math> and prime factorize. <math>936 = 12 \cdot 78 = 2^3 \cdot 3^2 \cdot 13</math>, so <math>(n+1)^2(n+2) = 936 \cdot 2 = 2^4 \cdot 3^2 \cdot 13</math>. Clearly <math>13</math> cannot be anything squared and <math>2^4 \cdot 3^2</math> is a perfect square, so <math>n+2 = 13</math> and <math>n = 11 = \boxed{A}</math>
  
  
 +
 +
~KingRavi
  
 
==Solution 2==
 
==Solution 2==
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~Troublemaker
 
~Troublemaker
 +
 +
==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>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>
 +
\sum_{b+c\leq n} (b+2d+1)=\sum_{b+c+d=n} (b+2d+1).
 +
</cmath>
 +
By symmetry, <math> \sum_{b+c+d=n} d = \sum_{b+c+d=n} c</math>. Therefore,
 +
<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>
 +
 +
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.
 +
 +
<math>P=1^{k_1}\cdot 2^{k_2} \cdot 3^{k_3}\cdot 4^{k_4}\cdot 5^{k_5} \cdot 6^{k_6}</math>, with <math>k_1+k_2+...+k_6=n</math>.
 +
 +
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 <math>k_1</math>:
 +
 +
case 1(<math>k_1 <= k_6</math>): 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 <math>"3456"</math>, 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 <math>C^3_{n+3}</math> ways.
 +
 +
If "2"s are no less than remaining "6"s: all "6" cancels with "2", so the product consist of only <math>"2345"</math>. Again, there are <math>C^3_{n+3}</math> ways.
 +
 +
case 2(<math>k_1>=k_6</math>): 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 <math>"1235"</math>; there are, again, <math>C^3_{n+3}</math> 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:  <math>"2345"</math>, <math>"3456"</math> and <math>"1235"</math>, each with <math>C^3_{n+3}</math> ways. There are overlaps: <math>3^{k_3} 4^{k_4} 5^{k_5}</math>, which has <math>C^2_{n+2}</math> ways, are counted in both <math>"2345"</math> and <math>"3456"</math>, and <math>2^{k_2} 3^{k_3} 5^{k_5}</math>  are counted in both <math>"2345"</math> and <math>"1235"</math>.
 +
 +
So, the final number is:
 +
<cmath>
 +
3C^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 = \boxed{\textbf{(A) 11}}</math>.
 +
 +
~FireSirius
  
 
==Video Solution 1 by OmegaLearn==
 
==Video Solution 1 by OmegaLearn==
 
https://youtu.be/FZG1j95owTo
 
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)
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2023|ab=B|num-b=22|num-a=24}}
 
{{AMC12 box|year=2023|ab=B|num-b=22|num-a=24}}
 +
 +
[[Category:Intermediate Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

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)

See Also

2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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All AMC 12 Problems and Solutions

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