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 | + | 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>\displaystyle 100^{\mbox{th}}</math> term of this sequence. |
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== Solution == | == Solution == | ||
− | + | 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>\displaystyle 1100100</math>. However, we must change it back to base 3 for the answer, which is <math>\displaystyle 3^6 + 3^5 + 3^2 = 729 + 243 + 9 = 981</math>. | |
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== See also == | == See also == | ||
− | + | {{AIME box|year=1986|num-b=6|num-a=8}} | |
− | + | [[Category:Intermediate Number Theory Problems]] |
Revision as of 20:34, 23 March 2007
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
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 3 for the answer, which is .
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 |