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[1] For example, the expression "5 mod 2" evaluates to 1, because 5 divided by 2 has a quotient of 2 and a remainder of 1, while "9 mod 3" would evaluate to 0, because 9 divided by 3 has a quotient of 3 and a remainder of 0. Although typically performed with a and n both being integers, many computing systems now allow other types of numeric ...
In fact, x ≡ b m n −1 m + a n m −1 n (mod mn) where m n −1 is the inverse of m modulo n and n m −1 is the inverse of n modulo m. Lagrange's theorem : If p is prime and f ( x ) = a 0 x d + ... + a d is a polynomial with integer coefficients such that p is not a divisor of a 0 , then the congruence f ( x ) ≡ 0 (mod p ) has at most d ...
Modulo is a mathematical jargon that was introduced into mathematics in the book Disquisitiones Arithmeticae by Carl Friedrich Gauss in 1801. [3] Given the integers a, b and n, the expression "a ≡ b (mod n)", pronounced "a is congruent to b modulo n", means that a − b is an integer multiple of n, or equivalently, a and b both share the same remainder when divided by n.
If G is simple, and |G| = 30, then n 3 must divide 10 ( = 2 · 5), and n 3 must equal 1 (mod 3). Therefore, n 3 = 10, since neither 4 nor 7 divides 10, and if n 3 = 1 then, as above, G would have a normal subgroup of order 3, and could not be simple. G then has 10 distinct cyclic subgroups of order 3, each of which has 2 elements of order 3 ...
If p is an odd prime and p − 1 = 2 s d with s > 0 and d odd > 0, then for every a coprime to p, either a d ≡ 1 (mod p) or there exists r such that 0 ≤ r < s and a 2 r d ≡ −1 (mod p). This result may be deduced from Fermat's little theorem by the fact that, if p is an odd prime, then the integers modulo p form a finite field , in which ...
Let A 1 be the set whose elements are the numbers b 1, ab 1, a 2 b 1, ..., a k − 1 b 1 reduced modulo p. Then A 1 has k distinct elements because otherwise there would be two distinct numbers m , n ∈ {0, 1, ..., k − 1 } such that a m b 1 ≡ a n b 1 (mod p ) , which is impossible, since it would follow that a m ≡ a n (mod p ) .
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However, the linear congruence 4x ≡ 6 (mod 10) has two solutions, namely, x = 4 and x = 9. The gcd(4, 10) = 2 and 2 does not divide 5, but does divide 6. Since gcd(3, 10) = 1, the linear congruence 3x ≡ 1 (mod 10) will have solutions, that is, modular multiplicative inverses of 3 modulo 10 will exist. In fact, 7 satisfies this congruence (i ...