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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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| − | There must be some polynomial <math>Q(x)</math> such that <math>P(x)-a=(x-1)(x-3)(x-5)(x-7)Q(x)</math>
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| − | Then, plugging in values of <math>2,4,6,8,</math> we get
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| − | <math>P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a</math>
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| − | <math>P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = -15Q(4) = -2a</math>
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| − | <math>P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a</math>
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| − | <math>P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = -15Q(8) = -2a</math>
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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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| − | == See also ==
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| − | {{AMC10 box|year=2010|ab=B|num-b=24|after=Last question}}
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| − | [[Category:Intermediate Algebra Problems]] | |
