Difference between revisions of "1986 AIME Problems/Problem 7"

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== See also ==
 
== See also ==
 
{{AIME box|year=1986|num-b=6|num-a=8}}
 
{{AIME box|year=1986|num-b=6|num-a=8}}
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* [[AIME Problems and Solutions]]
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* [[American Invitational Mathematics Examination]]
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* [[Mathematics competition resources]]
  
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]

Revision as of 13:49, 6 May 2007

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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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