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

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 $100^{\mbox{th}}$ term of this sequence.

Solution

Solution 1

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 $1100100$. However, we must change it back to base 10 for the answer, which is $3^6 + 3^5 + 3^2 = 729 + 243 + 9 = \boxed {981}$.

Solution 2

Notice that the first term of the sequence is $1$, the second is $3$, the fourth is $9$, and so on. Thus the $64th$ term of the sequence is $729$. Now out of $64$ terms which are of the form $729$ + $'''S'''$, $32$ of them include $243$ and $32$ do not. The smallest term that includes $243$, i.e. $972$, is greater than the largest term which does not, or $854$. So the $95$th term will be $972$, then $973$, then $975$, then $976$, and finally $\boxed{981}$

Solution 3

After the nth power of $3$ in the sequence, the number of terms after that power but before the n+1th power of $3$ is equal to the number of terms before the nth power because those terms after the nth power are just the nth power plus all the distinct combinations of powers of $3$ before it, which is just all the terms before it. Adding the powers of $3$ and the terms that come after them, we see that the $100$th term is after $729$, which is the $64$th term. Also, note that the kth term after the nth power of $3$ is equal to the power plus the kth term in the entire sequence. Thus, the $100$th term is $729$ plus the $36$th term. Using the same logic, the $36$th term is $243$ plus the $4$th term, 9. We now have 7 $29$+ $243$+ $9$= $\boxed{981}$

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