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Long division is the standard algorithm used for pen-and-paper division of multi-digit numbers expressed in decimal notation. It shifts gradually from the left to the right end of the dividend, subtracting the largest possible multiple of the divisor (at the digit level) at each stage; the multiples then become the digits of the quotient, and the final difference is then the remainder.
Computing the remainder then consists of subtracting multiples of the generator polynomial (). This is just like decimal long division, but even simpler because the only possible multiples at each step are 0 and 1, and the subtractions borrow "from infinity" instead of reducing the upper digits. Because we do not care about the quotient, there ...
E.g.: x**2 + 3*x + 5 will be represented as [1, 3, 5] """ out = list (dividend) # Copy the dividend normalizer = divisor [0] for i in range (len (dividend)-len (divisor) + 1): # For general polynomial division (when polynomials are non-monic), # we need to normalize by dividing the coefficient with the divisor's first coefficient out [i ...
A further problem occurs when a simple linear congruential PRNG is used with the divide-and-take-remainder method of range reduction described above. The problem here is that the low-order bits of a linear congruential PRNG with modulo 2 e are less random than the high-order ones: [6] the low n bits of the generator themselves have a period of ...
Classical modular multiplication reduces the double-width product ab using division by N and keeping only the remainder. This division requires quotient digit estimation and correction. The Montgomery form, in contrast, depends on a constant R > N which is coprime to N, and the only division necessary in Montgomery multiplication is division by R.
Modular exponentiation is the remainder when an integer b (the base) is raised to the power e (the exponent), and divided by a positive integer m (the modulus); that is, c = b e mod m. From the definition of division, it follows that 0 ≤ c < m. For example, given b = 5, e = 3 and m = 13, dividing 5 3 = 125 by 13 leaves a remainder of c = 8.
Divide the highest term of the remainder by the highest term of the divisor (x 2 ÷ x = x). Place the result (+x) below the bar. x 2 has been divided leaving no remainder, and can therefore be marked as used. The result x is then multiplied by the second term in the divisor −3 = −3x. Determine the partial remainder by subtracting 0x − ...
None of the preceding remainders r N−2, r N−3, etc. divide a and b, since they leave a remainder. Since r N −1 is a common divisor of a and b , r N −1 ≤ g . In the second step, any natural number c that divides both a and b (in other words, any common divisor of a and b ) divides the remainders r k .