Difference between revisions of "1986 AIME Problems/Problem 7"
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=== Solution 1 === | === 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 | + | 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 = 981</math>. |
=== Solution 2 === | === Solution 2 === |
Revision as of 16:39, 24 March 2013
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
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 |