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Difference between revisions of "2008 AIME I Problems/Problem 9"

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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==
−
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 it 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 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.
  
 
== See also ==
 
== See also ==

Revision as of 21:54, 24 January 2020

Problem

Ten identical crates each of dimensions $3$ ft $\times$ $4$ ft $\times$ $6$ 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 $41$ 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 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.

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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