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Harvard-MIT Mathematics Tournament

Combinatorics Difficulty 5.1 AIME, harder Prove it United States

Problem:
Let SS be the smallest subset of the integers with the property that 0∈S0 \in S and for any x∈Sx \in S, we have 3x∈S3x \in S and 3x+1∈S3x+1 \in S. Determine the number of non-negative integers in SS less than 20082008.

Solution

Solution:
Write the elements of SS in their ternary expansion (i.e. base 33). Then the second condition translates into: if d1d2⋯dk‾∈S\overline{d_{1} d_{2} \cdots d_{k}} \in S, then d1d2⋯dk0‾\overline{d_{1} d_{2} \cdots d_{k} 0} and d1d2⋯dk1‾\overline{d_{1} d_{2} \cdots d_{k} 1} are also in SS. It follows that SS is the set of nonnegative integers whose ternary representation contains only the digits 00 and 11.

Since 2⋅36<2008<372 \cdot 3^{6} < 2008 < 3^{7}, there are 27=1282^{7} = 128 such elements less than 20082008. Therefore, there are 128128 such non-negative elements.

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