proposed
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proposed
The powers of 2 having at least 2 digits are 16, 32, 64, 128, 256, ...; discarding all but their first 2 digits yields the sequence 16, 32, 64, 12, 25, ..., in which all 90 of the possible 2-digit numbers (10 through 99) eventually appear; the last to appear is 97, which appears (at 2^279 = 9.713...*10^83), so a(2) = 279.
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editing
reviewed
approved
proposed
reviewed
editing
proposed
The powers of 2 having at least 2 digits are 16, 32, 64, 128, 256, ...; discarding all but their first 2 digits yields the sequence 16, 32, 64, 12, 25, ..., in which all 90 of the possible 2-digit numbers (10 through 99) eventually appear; the last to appear is 97, which appears at 2^279, = 9.713...*10^83), so a(2) = 279.
The powers of 2 having at least 2 digits are 16, 32, 64, 128, 256, ...; discarding all but their first 2 digits yields the sequence 16, 32, 64, 12, 25, ..., in which all 90 of the possible 2-digit numbers (10 through 99) eventually appear; the last to appear is 97, which appears at 2^279, so a(2) = 279.
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editing
reviewed
approved
proposed
reviewed