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Some calculators have a mod() function button, and many programming languages have a similar function, expressed as mod(a, n), for example. Some also support expressions that use "%", "mod", or "Mod" as a modulo or remainder operator, such as a % n or a mod n. For environments lacking a similar function, any of the three definitions above can ...
Modular multiplicative inverses are used to obtain a solution of a system of linear congruences that is guaranteed by the Chinese Remainder Theorem. For example, the system X ≡ 4 (mod 5) X ≡ 4 (mod 7) X ≡ 6 (mod 11) has common solutions since 5,7 and 11 are pairwise coprime. A solution is given by
Integer multiplication respects the congruence classes, that is, a ≡ a' and b ≡ b' (mod n) implies ab ≡ a'b' (mod n). This implies that the multiplication is associative, commutative, and that the class of 1 is the unique multiplicative identity. Finally, given a, the multiplicative inverse of a modulo n is an integer x satisfying ax ≡ ...
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.
This is the group of units of the ring Z n; it has φ(n) elements, φ being Euler's totient function, and is denoted as U(n) or U(Z n). As a consequence of Lagrange's theorem, the order of a (mod n) always divides φ(n). If the order of a is actually equal to φ(n), and therefore as large as possible, then a is called a primitive root modulo n.
c = b e mod m = d −e mod m, where e < 0 and b ⋅ d ≡ 1 (mod m). Modular exponentiation is efficient to compute, even for very large integers. On the other hand, computing the modular discrete logarithm – that is, finding the exponent e when given b , c , and m – is believed to be difficult.
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The discrete logarithm is just the inverse operation. For example, consider the equation 3 k ≡ 13 (mod 17). From the example above, one solution is k = 4, but it is not the only solution. Since 3 16 ≡ 1 (mod 17)—as follows from Fermat's little theorem—it also follows that if n is an integer then 3 4+16n ≡ 3 4 × (3 16) n ≡ 13 × 1 n ...