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The Chinese remainder theorem is widely used for computing with large integers, as it allows replacing a computation for which one knows a bound on the size of the result by several similar computations on small integers. The Chinese remainder theorem (expressed in terms of congruences) is true over every principal ideal domain.
The polynomial remainder theorem may be used to evaluate () by calculating the remainder, . Although polynomial long division is more difficult than evaluating the function itself, synthetic division is computationally easier. Thus, the function may be more "cheaply" evaluated using synthetic division and the polynomial remainder theorem.
The rings for which such a theorem exists are called Euclidean domains, but in this generality, uniqueness of the quotient and remainder is not guaranteed. [8] Polynomial division leads to a result known as the polynomial remainder theorem: If a polynomial f(x) is divided by x − k, the remainder is the constant r = f(k). [9] [10]
This is the form of the remainder term mentioned after the actual statement of Taylor's theorem with remainder in the mean value form. The Lagrange form of the remainder is found by choosing () = + and the Cauchy form by choosing () =. Remark.
The Chinese remainder theorem appears as an exercise [16] in Sunzi Suanjing (between the third and fifth centuries). [17] (There is one important step glossed over in Sunzi's solution: [note 4] it is the problem that was later solved by Āryabhaṭa's Kuṭṭaka – see below.)
Divide the highest term of the remainder by the highest term of the divisor (3x ÷ x = 3). Place the result (+3) below the bar. 3x has been divided leaving no remainder, and can therefore be marked as used. The result 3 is then multiplied by the second term in the divisor −3 = −9. Determine the partial remainder by subtracting −4 − (− ...
The monkey and the coconuts is the best known representative of a class of puzzle problems requiring integer solutions structured as recursive division or fractionating of some discretely divisible quantity, with or without remainders, and a final division into some number of equal parts, possibly with a remainder. The problem is so well known ...
Hermite interpolation problems are those where not only the values of the polynomial p at the nodes are given, but also all derivatives up to a given order. This turns out to be equivalent to a system of simultaneous polynomial congruences, and may be solved by means of the Chinese remainder theorem for polynomials.