Скачать или смотреть 12𝑥 ≡ 6 (mod 21) | Abstract Algebra | Number Theory | Discrete Math

In this video we solve the congruence 12x≡6(mod21) step by step. We turn the congruence into an equation, reduce by the greatest common divisor, use the existence criterion gcd(a,m) divides b, and find a modular inverse to get every solution. After dividing both sides by 3 we get 4x≡2(mod7), note that gcd(4,7)=1, observe 4^(−1)≡2(mod7), and conclude the general solution x≡4(mod7). We also check sample values back in the original modulus 21 to verify the result. Perfect for modular arithmetic, congruences, Euclid’s algorithm, gcd, and linear Diophantine equations practice.

Proof of: gcd(𝑎, 𝑚) divides 𝑏 ⇔ 𝑎𝑥 ≡ 𝑏 (mod 𝑚) has solutions can be found in this video:    • 8𝑥 ≡ 3 (mod 10) | Abstract Algebra | Numbe...  

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Chapters:
00:00 Introduction
00:27 From congruence to equation and reducing by gcd
01:50 Solution Existence
02:30 Simplifying to 4x≡2(mod7)
03:20 Modular Inverse of 4 mod 7
04:40 General Solution x≡4(mod7) and Verification
07:40 Thanks for Watching

Audio dubbing is available in multiple languages, including French, German, Hindi, Indonesian, Italian, Japanese, Korean, Dutch, Portuguese, and Spanish.

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