Difference between revisions of "2010 AMC 10B Problems/Problem 25"
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Thus, the least value of <math>a</math> must be the <math>lcm(15,9,15,105)</math>. | Thus, the least value of <math>a</math> must be the <math>lcm(15,9,15,105)</math>. | ||
Solving, we receive <math>315</math>, so our answer is <math> \boxed{\textbf{(B)}\ 315} </math>. | Solving, we receive <math>315</math>, so our answer is <math> \boxed{\textbf{(B)}\ 315} </math>. | ||
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+ | == Critique == | ||
+ | |||
+ | The above solution is incomplete. What is really proven is that 315 is a factor of <math>a</math>, if such an <math>a</math> exists. That only rules out answer A. | ||
+ | |||
+ | To prove that the answer is correct, one could exhibit a polynomium that satisfies the requirements with <math>a=315</math>. Here's one: <math>P(x) = -8x^7 + 252 x^6 -3248x^5 + 22050x^4 -84392x^3 + 179928x^2-194592x+80325</math>. | ||
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+ | You get that from the <math>8\times8</math> matrix <math>M_{i,j} = i^{j-1}</math> and <math>y^T=(315,-315,\ldots,315,-315)</math> and computing <math>c=M^{-1}y</math> which comes out as the all-integer coefficients above. | ||
== See also == | == See also == |
Revision as of 18:20, 7 February 2015
Contents
Problem
Let , and let be a polynomial with integer coefficients such that
, and
.
What is the smallest possible value of ?
Solution
We observe that because , if we define a new polynomial such that , has roots when ; namely, when .
Thus since has roots when , we can factor the product out of to obtain a new polynomial such that .
Then, plugging in values of we get
Thus, the least value of must be the . Solving, we receive , so our answer is .
Critique
The above solution is incomplete. What is really proven is that 315 is a factor of , if such an exists. That only rules out answer A.
To prove that the answer is correct, one could exhibit a polynomium that satisfies the requirements with . Here's one: .
You get that from the matrix and and computing which comes out as the all-integer coefficients above.
See also
2010 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last question | |
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 10 Problems and Solutions |
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