Solve the simultaneous congruences

2x = 1 (mod7)

x= 3(mod5)

x= 3(mod8)

those aren't supposed to be equal signs- they are three slashes(congruences).

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- Nov 7th 2007, 04:33 AManncarcongruences
Solve the simultaneous congruences

2x = 1 (mod7)

x= 3(mod5)

x= 3(mod8)

those aren't supposed to be equal signs- they are three slashes(congruences). - Nov 7th 2007, 05:32 AMangel.white

I would do the first one, but I'm out of time, gotta get ready for work, and wouldn't want to steer you wrong.

For the second one, you can see that 3=5*0+3, so when x = 3, it is congruent to 3(mod5). So find the next in the progression by just adding 5, so x=3+5n for any integer n (this creates the series ...,3,8,13...)

For the third one, you can see that 3=8*0+3, so when x=3 it is congruent to 3(mod8). So find the next progression by just adding 8, so x=3+8n for any integer n (this creates the series ...3,11,19...) - Nov 7th 2007, 05:34 AMtopsquark
See here.

First let's solve the first congruence for x:

So we need to solve

The modulos are pairwise coprime, so we may state a solution exists by the Chinese Remainder Theorem.

Use the extended Euclidean algorithm to find integer r1 and s1 such that

One such pair is and .

Now do the same for

<-- r2 = -17 and s2 = 3

and

<-- r3 = -13 and s3 = 3

Now, define , , and .

Then a solution to the system will be:

which you may verify is a correct solution.

Edit: Whoops! I just noticed. This answer will be correct a modulo 5 x 7 x 8 = 280. So we have a smaller answer which is equivalent to . So we can use x = 163 as the solution.

-Dan

(See, you can teach an old dog new tricks!) - Nov 7th 2007, 09:14 AMangel.white
Oh, my apologies, I looked at it as three independent equations, just ignore my earlier post.