2023 AMC 12B Problems/Problem 23

Problem

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

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

Solution1

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)

Solution2(Informal)

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

Let $n=1$ , 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≤n,y≤n,z≤n,we can get (2+1)(1+1)(1+1)=12 possible values,(possible values of real situation)/(possible values of ideal situation)= $6/12=0.5$ .

Let $n=2$ ,we get $(x,y,z)=<br /> (0,0,0),(0,0,1),(0,0,2),(0,1,0),(0,1,1),<br /> (1,0,0),(1,0,1),(1,1,0),(1,1,1),(1,2,0),<br /> (2,0,0),(2,0,1),(2,1,0),(2,2,0),<br /> (3,0,0),(3,1,0),(4,0,0)$ 17 possible values. possible values of ideal situation= $5*3*3=45$ , $17/45$ ≈ $0.378$ .

Now we can Predict the trend of the product if n becoming bigger,the quotient of (possible values of real situation)/(possible values of ideal situation) will be smaller and smaller. Let $n=3$ ,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

Video Solution 1 by OmegaLearn

https://youtu.be/FZG1j95owTo

2023 AMC 12B (Problems • Answer Key • Resources)
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2023 AMC 12B Problems/Problem 23

Contents

Problem

Solution1

Solution2(Informal)

Video Solution 1 by OmegaLearn

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