Difference between revisions of "2010 AMC 10B Problems/Problem 25"

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== Problem ==
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#REDIRECT [[2010 AMC 12B Problems/Problem 21]]
Let <math>a > 0</math>, and let <math>P(x)</math> be a polynomial with integer coefficients such that
 
 
 
<center>
 
<math>P(1) = P(3) = P(5) = P(7) = a</math>, and<br/>
 
<math>P(2) = P(4) = P(6) = P(8) = -a</math>.
 
</center>
 
 
 
What is the smallest possible value of <math>a</math>?
 
 
 
<math>\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!</math>
 
 
 
== Solution ==
 
We observe that because <math>P(1) = P(3) = P(5) = P(7) = a</math>, if we define a new polynomial <math>R(x)</math> such that <math>R(x) = P(x) - a</math>, <math>R(x)</math> has roots when <math>P(x) = a</math>; namely, when <math>x=1,3,5,7</math>.
 
 
 
Thus since <math>R(x)</math> has roots when <math>x=1,3,5,7</math>, we can factor the product <math>(x-1)(x-3)(x-5)(x-7)</math> out of <math>R(x)</math> to obtain a new polynomial <math>Q(x)</math> such that <math>(x-1)(x-3)(x-5)(x-7)(Q(x)) = R(x) = P(x) - a</math>.
 
 
 
Then, plugging in values of <math>2,4,6,8,</math> we get
 
 
 
<cmath>P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a</cmath>
 
<cmath>P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = 9Q(4) = -2a</cmath>
 
<cmath>P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a</cmath>
 
<cmath>P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = 105Q(8) = -2a</cmath>
 
 
 
<math>-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).</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>.
 
 
 
== 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>.
 
 
 
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 ==
 
{{AMC10 box|year=2010|ab=B|num-b=24|after=Last question}}
 
 
 
[[Category:Intermediate Algebra Problems]]
 
{{MAA Notice}}
 

Latest revision as of 15:26, 17 June 2020

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