2008 AIME I Problems/Problem 9
Contents
Problem
Ten identical crates each of dimensions ft ft 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 be the probability that the stack of crates is exactly ft tall, where and are relatively prime positive integers. Find .
Solution
Only the heights matter, and each crate is either 3, 4, or 6 feet tall with equal probability. We have the following:
Subtracting 3 times the second from the first gives , or . The last doesn't work, obviously. This gives the three solutions . In terms of choosing which goes where, the first two solutions are analogous.
For , we see that there are ways to stack the crates. For , there are . Also, there are total ways to stack the crates to any height.
Thus, our probability is . Our answer is the numerator, .
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.
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. 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.
Case 1: 0 zeroes, 0 threes, 11 ones Impossible, because there are only ten available spots.
Case 2: 1 zero, 1 three, 8 ones This is just 10! over 8!, so there are 90 possibilities.
Case 3: 3 zeroes, 2 threes, 5 ones This is just 10! over 3!, 2!, and 5! This gives 2520 possibilities.
Case 4: 5 zeroes, 3 threes, and 2 ones. This is the same as case 3, so also 2520 possibilities.
5130 has three powers of 3, so 5130 divided by 27 is .
-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 | ||
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