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A fifth degree polynomial $P(x)$ with integral coefficients takes on values $0,1,2,3,4$ at $x=0,1,2,3,4$, respectively.

Which of the following is a possible value for $P(5)$?

A) $5$

B) $24$

C) $125$

D)None of the above

  • 1
    Welcome to math.SE, bazinga: since you are a new user, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level.2012-06-25
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    Hello draks. I've thought of some possible approaches like putting 0, 1, 2, 3, 4 and making linear equations and solve the system but its quite long. I think there's a shorter way.2012-06-25
  • 0
    This is not my homework problem, I like to do math in summer vacations. Thank you for your concern.2012-06-25

3 Answers 3

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Any polynomial of degree at most five, fulfilling the given interpolation conditions, has the form \[ P(x) = x + a\prod_{i=0}^4 (x-i) \] for some $a \in \mathbb Z$ (we can see this if we write $P$ in the Newton basis, for example). So $P(5) = 5 + 5!a = 5 + 120a$. Now A) corresponds to $a = 0$, B) to $a = \frac{19}{120}$ and C) to $a=1$. As we want $P$ to be of fifth degree, we need $a \ne 0$, and as $\frac{19}{120} \not\in\mathbb Z$, the correct answer is C), giving $P(x) = x + \prod_{i=0}^4(x-i)$.

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For example, $p\left(x\right)=x+x(x-1)(x-2)(x-3)(x-4)$.

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Hint $\ $ Let $\rm\:g(x) = f(x)-x.\:$ Then

$\begin{eqnarray} &&\rm g(0)=0,\ g(1)=0,\ldots,\ g(4) = 0\\ &\iff&\rm\quad x\:|\:g,\quad\ x\!-\!1\:|\:g,\ \ldots,\ x\!-4\:|\:g \\ &\iff&\rm\quad x\,(x\!-\!1)\,\cdots\,(x\!-4)\:|\:g = f-x\\ &\iff&\rm f = x + c\, x\,(x\!-\!1)\,\cdots\,(x\!-4)\quad c\ne 0,\ deg\ c = 0\ \ by\ \ deg\ f = 5\\ &\Rightarrow&\rm f(5) = 5 + c\,5! = 5 + 120\,c\qquad excludes\ A,B,\ includes\ C \end{eqnarray}$