1986 AIME Problems/Problem 7

Revision as of 19:34, 23 March 2007 by Azjps (talk | contribs) (solution)

Problem

The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $\displaystyle 100^{\mbox{th}}$ term of this sequence.

Solution

Rewrite all of the terms in base 3. Since the numbers are sums of distinct powers of 3, in base 3 each number is a sequence of 1s and 0s (if there is a 2, then it is no longer the sum of distinct powers of 3). Therefore, we can recast this into base 2 (binary) in order to determine the 100th number. $100$ is equal to $64 + 32 + 4$, so in binary form we get $\displaystyle 1100100$. However, we must change it back to base 3 for the answer, which is $\displaystyle 3^6 + 3^5 + 3^2 = 729 + 243 + 9 = 981$.

See also

1986 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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All AIME Problems and Solutions
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