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

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=== Solution 5 ===
 
=== Solution 5 ===
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The number of terms <math>3^n</math> produces includes each power of 3 (<math>1, 3^1, ..., 3^n</math>), the sums of two power of 3s(ex. <math>3^1 + 1</math>), three power of 3s (ex. <math>3^1 + 1 + 3^n</math>), all the way to the sum of them all. Since there are <math>n+1</math> powers of 3, the one number sum gives us <math>{n+1\choose 1} </math> terms, the two number <math>{n+1\choose 2} </math> terms, all the way to the sum of all the powers which gives us <math>{n+1\choose n}</math> terms , which is <math>2^{n+1} - 1</math> according to the theorem <center><math>\sum_{k=0}^n{{n}\choose{k}} = 2^n</math>.</center>
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The number of terms <math>3^n</math> produces includes each power of 3 (<math>1, 3^1, ..., 3^n</math>), the sums of two power of 3s(ex. <math>3^1 + 1</math>), three power of 3s (ex. <math>3^1 + 1 + 3^n</math>), all the way to the sum of them all. Since there are <math>n+1</math> powers of 3, the one number sum gives us <math>{n+1\choose 1} </math> terms, the two number <math>{n+1\choose 2} </math> terms, all the way to the sum of all the powers which gives us <math>{n+1\choose n+1}</math> terms , which is <math>2^{n+1} - 1</math> according to the theorem <center><math>\sum_{k=0}^n{{n}\choose{k}} = 2^n</math>.</center>
 
Since <math>2^6</math> is the greatest power <math><100</math>, then <math>n=5</math> and the sequence would look like {<math>3^0, ..., 3^5</math>}, where <math>3^5</math> or <math>243</math> would be the <math>2^5 - 1 = 63</math>rd number. The next largest power <math>729</math> would be the 64th number. However, its terms contributed extends beyond 100, so we break it to smaller pieces.  
 
Since <math>2^6</math> is the greatest power <math><100</math>, then <math>n=5</math> and the sequence would look like {<math>3^0, ..., 3^5</math>}, where <math>3^5</math> or <math>243</math> would be the <math>2^5 - 1 = 63</math>rd number. The next largest power <math>729</math> would be the 64th number. However, its terms contributed extends beyond 100, so we break it to smaller pieces.  
 
Noting that <math>729</math> plus any combination of lower powers <math>{1, 3^1 . . .3^4}</math> is < <math>729 + 243</math>, we can add all those terms(<math>2^5 - 1 = 31</math>) into our sequence:
 
Noting that <math>729</math> plus any combination of lower powers <math>{1, 3^1 . . .3^4}</math> is < <math>729 + 243</math>, we can add all those terms(<math>2^5 - 1 = 31</math>) into our sequence:

Revision as of 22:46, 21 December 2022

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 $972 + 9 = \boxed{981}$

Solution 5

The number of terms $3^n$ produces includes each power of 3 ($1, 3^1, ..., 3^n$), the sums of two power of 3s(ex. $3^1 + 1$), three power of 3s (ex. $3^1 + 1 + 3^n$), all the way to the sum of them all. Since there are $n+1$ powers of 3, the one number sum gives us ${n+1\choose 1}$ terms, the two number ${n+1\choose 2}$ terms, all the way to the sum of all the powers which gives us ${n+1\choose n+1}$ terms , which is $2^{n+1} - 1$ according to the theorem

$\sum_{k=0}^n{{n}\choose{k}} = 2^n$.

Since $2^6$ is the greatest power $<100$, then $n=5$ and the sequence would look like {$3^0, ..., 3^5$}, where $3^5$ or $243$ would be the $2^5 - 1 = 63$rd number. The next largest power $729$ would be the 64th number. However, its terms contributed extends beyond 100, so we break it to smaller pieces. Noting that $729$ plus any combination of lower powers ${1, 3^1 . . .3^4}$ is < $729 + 243$, we can add all those terms($2^5 - 1 = 31$) into our sequence:

{$3^0, ..., 3^5, 729, 729 + 1, .... 729 + (1 + 3^1 + . . . +3^4)$}

Our sequence now has $63 + 1 + 31 = 95$ terms. The remaining $5$ would just be the smallest sums starting with $729 + 243$ or $972$:

[$972$, $972 + 1$, $972 + 3$, $972+1+3$, $972 + 9$]

Hence the 100th term would be $972 + 9 = \boxed{981}$. ~SoilMilk

See also

1986 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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