1986 AIME Problem 7

Attempt Problem 7 of the 1986 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1986 AIME solutions, or check the answer key.

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7.

The increasing sequence 1,1, 3,3, 4,4, 9,9, 10,10, 12,12, 13,13, \ldots consists of all those positive integers which are powers of 33 or sums of distinct powers of 3.3. Find the 100100th term of this sequence.

Answer: 981
Concepts:number baseplace value
Difficulty rating: 1970
Small Hint:

These are precisely the numbers whose base-33 digits are all 00 or 11

Big Hint:

Write the index in base 2,2, then reinterpret those same digits in base 33

Solution:

A sum of distinct powers of 33 has only 00’s and 11’s in base 3.3. As these strings increase, they occur in the same order as binary numerals with the same digit strings. Since 100=11001002, 100=1100100_2, the 100100th positive term is 11001003=36+35+32=729+243+9=981. \begin{aligned} 1100100_3 &=3^6+3^5+3^2\\ &=729+243+9\\ &=981. \end{aligned}

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