Difference between revisions of "1998 AIME Problems/Problem 13"
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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 =\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>. | 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 =\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>. | ||
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| + | == Video Solution == | ||
| + | https://youtu.be/cpzXYDvkZIg?si=6pv_9cwfpW0bhBId | ||
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| + | ~MathProblemSolvingSkills.com | ||
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== See also == | == See also == | ||
Revision as of 18:02, 20 October 2023
Contents
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 
where 
and 
are integers, find 
Solution
We note that the number of subsets (for now, including the empty subset, which we will just define to have a power sum of zero) 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, 
.
Video Solution
https://youtu.be/cpzXYDvkZIg?si=6pv_9cwfpW0bhBId
~MathProblemSolvingSkills.com
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
