2019 AIME II Problem 2

Below is the professionally curated solution for Problem 2 of the 2019 AIME II, 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.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:recursive probabilityrecursion

Difficulty rating: 2270

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.

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.p_k = \tfrac{1}{2}p_{k-1} + \tfrac{1}{2}p_{k-2}, \qquad p_1 = 1, \quad p_2 = \tfrac{1}{2}.

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.

← Problem 1Full ExamProblem 3

Problem 2 in Other Years