Question: Solve the equation 27y = 12 Solution: We Solving a Modular Congruence Now, we Question: Solve the congruence 27y ≡ 10 (mod 4) Note: We
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A simple consequence is this: Any number is congruent mod n to its remainder when Here is another approach: Start with the equation 5x ≡ 1 mod 12
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Solving the congruence ax ≡ b (mod m) is equivalent to solving the linear diophantine equation ax − my = b Since we already know how to solve linear
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Question: Solve the equation 27y = 12 Solution: We Solving a Modular Congruence Now, we Question: Solve the congruence 27y ≡ 10 (mod 4) Note: We
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That is, find sa ` tm “ 1, where s and t are integers Reducing both sides of this equation modulo m tells us that s is an inverse of a modulo m Example 2 Find an
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are not congruent mod (n), that is, that they leave different remainders when the congruence f(x) == 0 mod (n) has no solutions x, then the equation f(x) = 0
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Solving the congruence 42x ≡ 12 (mod 90) is equivalent to solving the equation 42x = 12+90q for integers x and q This reduces to 7x = 2+15q, or 7x ≡ 2 (mod
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30 mai 2015 · Two integers are said to be equivalent (or congruent) modulo a if their A linear congruence equation is a congruence that has a variable
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The equation ax ≡ b (mod n) has a solution if and only if d = gcd(a, n) divides b 6 Page 7 Proof The proof follows by writing out the definitions carefully and
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When we solve a linear equation ax ≡ b (mod n) but gcd(a, n) > 1, if gcd(a, where n = n1n2 ···nk, that lifts simultaneously all of the congruence classes listed
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CSE 311: Foundations of Computing I
Solving Modular Equivalences
Solving a Normal Equation
First, we discuss an analogous type of question when using normal arithmetic.Question:Solve the equation27y= 12.
Solution:We divide both sides by 27 to gety=1227
Solution:We multiply both sides by 1/27 to gety=1227 These solutions are two ways of saying the same thing.Solving a Modular Congruence
Now, we consider a congruence instead:
Question:Solve the congruence27y10 (mod 4).
Note:We can"t just divide both sides. For example, consider510 (mod 5). If we were to divide both sides by 5, we would get12 (mod 5)which is definitely false. Another way of looking at this would be to ask the question What is 15 mod5? It really doesn"t make any sense, because remainders should always be integers. So, instead, we need to create machinery to multiply by whatever thecorrectinverse is mod a number.Inverses
Ifxy= 1, we say thatyis the "multiplicative inverse ofx". We have a similar idea modm: Ifxy1 (modm), we sayyis the "multiplicative inverse ofxmodulo m". How do we compute the multiplicative inverse ofxmodulom? By definition,xy1 (modm)iffxy+tm= 1for somet2Z. We know by Bezout"s Theorem that we can findyandtsuch thatxy+tm= gcd(x;m). Said another way: Ifgcd(x;m) = 1, then we can find a multiplicative inverse!To actually compute the multiplicative inverse, we use the Extended Euclidean Algorithm. For example,
consider the equation we were trying to solve above:27y10 (mod 4). First, we find the multiplicative inverse of 27 modulo 4. That is, we find aysuch that27y1 (mod 4). To do this, we first note that thegcd(27;4) = gcd(4;3) = gcd(3;1) = gcd(1;0) = 1, which means an inverse does exist!