Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 2)
(→‎Solution 5)
 
(14 intermediate revisions by 3 users not shown)
Line 2: Line 2:
 
The increasing [[sequence]] <math>1,3,4,9,10,12,13\cdots</math> consists of all those positive [[integer]]s which are [[exponent|powers]] of 3 or sums of distinct powers of 3. Find the <math>100^{\mbox{th}}</math> term of this sequence.
 
The increasing [[sequence]] <math>1,3,4,9,10,12,13\cdots</math> consists of all those positive [[integer]]s which are [[exponent|powers]] of 3 or sums of distinct powers of 3. Find the <math>100^{\mbox{th}}</math> term of this sequence.
  
−
== Solution ==
+
== Solutions ==
 
=== Solution 1 ===
 
=== Solution 1 ===
  
Line 14: Line 14:
  
 
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>
 +
 +
=== Solution 4 ===
 +
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>
 +
 +
=== Solution 5  ===
 +
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. Summing all these up gives us <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.
 +
Noting that <math>729</math> plus any combination of lower powers <math>{1, 3^1 . . .3^4}</math> is < <math>729 + 243</math>, so we can add all those terms(<math>2^5 - 1 = 31</math>) into our sequence:
 +
<center>{<math>3^0, ..., 3^5, 729, 729 + 1, .... 729 + (1 + 3^1 + . . . +3^4)</math>}</center>
 +
Our sequence now has <math>63 + 1 + 31 = 95</math> terms. The remaining <math>5</math> would just be the smallest sums starting with <math>729 + 243</math> or <math>972</math>:
 +
 +
<center>[<math>972</math>, <math>972 + 1</math>, <math>972 + 3</math>, <math>972+1+3</math>, <math>972 + 9</math>]</center>
 +
Hence the 100th term would be <math>972 + 9 = \boxed{981}</math>. ~SoilMilk
 +
 +
<math>{}^{*}</math>Note that there isn't a <math>{n+1\choose 0}</math> as choosing 0 numbers will not give you a term.
  
 
== See also ==
 
== See also ==

Latest revision as of 23:00, 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. Summing all these up gives us $2^{n+1} - 1 ^ {*}$ 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 < $729 + 243$, so 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

${}^{*}$Note that there isn't a ${n+1\choose 0}$ as choosing 0 numbers will not give you a term.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_7&oldid=184415"