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

Revision as of 21:46, 2 April 2018

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 $\boxed{981}$

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 $729+243+9=\boxed{981}$

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 === Solution 3 ===
-After the nth power of <math>3</math> in the sequence, the number of terms after that power but before the n+1th power of <math>3</math> is equal to the number of terms before the nth power because those terms after the nth power are just the nth power plus all the distinct combinations of powers of <math>3</math> 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 kth term after the nth power of <math>3</math> is equal to the power plus the kth 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, 9. 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>
 == 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
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