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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 ...
Time-keeping on this clock uses arithmetic modulo 12. Adding 4 hours to 9 o'clock gives 1 o'clock, since 13 is congruent to 1 modulo 12. In mathematics, modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" when reaching a certain value, called the modulus.
Even without knowledge that we are working in the multiplicative group of integers modulo n, we can show that a actually has an order by noting that the powers of a can only take a finite number of different values modulo n, so according to the pigeonhole principle there must be two powers, say s and t and without loss of generality s > t, such that a s ≡ a t (mod n).
On the other hand, the primes 3, 7, 11, 19, 23 and 31 are all congruent to 3 modulo 4, and none of them can be expressed as the sum of two squares. This is the easier part of the theorem, and follows immediately from the observation that all squares are congruent to 0 (if number squared is even) or 1 (if number squared is odd) modulo 4.
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 ) .
Then () = means that the order of the group is 8 (i.e., there are 8 numbers less than 20 and coprime to it); () = means the order of each element divides 4, that is, the fourth power of any number coprime to 20 is congruent to 1 (mod 20). The set {3,19} generates the group, which means that every element of (/) is of the form 3 a × 19 b (where ...
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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 ...