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Difference between revisions of "2008 AIME I Problems/Problem 9"
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== Problem ==
 
== Problem ==
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Ten identical crates each of dimensions <math>3</math> ft <math>\times</math> <math>4</math> ft <math>\times</math> <math>6</math> ft.  The first crate is placed flat on the floor.  Each of the remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random.  Let <math>\frac {m}{n}</math> be the probability that the stack of crates is exactly <math>41</math> ft tall, where <math>m</math> and <math>n</math> are relatively prime positive integers.  Find <math>m</math>.
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Ten identical crates each of dimensions <math>3\mathrm{ft}\times 4\mathrm{ft}\times 6\mathrm{ft}</math>.  The first crate is placed flat on the floor.  Each of the remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random.  Let <math>\frac {m}{n}</math> be the probability that the stack of crates is exactly <math>41\mathrm{ft}</math> tall, where <math>m</math> and <math>n</math> are relatively prime positive integers.  Find <math>m</math>.
  
 
== Solution ==
 
== Solution ==
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Thus, our probability is <math>\dfrac{10\cdot9\cdot8\cdot7 + 90}{3^{10}} = \dfrac{10\cdot8\cdot7 + 10}{3^{8}} = \dfrac{570}{3^8} = \dfrac{190}{3^{7}}</math>. Our answer is the numerator, <math>\boxed{190}</math>.
 
Thus, our probability is <math>\dfrac{10\cdot9\cdot8\cdot7 + 90}{3^{10}} = \dfrac{10\cdot8\cdot7 + 10}{3^{8}} = \dfrac{570}{3^8} = \dfrac{190}{3^{7}}</math>. Our answer is the numerator, <math>\boxed{190}</math>.
 +
 
==1 Min Solution==
 
==1 Min Solution==
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It would be helpful for this solution to be reformatted. To start with, let us observe the three numbers. Note that 3 and 6 are both divisible by 3, so the number of 4-crates must be congruent to 41 mod 3, which is also congruent to 2 mod 3. Our solutions for the number of 4-crates will repeat mod 3, so if x is a solution, so is x+3. By inspection, we have that 2 is solution, and so are 5 and 8. Each solution splits into its own case.We must solve the equation 41-4*(z)=6x+3y, simultaneously with x+y=10-z. Note that we already know the possible values of z. Solving these(it's AIME 9, you should be able to do this and if anyone feels like they want to write a rundown of this please go ahead), we get the solution  sets {8,1,1},{5,2,3},and {2,3,5}. We can count the number of possible arrangements for each solution by taking 10 choose z and then multiplying by 10-z choose x(the solution sets, for the sake of consistency, are in the form z,x,y). Summing the results for all the solutions gives us 5130. Finally, to calculate the probability we must determine our denominator. Since we have 3 ways to arrange each block, our denominator is 3^10. 5130/3^10=190/3^7. The answer is m= 190.
+
It would be helpful for this solution to be reformatted. To start with, let us observe the three numbers. Note that <math>3</math> and <math>6</math> are both divisible by <math>3</math>, so the number of <math>4</math>-crates must be congruent to <math>41\bmod{3}</math>, which is also congruent to <math>2\bmod{3}</math>. Our solutions for the number of <math>4</math>-crates will repeat mod <math>3</math>, so if <math>x</math> is a solution, so is <math>x+3</math>. By inspection, we have that <math>2</math> is solution, and so are <math>5</math> and <math>8</math>. Each solution splits into its own case.We must solve the equation <math>41-4z=6x+3y</math>, simultaneously with <math>x+y=10-z</math>. Note that we already know the possible values of <math>z</math>. Solving these (it's AIME <math>9</math>, you should be able to do this and if anyone feels like they want to write a rundown of this please go ahead), we get the solution  sets <math>\{8,1,1\},\{5,2,3\},and \{2,3,5}</math>. We can count the number of possible arrangements for each solution by taking <math>\dbinom{10}{z}</math> and then multiplying by <math>\dbinom{10-z}{x}</math> (the solution sets, for the sake of consistency, are in the form <math>z,x,y</math>). Summing the results for all the solutions gives us <math>5130</math>. Finally, to calculate the probability we must determine our denominator. Since we have <math>3</math> ways to arrange each block, our denominator is <math>3^{10}</math>. <math>\frac{5130}{3^{10}}=\frac{190}{3^7}</math>. The answer is <math>m=\boxed{190}</math>.
  
 
==Solution 2 (a more direct approach)==
 
==Solution 2 (a more direct approach)==
−
Let's make two observations. We are trying to find the number of ways we can add 3,4, and 6 to get 41, and the total number of (non-distinct) sums possible is 3^10.
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Let's make two observations. We are trying to find the number of ways we can add <math>3</math>s,4<math>s, and </math>6<math>s to get </math>41<math>, and the total number of (non-distinct) sums possible is </math>3^{10}<math>.
−
Then we just use casework to easily and directly solve for the number of ways to get 41. To begin, the minimum sum is produced with 10 threes, so WLOG we can solve for the number of ways to get 11 with 0, 1, and 3.  
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Then we just use casework to easily and directly solve for the number of ways to get </math>41<math>. To begin, the minimum sum is produced with </math>10<math> threes, so WLOG we can solve for the number of ways to get </math>11<math> with </math>0<math>s, </math>1<math>s, and </math>3<math>s.  
  
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Case 1: 0 zeroes, 0 threes, 11 ones
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Case I: </math>0<math> zeroes, </math>0<math> threes, </math>11<math> ones
 
Impossible, because there are only ten available spots.
 
Impossible, because there are only ten available spots.
  
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Case 2: 1 zero, 1 three, 8 ones
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Case II: </math>1<math> zero, </math>1<math> three, </math>8<math> ones
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This is just 10! over 8!, so there are 90 possibilities.
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This is just </math>\frac{10!}{8!}<math>, so there are </math>90<math> possibilities.
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Case 3: 3 zeroes, 2 threes, 5 ones
 
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This is just 10! over 3!, 2!, and 5! This gives 2520 possibilities.
 
  
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Case 4: 5 zeroes, 3 threes, and 2 ones.
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Case III: </math>3<math> zeroes, </math>2<math> threes, </math>5<math> ones
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This is the same as case 3, so also 2520 possibilities.
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This is just </math>\frac{10!}{3!2!5!}<math>. This gives </math>2520<math> possibilities.
  
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<math>90+2520+2520=5130</math>
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Case IV: 5 zeroes, 3 threes, and 2 ones.
 +
This is the same as case </math>3<math>, so also </math>2520<math> possibilities.
  
−
5130 has three powers of 3, so 5130 divided by 27 is <math>\boxed{190}</math>.
+
</math>90+2520+2520=5130<math>
  
 +
</math>5130<math> has three powers of </math>3<math>, so </math>5130<math> divided by </math>27<math> is </math>\boxed{190}$.
  
 
-jackshi2006
 
-jackshi2006

Revision as of 14:15, 17 December 2020

Problem

Ten identical crates each of dimensions $3\mathrm{ft}\times 4\mathrm{ft}\times 6\mathrm{ft}$. The first crate is placed flat on the floor. Each of the remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random. Let $\frac {m}{n}$ be the probability that the stack of crates is exactly $41\mathrm{ft}$ tall, where $m$ and $n$ are relatively prime positive integers. Find $m$.

Solution

Only the heights matter, and each crate is either 3, 4, or 6 feet tall with equal probability. We have the following:

\begin{align*}3a + 4b + 6c &= 41\\ a + b + c &= 10\end{align*}

Subtracting 3 times the second from the first gives $b + 3c = 11$, or $(b,c) = (2,3),(5,2),(8,1),(11,0)$. The last doesn't work, obviously. This gives the three solutions $(a,b,c) = (5,2,3),(3,5,2),(1,8,1)$. In terms of choosing which goes where, the first two solutions are analogous.

For $(5,2,3),(3,5,2)$, we see that there are $2\cdot\dfrac{10!}{5!2!3!} = 10\cdot9\cdot8\cdot7$ ways to stack the crates. For $(1,8,1)$, there are $2\dbinom{10}{2} = 90$. Also, there are $3^{10}$ total ways to stack the crates to any height.

Thus, our probability is $\dfrac{10\cdot9\cdot8\cdot7 + 90}{3^{10}} = \dfrac{10\cdot8\cdot7 + 10}{3^{8}} = \dfrac{570}{3^8} = \dfrac{190}{3^{7}}$. Our answer is the numerator, $\boxed{190}$.

1 Min Solution

It would be helpful for this solution to be reformatted. To start with, let us observe the three numbers. Note that $3$ and $6$ are both divisible by $3$, so the number of $4$-crates must be congruent to $41\bmod{3}$, which is also congruent to $2\bmod{3}$. Our solutions for the number of $4$-crates will repeat mod $3$, so if $x$ is a solution, so is $x+3$. By inspection, we have that $2$ is solution, and so are $5$ and $8$. Each solution splits into its own case.We must solve the equation $41-4z=6x+3y$, simultaneously with $x+y=10-z$. Note that we already know the possible values of $z$. Solving these (it's AIME $9$, you should be able to do this and if anyone feels like they want to write a rundown of this please go ahead), we get the solution sets $\{8,1,1\},\{5,2,3\},and \{2,3,5}$ (Error compiling LaTeX. Unknown error_msg). We can count the number of possible arrangements for each solution by taking $\dbinom{10}{z}$ and then multiplying by $\dbinom{10-z}{x}$ (the solution sets, for the sake of consistency, are in the form $z,x,y$). Summing the results for all the solutions gives us $5130$. Finally, to calculate the probability we must determine our denominator. Since we have $3$ ways to arrange each block, our denominator is $3^{10}$. $\frac{5130}{3^{10}}=\frac{190}{3^7}$. The answer is $m=\boxed{190}$.

Solution 2 (a more direct approach)

Let's make two observations. We are trying to find the number of ways we can add $3$s,4$s, and$6$s to get$41$, and the total number of (non-distinct) sums possible is$3^{10}$. Then we just use casework to easily and directly solve for the number of ways to get$41$. To begin, the minimum sum is produced with$10$threes, so WLOG we can solve for the number of ways to get$11$with$0$s,$1$s, and$3$s.

Case I:$ (Error compiling LaTeX. Unknown error_msg)0$zeroes,$0$threes,$11$ones Impossible, because there are only ten available spots.

Case II:$ (Error compiling LaTeX. Unknown error_msg)1$zero,$1$three,$8$ones This is just$\frac{10!}{8!}$, so there are$90$possibilities.

Case III:$ (Error compiling LaTeX. Unknown error_msg)3$zeroes,$2$threes,$5$ones This is just$\frac{10!}{3!2!5!}$. This gives$2520$possibilities.

Case IV: 5 zeroes, 3 threes, and 2 ones. This is the same as case$ (Error compiling LaTeX. Unknown error_msg)3$, so also$2520$possibilities.$90+2520+2520=5130$$ (Error compiling LaTeX. Unknown error_msg)5130$has three powers of$3$, so$5130$divided by$27$is$\boxed{190}$.

-jackshi2006

Video Solution (Very fast)

https://www.youtube.com/watch?v=qeSY_ISSX6M&t=33s

~fidgetboss_4000

See also

2008 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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