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

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=== Solution 2 ===
 
=== Solution 2 ===
  
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Notice that the first term of the sequence is <math>1</math>, the second is <math>3</math>, the fourth is <math>9</math>, and so on. Thus the <math>64th</math> term of the sequence is <math>729</math>. Now out of <math>64</math> terms which are of the form <math>729</math> + <math>'''S'''</math>, <math>32</math> of them include <math>243</math> and <math>32</math> do not. The smallest term that includes <math>243</math>, i.e. <math>972</math>, is greater than the largest term which does not, or <math>854</math>. So the <math>95</math>th term will be <math>972</math>, then <math>973</math>, then <math>975</math>, then <math>976</math>, and finally <math>981</math>
+
Notice that the first term of the sequence is <math>1</math>, the second is <math>3</math>, the fourth is <math>9</math>, and so on. Thus the <math>64th</math> term of the sequence is <math>729</math>. Now out of <math>64</math> terms which are of the form <math>729</math> + <math>'''S'''</math>, <math>32</math> of them include <math>243</math> and <math>32</math> do not. The smallest term that includes <math>243</math>, i.e. <math>972</math>, is greater than the largest term which does not, or <math>854</math>. So the <math>95</math>th term will be <math>972</math>, then <math>973</math>, then <math>975</math>, then <math>976</math>, and finally <math>\boxed{981}</math>
  
 
== See also ==
 
== See also ==

Revision as of 15:41, 24 March 2013

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 $\boxed{981}$

See also

1986 AIME (Problems • Answer Key • Resources)
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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