
#1
Dec1604, 07:08 PM

P: 6

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? 



#2
Dec1604, 10:50 PM

Sci Advisor
HW Helper
P: 11,863

I found x=47.What did u find?? Daniel. 



#3
Dec1804, 06:05 PM

P: 6

i got x== 2(mod 35).




#4
Dec1804, 06:11 PM

P: 6

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) l=5m x= 7 (5m) +2= 35m +2 or x== 2 (mod 35) 



#5
Dec1804, 06:49 PM

Sci Advisor
HW Helper
P: 11,863

Here's how i see it: x==7(mod 10)=>x=10a+7,a={0,1,...}(1) x==2(mod 5)=>x=5b+2,b={0,1,...}(2) x==12(mod 35)=>x=35c+12,c={0,1,...}(3) You wanna find "x" and hence "a","b","c". From (1) and (2)=>b=2a+1,a={0,1,...}(4) From (1) and (3)=>a=(7c+1)/2,c={1,3,5,...}(5) From (4) and (5)=>b=7c+2,c={1,3,5,...}(6) From (3),(5) and (6) u can write the solution to the problem [tex] x=35c+12 [/tex](7) ,where "c" can take only positive odd values (eq.(5)/(6)). So [tex] x\in\{47,117,187,...\}[/tex] Daniel. PS.I solved the problem in the positive numbers set,for the set of integers it can be generalized straightforwardly. 


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