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Algebra Difficulty 4.5 AIME Find the answer Italy

Problem:

The polynomial P(x)P(x), of degree 42, takes the value 0 at the first 21 odd prime numbers and at their reciprocals (recall that the reciprocal of a positive integer nn is the rational number 1/n1 / n). What is the value of the ratio P(2)/P(1/2)P(2) / P(1 / 2)?

Pick one

Solution

Solution:

The answer is (E)\mathbf{(E)}. Observe that the expression Q(x)=P(x)x42P(1/x)Q(x) = P(x) - x^{42} P(1 / x) is a polynomial, since the monomial x42x^{42} simplifies the denominator of P(1/x)P(1 / x). Moreover, it has degree at most 42, and if rr is one of the first 21 odd prime numbers, Q(x)Q(x) vanishes at rr and at 1/r1 / r: indeed, we have
Q(r)=P(r)r42P(1/r)=0Q(1/r)=P(1/r)(1/r)42P(r)=0, Q(r) = P(r) - r^{42} P(1 / r) = 0 \quad Q(1 / r) = P(1 / r) - (1 / r)^{42} P(r) = 0,
where we used the fact that P(r)=P(1/r)=0P(r) = P(1 / r) = 0 by hypothesis. Finally, Q(x)Q(x) vanishes at 1, because Q(1)=P(1)P(1)=0Q(1) = P(1) - P(1) = 0. Since Q(x)Q(x) vanishes at at least 43 distinct values of xx but has degree at most 42 we obtain that Q(x)Q(x) is the constant polynomial 0, hence P(x)=x42P(1/x)P(x) = x^{42} P(1 / x) holds for every xx. We therefore have
P(2)P(1/2)=242P(1/2)P(1/2)=242=421. \frac{P(2)}{P(1 / 2)} = \frac{2^{42} P(1 / 2)}{P(1 / 2)} = 2^{42} = 4^{21}.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.