Problem 5

Two squirrels, Bushy and Jumpy, have collected

2021

walnuts for the winter. Jumpy numbers the walnuts from

1

through

2021

and digs

2021

little holes in a circular pattern in the ground around their favourite tree. The next morning Jumpy notices that Bushy had placed one walnut into each hole, but had paid no attention to the numbering. Unhappy, Jumpy decides to reorder the walnuts by performing a sequence of

2021

moves. In the

k

-th move, Jumpy swaps the positions of the two walnuts adjacent to walnut

k

. Prove that there exists a value of

k

such that, on the

k

-th move, Jumpy swaps some walnuts

a

and

b

with

a<k<b

Step 5 of 5: Reach a contradiction after all

2021

moves

Previous step Next step

\exists\ \text{frozen black block of length } 2\ \text{forever}

Detailed analysis

Starting from the even block of length $2020$ , each split produces one even and one odd sub-block, and every even sub-block eventually reduces (through further splits) to a frozen block of length exactly $2$ that can never turn fully red. So at least one pair of black walnuts survives forever. But move $k$ colours walnut $k$ red for every $k=1,\ldots,2021$ , so after all $2021$ moves every walnut is red — a contradiction. Hence the assumption fails, and some move $k$ must swap walnuts $a<k<b$ .