Thread: Math Q&A Game
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Dec5-07, 01:54 AM
P: 163
Quote Quote by AKG View Post
Or maybe a better one to do is this: first evaluate p(1), that's the sum of all the coefficients. Then pick a prime P > p(1), and evaluate p(P). Then if you consider p(P) mod P, p(P) mod P, ..., p(P) mod Pd or something you can recover the coefficients.
That's essentially it, but there's still some extra math required if you pick a prime. Instead, let [tex]n=\lceil \log_{10} p(1) \rceil[/tex]. Then evaluate [tex]p(10^n)[/tex]. The k-th coefficient can be recovered by simply reading off the k-th block of n digits.


p(x) = 99 x^2 + 23.
p(1) = 99+23 = 122.
n = 3

p(1000) = 99 (1000)^2 + 23 = 099, 000, 023