Difference between revisions of "1998 AIME Problems/Problem 13"
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== Problem == | == Problem == | ||
+ | If <math>\{a_1,a_2,a_3,\ldots,a_n\}</math> is a [[set]] of [[real numbers]], indexed so that <math>a_1 < a_2 < a_3 < \cdots < a_n,</math> its complex power sum is defined to be <math>a_1i + a_2i^2 + a_3i^3 + \cdots + a_ni^n,</math> where <math>i^2 = - 1.</math> Let <math>S_n</math> be the sum of the complex power sums of all nonempty [[subset]]s of <math>\{1,2,\ldots,n\}.</math> Given that <math>S_8 = - 176 - 64i</math> and <math>S_9 = p + qi,</math> were <math>p</math> and <math>q</math> are integers, find <math>|p| + |q|.</math> | ||
== Solution == | == Solution == | ||
+ | We note that the number of subsets (for now, including the empty subset) with <math>9</math> in it is equal to the number of subsets without a <math>9</math>. To easily see this, take all possible subsets of <math>\{1,2,\ldots,8\}</math>. Since the sets are ordered, a <math>9</math> must go at the end; hence we can just append a <math>9</math> to any of those subsets to get a new one. | ||
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+ | Now that we have drawn that [[bijection]], we can calculate the complex power sum recursively. Since appending a <math>9</math> to a subset doesn't change anything about that subset's complex power sum besides adding an additional term, we have that <math>S_9 = 2S_8 + T_9</math>, where <math>T_9</math> refers to the sum of all of the <math>9i^x</math>. | ||
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+ | It a subset of size 1 has a 9, then its power sum must be <math>9i</math>, and there is only <math>1</math> of these such subsets. There are <math>{8\choose1}</math> with <math>9\cdot i^2</math>, <math>{8\choose2}</math> with <math>9\cdot i^3</math>, and so forth. So <math>T_9 = \displaystyle\sum_{k=0}^{8} 9{8\choose{k}}i^{k+1}</math>. This is exactly the [[binomial expansion]] of <math>9i \cdot (1+i)^8</math>. We can use [[De Moivre's Theorem]] to calculate the power: <math>\sqrt{2}^8\cos{8\cdot45} = 16</math>. Hence <math>T_9 = 16\cdot9i = 144i</math>, and <math>S_9 = 2S_8 + 144i = 2(-176 -64i) + 144i = -352 + 16i</math>. Thus, <math>|p| + |q| = |-352| + |16| = 368</math>. | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=1998|num-b=12|num-a=14}} | |
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+ | [[Category:Intermediate Complex Numbers Problems]] |
Revision as of 21:44, 8 September 2007
Problem
If is a set of real numbers, indexed so that its complex power sum is defined to be where Let be the sum of the complex power sums of all nonempty subsets of Given that and were and are integers, find
Solution
We note that the number of subsets (for now, including the empty subset) with in it is equal to the number of subsets without a . To easily see this, take all possible subsets of . Since the sets are ordered, a must go at the end; hence we can just append a to any of those subsets to get a new one.
Now that we have drawn that bijection, we can calculate the complex power sum recursively. Since appending a to a subset doesn't change anything about that subset's complex power sum besides adding an additional term, we have that , where refers to the sum of all of the .
It a subset of size 1 has a 9, then its power sum must be , and there is only of these such subsets. There are with , with , and so forth. So . This is exactly the binomial expansion of . We can use De Moivre's Theorem to calculate the power: . Hence , and . Thus, .
See also
1998 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |