Difference between revisions of "1986 AIME Problems/Problem 7"
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== Problem == | == Problem == | ||
| + | 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 === | ||
| + | |||
| + | 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. <math>100</math> is equal to <math>64 + 32 + 4</math>, so in binary form we get <math>1100100</math>. However, we must change it back to base 10 for the answer, which is <math>3^6 + 3^5 + 3^2 = 729 + 243 + 9 = \boxed {981}</math>. | ||
| + | |||
| + | === Solution 2 === | ||
| + | |||
| + | 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>96</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> | ||
| + | |||
| + | === Solution 3 === | ||
| + | |||
| + | 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 == | ||
| − | * [[ | + | {{AIME box|year=1986|num-b=6|num-a=8}} |
| + | * [[AIME Problems and Solutions]] | ||
| + | * [[American Invitational Mathematics Examination]] | ||
| + | * [[Mathematics competition resources]] | ||
| + | |||
| + | [[Category:Intermediate Number Theory Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 23:00, 21 December 2022
Contents
Problem
The increasing sequence 
consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the 
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. 
is equal to 
, so in binary form we get 
. However, we must change it back to base 10 for the answer, which is 
.
Solution 2
Notice that the first term of the sequence is 
, the second is 
, the fourth is 
, and so on. Thus the 
term of the sequence is 
. Now out of 
terms which are of the form + 
, 
of them include 
and do not. The smallest term that includes , i.e. 
, is greater than the largest term which does not, or 
. So the 
th term will be , then 
, then 
, then 
, and finally 
Solution 3
After the 
th power of 3 in the sequence, the number of terms after that power but before the 
th power of 3 is equal to the number of terms before the th power, because those terms after the th power are just the 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 and the terms that come after them, we see that the th term is after , which is the th term. Also, note that the 
th term after the th power of 3 is equal to the power plus the th term in the entire sequence. Thus, the th term is plus the 
th term. Using the same logic, the th term is plus the 
th term, . We now have 
Solution 4
Writing out a few terms of the sequence until we reach the next power of 3 (27), we see that the 
term is equal to 
. From here, we can ballpark the range of the 100th term. The 64th term is 
= and the 128th term is 
= 
. Writing out more terms of the sequence until the next power of 3 again (81) we can see that the (
+
)/2 term is equal to + 
. From here, we know that the 96th term is + 
= . From here, we can construct the 100th term by following the sequence in increasing order. The 97th term is 
, the 98th term is 
, the 99th term is 
, and finally the 100th term is 
Solution 5
The number of terms produces includes each power of 3 (
), the sums of two power of 3s(ex. 
), three power of 3s (ex. 
), all the way to the sum of them all. Since there are 
powers of 3, the one number sum gives us 
terms, the two number 
terms, all the way to the sum of all the powers which gives us 
terms. Summing all these up gives us 
according to the theorem
.Since 
is the greatest power 
, then 
and the sequence would look like {
}, where or would be the 
rd number. The next largest power would be the 64th number. However, its terms contributed extends beyond 100, so we break it to smaller pieces.
Noting that plus any combination of lower powers 
is < 
, so we can add all those terms(
) into our sequence:
}Our sequence now has 
terms. The remaining 
would just be the smallest sums starting with or :
, 
, 
, 
, 
]Hence the 100th term would be . ~SoilMilk
Note that there isn't a 
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 | ||
- AIME Problems and Solutions
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- Mathematics competition resources
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