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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. The modern approach to modular arithmetic was developed by Carl Friedrich Gauss in his book Disquisitiones ...
Sunzi's original formulation: x ≡ 2 (mod 3) ≡ 3 (mod 5) ≡ 2 (mod 7) with the solution x = 23 + 105k, with k an integer In mathematics, the Chinese remainder theorem states that if one knows the remainders of the Euclidean division of an integer n by several integers, then one can determine uniquely the remainder of the division of n by the product of these integers, under the condition ...
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 operands.
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 3 x ≡ 1 (mod 10) will have solutions, that is, modular multiplicative inverses of 3 modulo 10 will exist.
The multiplicative order of a number a modulo n is the order of a in the multiplicative group whose elements are the residues modulo n of the numbers coprime to n, and whose group operation is multiplication modulo n. This is the group of units of the ring Zn; it has φ (n) elements, φ being Euler's totient function, and is denoted as U (n) or ...
The quadratic excess E ( p) is the number of quadratic residues on the range (0, p /2) minus the number in the range ( p /2, p) (sequence A178153 in the OEIS ). For p congruent to 1 mod 4, the excess is zero, since −1 is a quadratic residue and the residues are symmetric under r ↔ p − r.
The former are ≡ ±1 (mod 12) and the latter are all ≡ ±5 (mod 12). −3 is in rows 7, 13, 19, 31, 37, and 43 but not in rows 5, 11, 17, 23, 29, 41, or 47. The former are ≡ 1 (mod 3) and the latter ≡ 2 (mod 3). Since the only residue (mod 3) is 1, we see that −3 is a quadratic residue modulo every prime which is a residue modulo 3.
x ≡ 3 (mod 4) x ≡ 5 (mod 6) Now, for x ≡ 3 (mod 4) to be true, x = 3 + 4j for some integer j. Substitute this in the second equation 3+4j ≡ 5 (mod 6) since we are looking for a solution to both equations. Subtract 3 from both sides (this is permitted in modular arithmetic) 4j ≡ 2 (mod 6)