this problem Section 4 4, problem 4(a) These three congruences lift to a unique class modulo n = 3 · 5 · 7 = 105: x ≡ 52 (mod 105)
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Modular Arithmetic Practice Joseph Zoller September 13, 2015 Practice Problem Solutions 1 Given that 5x ≡ 6 (mod 8), find x [Solution: 6] 2 Find the last
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Solving Modular Equivalences Solving a Question: Solve the congruence 27y ≡ 10 (mod 4) How do we compute the multiplicative inverse of x modulo m?
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Now suppose that we wish to solve the congruence ax ≡ b mod n where d = gcd( a This is the content of the following theorem which generalizes this problem
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this problem Section 4 4, problem 4(a) These three congruences lift to a unique class modulo n = 3 · 5 · 7 = 105: x ≡ 52 (mod 105)
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We will apply the theorem from class that fully describes the solutions of linear congruences a) Solve 3x ≡ 2 (mod 7) Since (3,7) = 1 there is exactly one solution
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consider how to solve systems of simultaneous linear congruences Example We solve the system 2x ≡ 5 (mod 7); 3x ≡ 4 (mod 8) of two not the only way to solve such problems; the technique presented at the beginning of this lecture is
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way of efficiently solving the second problem use to solve such problems powers of 10 are congruent to 1 mod 3, since we can continue multiplying our
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The key to solving this problem is realizing that the times will repeat themselves modulo n that have a common remainder when divided by n, and we say two
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This is the well-known RSA problem: Given an RSA modulus N, a public exponent e and a ciphertext c ≡ me (mod N), find the corresponding plaintext m If me < N
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27 sept 2011 · (c) We check that gcd(4, 10) = 2 Now 2 6, so we can divide through by 2 and reduce to solving the congruence 2x ≡ 3 (mod 5)
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