Difference between revisions of "2009 AMC 12A Problems/Problem 15"
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<math>\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 49 \qquad \textbf{(D)}\ 97 \qquad \textbf{(E)}\ 98</math> | <math>\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 49 \qquad \textbf{(D)}\ 97 \qquad \textbf{(E)}\ 98</math> | ||
− | == Solution == | + | == Solution 1 == |
+ | We know that <math>i^x</math> cycles every <math>4</math> numbers so we group the sum in <math>4</math>s. | ||
+ | <cmath>i+2i^2+3i^3+4i^4=2-2i</cmath> | ||
+ | <cmath>5i^5+6i^6+7i^7+8i^8=2-2i</cmath> | ||
+ | |||
+ | We can postulate that every group of <math>4</math> is equal to <math>2-2i</math>. | ||
+ | For 24 groups we thus, get <math>48-48i</math> as our sum. | ||
+ | We know the solution must lie near | ||
+ | The next term is the <math>24*4+1=97</math>th term. This term is equal to <math>97i</math> (first in a group of <math>4</math>) and our sum is now <math>48+49i</math> so <math>n=97</math> is our answer | ||
+ | |||
+ | |||
+ | == Solution 2== | ||
Obviously, even powers of <math>i</math> are real and odd powers of <math>i</math> are imaginary. | Obviously, even powers of <math>i</math> are real and odd powers of <math>i</math> are imaginary. |
Revision as of 20:25, 7 March 2019
Contents
[hide]Problem
For what value of is ?
Note: here .
Solution 1
We know that cycles every numbers so we group the sum in s.
We can postulate that every group of is equal to . For 24 groups we thus, get as our sum. We know the solution must lie near The next term is the th term. This term is equal to (first in a group of ) and our sum is now so is our answer
Solution 2
Obviously, even powers of are real and odd powers of are imaginary. Hence the real part of the sum is , and the imaginary part is .
Let's take a look at the real part first. We have , hence the real part simplifies to . If there were an odd number of terms, we could pair them as follows: , hence the result would be negative. As we need the real part to be , we must have an even number of terms. If we have an even number of terms, we can pair them as . Each parenthesis is equal to , thus there are of them, and the last value used is . This happens for and . As is not present as an option, we may conclude that the answer is .
In a complete solution, we should now verify which of and will give us the correct imaginary part.
We can rewrite the imaginary part as follows: . We need to obtain . Once again we can repeat the same reasoning: If the number of terms were even, the left hand side would be negative, thus the number of terms is odd. The left hand side can then be rewritten as . We need parentheses, therefore the last value used is . This happens when or , and we are done.
See Also
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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