Difference between revisions of "1986 AIME Problems/Problem 7"
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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>95</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> | 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>95</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> | ||
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+ | === Solution 3 === | ||
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+ | 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 7<math>29</math>+<math>243</math>+<math>9</math>=<math>\boxed{981}</math> | ||
== See also == | == See also == |
Revision as of 21:53, 10 December 2016
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.
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. 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 nth power of in the sequence, the number of terms after that power but before the n+1th power of 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 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 kth term after the nth power of is equal to the power plus the kth 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, 9. We now have 7++=
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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