Difference between revisions of "2004 AMC 12A Problems/Problem 23"
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<math>\text {(A)} c_0 \qquad \text {(B)} c_{2003} \qquad \text {(C)} b_2b_3...b_{2004} \qquad \text {(D)} \sum_{k = 1}^{2004}{a_k} \qquad \text {(E)}\sum_{k = 1}^{2004}{c_k}</math> | <math>\text {(A)} c_0 \qquad \text {(B)} c_{2003} \qquad \text {(C)} b_2b_3...b_{2004} \qquad \text {(D)} \sum_{k = 1}^{2004}{a_k} \qquad \text {(E)}\sum_{k = 1}^{2004}{c_k}</math> | ||
− | == Solution == | + | == Solution 1== |
We have to evaluate the answer choices and use process of elimination: | We have to evaluate the answer choices and use process of elimination: | ||
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There is, however, no reason to believe that <math>\boxed{\mathrm{E}}</math> should be zero (in fact, that quantity is <math>P(1)</math>, and there is no evidence that <math>1</math> is a root of <math>P(x)</math>). | There is, however, no reason to believe that <math>\boxed{\mathrm{E}}</math> should be zero (in fact, that quantity is <math>P(1)</math>, and there is no evidence that <math>1</math> is a root of <math>P(x)</math>). | ||
+ | ==Solution 2(cheap method using answer choices)== | ||
+ | Rule out answer choices <math>A, B, D</math> as done in solution 1. Assume for the sake of contradiction that <math>\mathrm{(E)}</math> is <math>0</math>. Then <math>P(1)=0</math>, so there is sum <math>m\neq 1</math> such that <math>z_m=1+0i</math>, which implies at least one <math>b_i</math>, <math>i\neq 1</math> is <math>0</math> SO if <math>\sum_{k=1}^{2004}c_k=0</math>, then <math>\prod_{i=12}^{2004}b_i=0</math>. So we have that <math>\sum_{k=1}^{2004}c_k\neq 0</math>, because then that would rule out all the answer choices. Hence <math>\boxed{\mathrm{E}}</math> must be nonzero, so the answer is <math>\mathrm{(E)}</math>. | ||
+ | -vsamc | ||
== See also == | == See also == |
Latest revision as of 10:14, 17 March 2020
Problem
has real coefficients with and distinct complex zeroes , with and real, , and
Which of the following quantities can be a nonzero number?
Solution 1
We have to evaluate the answer choices and use process of elimination:
- : We are given that , so . If one of the roots is zero, then .
- : By Vieta's formulas, we know that is the sum of all of the roots of . Since that is real, , and , so .
- : All of the coefficients are real. For sake of contradiction suppose none of are zero. Then for each complex root , its complex conjugate is also a root. So the roots should pair up, but we have an odd number of imaginary roots! (Remember that .) This gives us the contradiction, and therefore the product is equal to zero.
- : We are given that . Since the coefficients are real, it follows that if a root is complex, its conjugate is also a root; and the sum of the imaginary parts of complex conjugates is zero. Hence the RHS is zero.
There is, however, no reason to believe that should be zero (in fact, that quantity is , and there is no evidence that is a root of ).
Solution 2(cheap method using answer choices)
Rule out answer choices as done in solution 1. Assume for the sake of contradiction that is . Then , so there is sum such that , which implies at least one , is SO if , then . So we have that , because then that would rule out all the answer choices. Hence must be nonzero, so the answer is . -vsamc
See also
2004 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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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