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sooo not hallal mod cheating
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Let a, b, and m be integers, and m ≥ 2. Prove that:
ab ≡ [ (a mod m) · (b mod m) ] (mod m). So I tried proof by cases:
Top answer:
a quick google search turned up several locations where it is proven as a corollary. Just search for
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Which Statements of congruence are true and which are false and why?
177 _= 17 (mod 8) 871 _= 713 (mod 29) 1322 _= 5294 (mod 12)
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To determine which statements of congruence are true and false, we need to apply the definition of
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Which Statements of congruence are true and which are false and why?
177 _= 17 (mod 8) 871 _= 713 (mod 29) 1322 _= 5294 (mod 12)
Top answer:
To determine if the given statements of congruence are true or false, we need to apply the
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Which Statements of congruence are true and which are false and why?
177 _= 17 (mod 8) 871 _= 713 (mod 29) 1322 _= 5294 (mod 12)
Top answer:
177-17 = 160 = 20*8 true 871-713 = 158 false 1322-5294 = -3972 = -331*12 true and so on
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Which two is true as i'm confused
A) 3+7 �ß 10 mod 15 17 + 9 �ß 4 mod 21 12 + 14 �ß 0 mod 26 B) 4+11 �ß 2 mod 13 9+7 �ß
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A) 3+7 ≡ 10 mod 15 3+7=10, 10÷15=0 remainder 10:TRUE. 17 + 9 ≡ 4 mod 21 17+9=26; 26÷21
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Prove that for the DSA:
(αe1 βe2 mod p) mod q = (αk mod p) mod q
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To prove this, we need to show that: (αe1 βe2 mod p) mod q = (αk mod p) mod q First, let's expand
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Find all numbers $r$ for which the system of congruences:
x == r mod 6 x == 9 mod 20 x == 4 mod 45 has a solution.
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as google will provide, there are many similar exercises online. You might start here:
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What are the 3 solutions? I'm stuck! 6x=15(mod 21)
a=6,m=21,b=15 d=gcd(6,21)=3 solns. 6x=15(mod 21) 2x=5(mod 7) 21=6(3)+3
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To find the solutions to the equation 6x ≡ 15 (mod 21), we can first simplify it by dividing both
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how do we find the least residue of
1789 (mod 4), (mod 10), (mod 101)
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If I remember modular arithmetic 1789(mod4) = 1 1789(mod10) = 9 1789(mod101) = 72 I don't know what
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how do we find the least residue of
1789 (mod 4), (mod 10), (mod 101)
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To find the least residue, we can use the modulus operator or the remainder when dividing. 1. Modulo
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If you are going to cheat, at least be smart about it. If you do it on a quick check, it's not necessarily cheating, you're
What is the theme of the passage below?
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