Recent content by oliver$

  1. O

    Using the Chinese Remainder Theorem to Solve Systems of Linear Congruences

    ok, so from x== 7 (mod 10) x== 2 (mod 5) x== 12 (mod 35) i got: 7 + 10t== 12 (mod 35), which reduces 10t==5 (mod 35) (10, 35)= 5 10r + 35s = 5 r=-3, s= 1 (-3)(5)/5 = -5 t= -5 (mod 7)== 2 (mod 7) t= 2 + 7l == 2 (mod 5) 7l== 0 (mod 5) (7,5)= 1 7r + 5s=0 r=0 s=0 x= 0 (mod 5)...
  2. O

    Using the Chinese Remainder Theorem to Solve Systems of Linear Congruences

    i got x== 2(mod 35).
  3. O

    Using the Chinese Remainder Theorem to Solve Systems of Linear Congruences

    ok, am i doing this right? here's the system: 3x== 1 (mod 10) 4x== 3 (mod 5) 3x== 1 (mod 35) (3,10), (4,5), (3,35) all are 1. it reduces to: x== -3 (mod 10) x== 12 (mod 35) x== 4 (mod 5) and then i use the CRT to solve this system? is that right?
  4. O

    Undergrad Are There Multiple Solutions to this System of Linear Congruences?

    the system: 24x + 11y= 4 (mod 35) 5x + 7y= -13 (mod 35) is solved to get: -113y= 111 (mod 35) 113x= 171 (mod 35) which gives: (17,8). should there not be more solutions?
  5. O

    Undergrad How Do You Solve 4x=7 (mod 45) Using Remodularization Concepts?

    i tried it your way, finding the multiplicative inverse, and got x=13. Then I tried my method with 4x= 0 (mod 45) to get x=0, which works with your method as well. i don't know where i got the formula. but thank you. i'd rather do the problem right than do it by fluke.
  6. O

    Undergrad How Do You Solve 4x=7 (mod 45) Using Remodularization Concepts?

    i'm not sure if i understand this concept at all. for finding all the solutions (mod 45) of 4x=7 (mod 45), we know that solutions exist, b/c the gcd (4, 45)=1 which divides 7. then 4x= [52,97,112,157] (mod 45*4) 4x= 52 (mod 180) and 4x=112(mod 180). these reduce to x=13 (mod 45) and...