Linear Congruence with Prime Modulus

Linear Congruence with Prime Modulus
This video discusses how to solve a system of linear congruences for the variable x.
The question is 4x=3 for modulus 5 modulus 7 modulus 11
The first issue in this video is to remove the 4x from the initial congruence for all 3, mod 5, mod 7, mod 11. Thus, leaving the x as a single x but the modulo stay the same. The remainder or the value that the value is congruent to is the first number we require for out solution.
Then we find the product of all the modulo. 3x5x7 =385.
Then we find the modulo concerning each of the 3 congruences. we have 7x11=77, 5x11=55 and 5x7=35. For each of these values we then find the multiplicative inverse of each of these number via the relevant modulo.
Then we find the product of all the relevant numbers to the first congruence and sum it to the product of all the relevant numbers to the second congruence and sum it to the product of all the relevant numbers to the third congruence.
With this total we then find the remainder that is congruent to mod 385.
This is a topic in Number Theory
   • The Liouville Lambda Function  
   • The Distribution of Prime Numbers  
   • The Liouville Function and  Divisors  
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