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

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After the <math>n</math>th power of 3 in the sequence, the number of terms after that power but before the <math>(n+1)</math>th power of 3 is equal to the number of terms before the <math>n</math>th power, because those terms after the <math>n</math>th power are just the <math>n</math>th power plus all the distinct combinations of powers of 3 before it, which is just all the terms before it. Adding the powers of <math>3</math> and the terms that come after them, we see that the <math>100</math>th term is after <math>729</math>, which is the <math>64</math>th term. Also, note that the <math>k</math>th term after the <math>n</math>th power of 3 is equal to the power plus the <math>k</math>th term in the entire sequence. Thus, the <math>100</math>th term is <math>729</math> plus the <math>36</math>th term. Using the same logic, the <math>36</math>th term is <math>243</math> plus the <math>4</math>th term, <math>9</math>. We now have <math>729+243+9=\boxed{981}</math>
 
After the <math>n</math>th power of 3 in the sequence, the number of terms after that power but before the <math>(n+1)</math>th power of 3 is equal to the number of terms before the <math>n</math>th power, because those terms after the <math>n</math>th power are just the <math>n</math>th power plus all the distinct combinations of powers of 3 before it, which is just all the terms before it. Adding the powers of <math>3</math> and the terms that come after them, we see that the <math>100</math>th term is after <math>729</math>, which is the <math>64</math>th term. Also, note that the <math>k</math>th term after the <math>n</math>th power of 3 is equal to the power plus the <math>k</math>th term in the entire sequence. Thus, the <math>100</math>th term is <math>729</math> plus the <math>36</math>th term. Using the same logic, the <math>36</math>th term is <math>243</math> plus the <math>4</math>th term, <math>9</math>. We now have <math>729+243+9=\boxed{981}</math>
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=== Solution 4 ===
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Writing out a few terms of the sequence until we reach the next power of 3 (27), we see that the <math>2^{nth}</math> term is equal to <math>3^n</math>. From here, we can ballpark the range of the 100th term. The 64th term is <math>3^6</math> = <math>729</math> and the 128th term is <math>3^7</math> = <math>2187</math>. Writing out more terms of the sequence until the next power of 3 again (81) we can see that the (<math>2^n</math>+<math>2^{n+1}</math>)/2 term is equal to <math>3^n</math> + <math>3^{n-1}</math>. From here, we know that the 96th term is <math>3^6</math> + <math>3^5</math> = <math>972</math>. From here, we can construct the 100th term by following the sequence in increasing order. The 97th term is <math>972 + 1 = 973</math>, the 98th term is <math>972 + 3 = 975</math>, the 99th term is <math>972 + 3 + 1 = 976</math>, and finally the 100th term is <math>972 + 9 = \boxed{981}</math>
  
 
== See also ==
 
== See also ==

Revision as of 09:17, 7 July 2021

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.

Solutions

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 $96$th term will be $972$, then $973$, then $975$, then $976$, and finally $\boxed{981}$

Solution 3

After the $n$th power of 3 in the sequence, the number of terms after that power but before the $(n+1)$th power of 3 is equal to the number of terms before the $n$th power, because those terms after the $n$th power are just the $n$th 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 $k$th term after the $n$th power of 3 is equal to the power plus the $k$th 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 $729+243+9=\boxed{981}$

Solution 4

Writing out a few terms of the sequence until we reach the next power of 3 (27), we see that the $2^{nth}$ term is equal to $3^n$. From here, we can ballpark the range of the 100th term. The 64th term is $3^6$ = $729$ and the 128th term is $3^7$ = $2187$. Writing out more terms of the sequence until the next power of 3 again (81) we can see that the ($2^n$+$2^{n+1}$)/2 term is equal to $3^n$ + $3^{n-1}$. From here, we know that the 96th term is $3^6$ + $3^5$ = $972$. From here, we can construct the 100th term by following the sequence in increasing order. The 97th term is $972 + 1 = 973$, the 98th term is $972 + 3 = 975$, the 99th term is $972 + 3 + 1 = 976$, and finally the 100th term is $972 + 9 = \boxed{981}$

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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