How to solve? 17x =9 (mod 276) Please, help
You could use your favorite way the euclidean way sometimes that can get pretty long though depending on the numbers
I break 276 into 3*4*23 Then it equivalents 17x = 9 mod 3 17x = 9 mod 4 17x =9 mod 23
then, it becomes x= 0 mod 3 x =1 mod4 x =13 mod 23 but can't get the right answer
oh, x = -13 mod 23
the euclidean way: \[276=17(16)+4 \\ 17=4(4)+1 \\ ... \\ 17-4(4)=1 \\ 17-4[276 - 17 \cdot 16]=1 \\ 17-4(276)+17(4 \cdot 16)=1 \\ -4(276)+17(4 \cdot 16+1)=1 \\ -4(276)+17(65)=1\]
oh, I got the right answer with it. Damn!! silly mistake.
and you can multiply both sides of my by 9
Yes, we can use Euclidean, but I would like to solve it by congruence.
By Euclidean, we establish 17x - 276 y =9 and solve diophantine, right?
yes I was call is solving the congurency equation by using the extended euclidean way
always*
\[276(-4 \cdot 9)+17(65 \cdot 9)=9 \\ x=65 \cdot 9 \mod 276\]
I need practice that way also. Tomorrow, I have test for both.
I don't think I ever practiced the other way much
it looks like you wrote the prime factorization for 276 and then setup 3 separate congruncey equations?
Yes, I did
the first one gives: 17x = 9 mod 3 2x = 0 mod 3 x=0 mod 3 second one gives: 17x=9 mod 4 x=1 mod 4 third one gives: 17x=9 mod 23 17x-23k=9 23=17(1)+6 17=6(2)+5 6=5(1)+1 --- 6-5=1 6-(17-6*2)=1 6(3)-17)=1 (23-17)(3)-17=1 23(3)-4(17)=1 So x=-4 mod 23 -- So once you get x=0 mod 3 from first one and x=1 mod 4 for second one and x=-4 mod 23 for last one what do you do?
For the last one 17x = 9 (mod 23) -6x = 9 (mod 23 -2x = 3 (mod 23) -2x = 26 (mod 23) x = -13 (mod 23)
I think there is a theorem somewhere that says you can do that one thing you did from line 2 to 3 but I can't remember what it said
anyways i will let you study for your test
just divide both sides by 3
i made a little mistake or forgot to multiply both sides by 9 again errr x=-4(9) mod 23 =-36 mod 23 which is the same as -13 mod 23
We can divide both sides by 3 iff (3,23) =1, if not, we can't do that.
k thanks for reminding me :)
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