2002 AIME II Problem 12

Below is the professionally curated solution for Problem 12 of the 2002 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AIME II solutions, or check the answer key.

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Concepts:lattice pathsrecursive countingbinomial probability

Difficulty rating: 2990

12.

A basketball player has a constant probability of .4.4 of making any given shot, independent of previous shots. Let ana_n be the ratio of shots made to shots attempted after nn shots. The probability that a10=.4a_{10} = .4 and an.4a_n \le .4 for all nn such that 1n91 \le n \le 9 is given to be paqbr/(sc),p^a q^b r / \left(s^c\right), where p,p, q,q, r,r, and ss are primes, and a,a, b,b, and cc are positive integers. Find (p+q+r+s)(a+b+c).(p + q + r + s)(a + b + c).

Solution:

Record the player's progress as a path through points (n,y),(n, y), where yy is the number of shots made after nn attempts. The condition an.4a_n \le .4 caps yy at 0.4n,\lfloor 0.4n \rfloor, which for n=1,,9n = 1, \ldots, 9 is 0,0,1,1,2,2,2,3,3,0, 0, 1, 1, 2, 2, 2, 3, 3, and a10=.4a_{10} = .4 means the path ends at (10,4).(10, 4).

Count the allowed paths by adding, at each point, the counts of its two predecessors (a miss keeps y,y, a make raises it by 11). The counts at the maximum allowed heights for n=3,,9n = 3, \ldots, 9 come out to 1,2,2,5,9,9,23,1, 2, 2, 5, 9, 9, 23, and the tenth shot must be a make, so 2323 shot sequences qualify. Each consists of 44 makes and 66 misses, so the probability is 23(25)4(35)6=243623510.23 \left(\tfrac{2}{5}\right)^4 \left(\tfrac{3}{5}\right)^6 = \frac{2^4 \, 3^6 \cdot 23}{5^{10}}.

Thus {p,q,r,s}={2,3,23,5}\{p, q, r, s\} = \{2, 3, 23, 5\} and (a,b,c)=(4,6,10),(a, b, c) = (4, 6, 10), giving (2+3+23+5)(4+6+10)=3320=660.(2 + 3 + 23 + 5)(4 + 6 + 10) = 33 \cdot 20 = 660.

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