2019 AIME II Problem 2

Attempt Problem 2 of the 2019 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 AIME II solutions, or check the answer key.

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2.

Lily pads 1,2,3,1, 2, 3, \ldots lie in a row on a pond. A frog makes a sequence of jumps starting on pad 1.1. From any pad kk the frog jumps to either pad k+1k + 1 or pad k+2k + 2 chosen randomly with probability 12\frac{1}{2} and independently of other jumps. The probability that the frog visits pad 77 is pq,\frac{p}{q}, where pp and qq are relatively prime positive integers. Find p+q.p + q.

Answer: 107
Concepts:recursive probabilityrecursion
Difficulty rating: 2270
Solution:

Let pkp_k be the probability that the frog visits pad k.k. The frog lands on pad kk in exactly one of two disjoint ways: it visits pad k1k - 1 and jumps +1+1 from there (if it jumps +2,+2, pad kk is skipped forever), or it skips pad k1k - 1 entirely, which requires visiting pad k2k - 2 and jumping +2+2 from it, landing on pad k.k. Hence pk=12pk1+12pk2,p1=1,p2=12. \begin{aligned} &p_k = \tfrac{1}{2}p_{k-1} + \tfrac{1}{2}p_{k-2}, \\ &\qquad p_1 = 1, \quad p_2 = \tfrac{1}{2}. \end{aligned}

Iterating: p3=34,p_3 = \frac{3}{4}, p4=58,p_4 = \frac{5}{8}, p5=1116,p_5 = \frac{11}{16}, p6=2132,p_6 = \frac{21}{32}, and p7=4364.p_7 = \frac{43}{64}. Since gcd(43,64)=1,\gcd(43, 64) = 1, the answer is 43+64=107.43 + 64 = 107.

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