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Problem 1599

National Olympiad, first round
Algebra Difficulty 6.1 Prove it HMMT February · United States

Michelle has a word with 2n2^{n} letters, where a word can consist of letters from any alphabet. Michelle performs a switcheroo on the word as follows: for each k=0,1,,n1k = 0, 1, \ldots, n-1, she switches the first 2k2^{k} letters of the word with the next 2k2^{k} letters of the word. For example, for n=3n = 3, Michelle changes
A B C D E F G HB A C D E F G HC D B A E F G HE F G H C D B A A\ B\ C\ D\ E\ F\ G\ H \rightarrow B\ A\ C\ D\ E\ F\ G\ H \rightarrow C\ D\ B\ A\ E\ F\ G\ H \rightarrow E\ F\ G\ H\ C\ D\ B\ A
in one switcheroo.

In terms of nn, what is the minimum positive integer mm such that after Michelle performs the switcheroo operation mm times on any word of length 2n2^{n}, she will receive her original word?

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

Solution:

Let m(n)m(n) denote the number of switcheroos needed to take a word of length 2n2^{n} back to itself. Consider a word of length 2n2^{n} for some n>1n > 1. After 2 switcheroos, one has separately performed a switcheroo on the first half of the word and on the second half of the word, while returning the (jumbled) first half of the word to the beginning and the (jumbled) second half of the word to the end.

After 2m(n1)2 \cdot m(n-1) switcheroos, one has performed a switcheroo on each half of the word m(n1)m(n-1) times while returning the halves to their proper order. Therefore, the word is in its proper order. However, it is never in its proper order before this, either because the second half precedes the first half (i.e. after an odd number of switcheroos) or because the halves are still jumbled (because each half has had fewer than m(n1)m(n-1) switcheroos performed on it).

It follows that m(n)=2m(n1)m(n) = 2 m(n-1) for all n>1n > 1. We can easily see that m(1)=2m(1) = 2, and a straightforward proof by induction shows that m=2nm = 2^{n}.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.