A b mod m. . ((a×b) mod M) I am learning about modular extraction and to efficiently calculate (a**b)%c. On our homework assignment, one of the questions is to prove the following (ab) mod m = [(a mod m Jul 11, 2025 · Your All-in-One Learning Portal: GeeksforGeeks is a comprehensive educational platform that empowers learners across domains-spanning computer science and programming, school education, upskilling, commerce, software tools, competitive exams, and more. Modular Arithmetic Remember: a ≡ b (mod m) means a and b have the same remainder when divided by m. modular exponentiation mod 7modular exponentiation mod 7 Mar 5, 2015 · You'll need to complete a few actions and gain 15 reputation points before being able to upvote. The following articles are prerequisites for this. BASIC PROPERTIES OF CONGRUENCES The letters a; b; c; d; k represent integers. The letters m; n represent positive integers. In simple terms, to compute (a/b) % M we instead calculate (a × b-1 ) % M where b-1 is the modular inverse of b modulo M ( i. Actually, you need to multiply A by B in power of f (P)-1 and f (P) for prime number is P-1. Compute the result of the modular multiplication of a and b under modulo M. The "mod m" in a ≡ b (mod m) is a note on the side of the equation indicating what we mean when we say " ≡ " Fact: These two uses of "mod" are quite related: a ≡ b (mod m) if and only if a mod m = b mod m. e. Learn how it works with addition, subtraction, multiplication, and division using rules. , b × b-1 ≡ 1 mod M ). Why is the formula (ab) (mod m) = (a (b (mod m))) (mod m) true? Although I am familiar with the concept of the modulus operator, I have little experience in modular algebra. a and b congruent modulo m, written only if a – b is divisible by m are said to be ≡ b (mod m), There is a small mistake in your explanation. So one can May 24, 2024 · What is modular arithmetic with examples. What's reputation and how do I get it? Instead, you can save this post to reference later. Rather, a ≡ b (mod m) asserts that a and b have the same remainder when divided by m. \ (a\) is congruent to \ (b\) modulo \ (m\) denoted as \ ( a \equiv b (mod \, n) \), if \ (a\) and \ (b\) have the remainder when they are divided by \ (n\), for \ (a, b \in \mathbb {Z}\). In fact there are five different definitions on the Wikipedia page alone! Although I would say that the statement you're trying to prove is practically part of the definition Jul 12, 2025 · Given three integers a, b, and M, where M is the modulus. That is, a = p m + r, b = q m + r, where 0 ≤ r < m is the common Relation between ”x ≡ b mod m” and ”x = b MOD m” ny solutions for x while x = b MOD m is an EQUALITY. Oct 23, 2021 · A good first step is to recall the formal definition of a mod m a mod m -- more specifically, to recall the formal definition previously given by whoever stated the theorem, because that will give specific steps to take. I am confused by the wikipedia article explaining the algorithm I found. Jul 12, 2025 · Q is the quotient B is the divisor R is the remainder The modulo operator (mod) helps us focus on the remainder: A mod B=R Example: 1 3 5 = 2 r e m a i n d e r 3 513 = 2 remainder 3 13mod 5=3 So, dividing 13 by 5 gives a remainder of 3 What is Modular Arithmetic? In modular arithmetic, numbers are reduced within a certain range, defined by the The congruence relation may be rewritten as a = k m + b, explicitly showing its relationship with Euclidean division. The notation a b (mod m) means that m Jul 14, 2025 · Instead, division is performed by multiplying the dividend by the modular multiplicative inverse of the divisor under a given modulus. I'm taking a cryptography class and we are going over proofs and modulus equations for hashing. Implementation to match this specification is left up to the reader. Upvoting indicates when questions and answers are useful. Can you help? I've defined the mod m o d operator to always return a value between 0 0 and m − 1 m 1, and noted that this may not match the % operator in some programming languages (see "terminology and notation" section). Jun 8, 2015 · I have to compute efficiently a^^b mod m for large values of a,b,m<2^32 where ^^ is the tetration operator: 2^^4=2^(2^(2^2)) m is not a prime number and not a power of ten. Let \ (m\) \ (\in\) \ (\mathbb {Z_+}\). However, the b here need not be the remainder in the division of a by m. Doesn't include zero a, b ∈ Z, m ∈ N. hkzero eajn xfcdp tbjxn xfg euek bmjsicet sef sdo ipx
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