HMMT 二月 2026 · 团队赛 · 第 2 题

2. [25] Jessica the jackrabbit wants to climb down a wall. The wall consists of 2026 horizontal layers n stacked vertically. The n th layer from the top is partitioned into 2 − 1 identical rectangular bricks arranged side by side. Jessica begins in the topmost layer, which contains a single brick. A move consists of Jessica going down one layer to a brick that shares a side with the brick she is currently on. ©2026 HMMT Jessica . . . Determine, with proof, the total number of distinct sequences of moves Jessica can take to reach the 2026 th layer. Proposed by: Isabella Zhu 2025 Answer: 3 2025 Solution 1: The answer is 3 . The key observation is that each brick on layer n shares a side with exactly 3 bricks on layer n + 1 , so Jessica always has 3 ways to move down one layer. Since she 2025 moves down 2025 layers, she can make 3 sequences of moves. To prove this, we directly show case-by-case that each brick shares a side with exactly 3 bricks below. n We label the position of the block as the following: fix n , where 1 ≤ n ≤ 2025 , and let N = 2 − 1 , so n +1 that 2 − 1 = 2 N + 1 . Let the left and right edges of the wall be at x = 0 and x = 1 respectively. Below is a diagram for n = 2 . 1 2 0 1 3 3 1 2 3 4 5 6 0 1 7 7 7 7 7 7 2 a a 2 a +1 Claim 1. We have < < for all 1 ≤ a ≤ N − 1 . 2 N +1 N 2 N +1 Proof. For the lower bound, write 2 a 2 a a < = . 2 N + 1 2 N N For the upper bound, note that 2 a + 1 a 2 a + 1 a 2( N − a ) N − a

⇐⇒ 1 − < 1 − ⇐⇒ < . 2 N + 1 N 2 N + 1 N 2 N + 1 N The last inequality holds because 2( N − a ) 2( N − a ) N − a < = . 2 N + 1 2 N N Claim 2. For all 1 ≤ n ≤ 2025 , each brick on layer n shares a side with exactly 3 bricks on layer n + 1 . th Proof. Consider the a brick on layer n borders exactly 3 bricks on layer n + 1 , namely bricks 2 a − 1 , 2 a , and 2 a + 1 . We split into cases based on a : 1 1 • If a = 1 , then the brick’s left edge is at 0 and its right edge is at . By the claim, is strictly N N 2 3 between and , which are the edges of brick 3 on layer n + 1 . Hence brick 1 on layer n 2 N +1 2 N +1 borders bricks 1 , 2 , and 3 on layer n + 1 . N − 1 N − 1 • If a = N , then the brick’s right edge is at 1 and its left edge is at . By the claim, is N N 2 N − 2 2 N − 1 strictly between and , which are the edges of brick 2 N − 1 on layer n + 1 . Hence brick 2 N +1 2 N +1 N on layer n borders bricks 2 N − 1 , 2 N , and 2 N + 1 on layer n + 1 . • If 1 < a < N , then: a − 1 2 a − 2 2 a − 1 – The brick’s left edge is at , which is strictly between and . These are the edges N 2 N +1 2 N +1 of brick 2 a − 1 on layer n + 1 . ©2026 HMMT a 2 a 2 a +1 – The brick’s right edge is at , which is strictly between and . These are the edges N 2 N +1 2 N +1 of brick 2 a + 1 on layer n + 1 . Hence brick a on layer n borders bricks 2 a − 1 , 2 a , and 2 a + 1 on layer n + 1 . With that, we’ve shown that every brick on layer n shares a side with exactly 3 bricks on layer n + 1 , as desired. This claim implies Jessica always has 3 ways to move down one layer. Since she moves down 2025 2025 layers, she has 3 possible sequences of moves. Solution 2: We present an alternative proof that each brick on layer n shares a side with exactly 3 bicks on layer n + 1 . Instead of working case-by-case, we show that, for each brick in layer n , the number of bricks on layer n + 1 sharing sides with that brick cannot be at most 2 nor at least 4 . n Similar to the first solution, fix n , where 1 ≤ n ≤ 2025 , and let N = 2 − 1 . Let the left and right edges of the wall be at x = 0 and x = N (2 N + 1) , respectively. In this setup, the vertical edges of the bricks on layer n are at x -coordinates that are multiples of 2 N + 1 , and the edges of the bricks on layer n + 1 are at x -coordinates that are multiples of N . In particular, all edges of bricks are at integer x -coordinates. Below is a diagram for n = 2 : 0 7 14 21 0 3 6 9 12 15 18 21 Consider any brick B on layer n , with left and right edges at A and A + 2 N + 1 , respectively. We’ll prove B borders exactly 3 bricks on layer n + 1 . • Suppose that B borders at most 2 bricks on layer n + 1 . The bottom edge of B has length 2 N + 1 . The bricks on layer n + 1 have length N , so they cover a length of at most 2 N on this edge, contradiction. • Suppose that B borders at least 4 bricks on layer n + 1 . Let E be the vertical border between 1 the first and second (from the left) of these bricks. E must have x -position strictly greater than 1 A . Since all borders between bricks are at integer coordinates, E ’s x -position is at least A + 1 . 1 Similarly, let E be the vertical border between the last and second-to-last of these bricks; E 2 2 must have x -position at most A + 2 N . By assumption, there must be at least 2 bricks between E and E . These bricks fit inside an 1 2 interval of length ( A + 2 N ) − ( A + 1) = 2 N − 1 . But this is a contradiction because 2 bricks have total length 2 N . We’ve reached a contradiction in both cases, so we’re done.