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

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Let <math>n=1</math>, we get <math>(x,y,z)=(0,0,0),(0,0,1),(0,1,0),(1,0,0),(1,1,0),(2,0,0)</math> 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)=<math>6/12=0.5</math>.
 
Let <math>n=1</math>, we get <math>(x,y,z)=(0,0,0),(0,0,1),(0,1,0),(1,0,0),(1,1,0),(2,0,0)</math> 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)=<math>6/12=0.5</math>.
  
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Let <math>n=2</math>,we get
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Let <math>n=2</math>,we get <math>(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),</math><br />
−
<math>(x,y,z)=<br />
+
<math>(2,0,0),(2,0,1),(2,1,0),(2,2,0),(3,0,0),(3,1,0),(4,0,0)</math> 17 possible values.
−
(0,0,0),(0,0,1),(0,0,2),(0,1,0),(0,1,1),<br />
 
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(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)</math> 17 possible values.
 
 
possible values of ideal situation=<math>5*3*3=45</math>,<math>17/45</math>≈<math>0.378</math>.
 
possible values of ideal situation=<math>5*3*3=45</math>,<math>17/45</math>≈<math>0.378</math>.
  

Revision as of 08:20, 16 November 2023

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 $(x,y,z)=(0,0,0),(0,0,1),(0,0,2),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1),(1,2,0),$
$(2,0,0),(2,0,1),(2,1,0),(2,2,0),(3,0,0),(3,1,0),(4,0,0)$ 17 possible values. 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


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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