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| − | == Problem ==
| + | #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
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| − | <center>
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| − | <math>P(1) = P(3) = P(5) = P(7) = a</math>, and<br/>
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| − | <math>P(2) = P(4) = P(6) = P(8) = -a</math>.
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| − | </center>
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| − | What is the smallest possible value of <math>a</math>?
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| − | <math>\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!</math>
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| − | == Solution ==
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| − | 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>.
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| − | 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>.
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| − | Then, plugging in values of <math>2,4,6,8,</math> we get
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| − | <cmath>P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a</cmath>
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| − | <cmath>P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = 9Q(4) = -2a</cmath>
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| − | <cmath>P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a</cmath>
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| − | <cmath>P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = 105Q(8) = -2a</cmath>
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| − | <math>-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).</math>
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| − | Thus, the least value of <math>a</math> must be the <math>lcm(15,9,15,105)</math>.
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| − | Solving, we receive <math>315</math>, so our answer is <math> \boxed{\textbf{(B)}\ 315} </math>.
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| − | == Critique ==
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| − | 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.
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| − | 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.
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| − | == Critique (cont.)==
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| − | First of all, the solution shows that <math>a</math> is a multiple of <math>315</math>, not a factor of <math>315</math>. Many people confuse the usage of the words 'factor' and 'multiple'. Secondly, even if <math>a</math> is a multiple of <math>315</math>, it does not mean that you can instantly get that the answer is <math>315</math> because we need to know that <math>a=315</math> is possible. After all, <math>a</math> is also a multiple of <math>1</math>, but <math>1</math> is definitely not the smallest possible number.
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| − | To complete the solution, we can let <math>a = 315</math>, and then try to find <math>Q(x)</math>. We know from the above calculation that <math>Q(2)=42, Q(4)=-70, Q(6)=42</math>, and <math>Q(8)=-6</math>. Then we can let <math>Q(x) = T(x)(x-2)(x-6)+42</math>, getting <math>T(4)=28, T(8)=-4</math>. Let <math>T(x)=L(x)(x-8)-4</math>, then <math>L(4)=-8</math>. Therefore, it is possible to choose <math>T(x) = -8(x-8)-4 = -8x + 60</math>, so the goal is accomplished. As a reference, the polynomial we get is
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| − | <cmath>P(x) = (x-1)(x-3)(x-5)(x-7)((-8x + 60)(x-2)(x-6)+42) + 315</cmath>
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| − | <cmath> = -8 x^7+252 x^6-3248 x^5+22050 x^4-84392 x^3+179928 x^2-194592 x+80325</cmath>
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| − | == See also ==
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| − | {{AMC10 box|year=2010|ab=B|num-b=24|after=Last question}}
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| − | Seealso
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| − | [[Category:Intermediate Algebra Problems]] | |
| − | {{MAA Notice}}
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