1. A strip for playing "hopscotch" consists of ten squares numbered consecutively $1,2, \ldots, 10$. Clarissa and Marissa start from the center of the first square, jump 9 times to the centers of the other squares so that they visit each square once, and end up at the tenth square. (Jumps forward and backward are allowed.) Each jump of Clarissa was for the same distance as the corresponding jump of Marissa. Does this mean that they both visited the squares in the same order? Alexey Tolpygo
Problem
Source: 45th International Tournament of Towns, Junior O-Level P1, Fall 2023
Tags: combinatorics
QueenArwen
04.10.2024 11:08
No.
Clarissa's order: $1\rightarrow8\rightarrow9\rightarrow7\rightarrow4\rightarrow6\rightarrow5\rightarrow3\rightarrow2\rightarrow10$
Marissa's order: $1\rightarrow8\rightarrow7\rightarrow9\rightarrow6\rightarrow4\rightarrow5\rightarrow3\rightarrow2\rightarrow10$
weedtaker567
04.10.2024 12:49
I unfortunately took a glance at the answer from the post above, which gave me a much easier time on this problem than it is supposed to be..
The answer is no.
Clarissa's path: $1 \rightarrow 3 \rightarrow 4 \rightarrow 2 \rightarrow 5 \rightarrow 7 \rightarrow 6 \rightarrow 8 \rightarrow 9 \rightarrow 10$.
Marissa's path: $1 \rightarrow 3 \rightarrow 2 \rightarrow 4 \rightarrow 7 \rightarrow 5 \rightarrow 6 \rightarrow 8 \rightarrow 9 \rightarrow 10$.
When I first tried this problem, I drew $6$ squares (as I didn't have enough space on my piece of paper), which actually meant I got the answer to be "Yes". This got me curious about the largest $n$ such that if the squares were to be numbered $1,2,...,n$, the answer would be "Yes". The construction on my original solution showed that the answer is "No" for all $n \ge 8$, and if I'm not mistaken then $n = 7$ is indeed the answer to this problem. The way I proved it is quite bashy though, which is why I didn't post it.