Question

Provide the complete solution for the following linear congruences.
3x-2 ≡ 7 (mod 11)
5 ≡ 4x-1 (mod 9)
Provide the complete solution for the following linear congruences.
3x-2 ≡ 7 (mod 11)
5 ≡ 4x-1 (mod 9)
@Mathematics

Answer 1

what does "a complete solution" look like?

Answer 2

first one is
\[x\equiv 3(\text{ mod }11)\] and second is
\[x\equiv 6 ( \text { mod } 9)\] so you need to solve them both?

Answer 3

well for
x ≡ 3 (mod 7)
x ≡ 5 (mod 12)

the solution is x ≡ 17 (mod 84)

Answer 4

it's a system of linear congruences so the x has to be for both

Answer 5

??

Answer 6

where does the 17 come from?

Answer 7

here is an example solution:

n ≡ 5 (mod 12)
n ≡ 11 (mod 18)

The first congruence implies n = 5+12x for some x 2 Z. Substitute this into the
second congruence to get 5 + 12x ≡ 11 (mod 18), or 12x ≡ 6 (mod 18). Since gcd(12, 18) = 6
which divides 6, there is a solution. We can use EEA or by inspection to see that x0 = 2
is one solution, hence by LCT 1, the complete solution to the second congruence is x ≡ 2
(mod 18/6) or x ≡ 2 (mod 3). So x = 2 + 3y for some y 2 Z. Substitute back into n to get
n = 5 + 12(2 + 3y) = 29 + 36y. So the complete solution is n ≡ 29 (mod 36).

Answer 8

those x 2 Z should be x is and element of the integers and y is an element of the integers

Answer 9

ok i think the first equation gives
\[x\equiv 3(\text{ mod }11)\] and the second gives
\[x\equiv 6 ( \text { mod } 9)\] does this look reasonable? then we can solve as above

Answer 10

I will show you what I've got so far, give me a second

Answer 11

\[x=3+11k\]
\[3+11k\equiv 6(\text { mod } 9)\]
\[11k\equiv 3 (\text { mod } 9)\] and by inspection
\[k=6\] works since
\[66\equiv 3( \text{ mod } 9)\]

Answer 12

http://dl.dropbox.com/u/6717478/part.png is what I have so far

Answer 13

it looks like you might have made a mistake in the line that says
\[5\equiv 4x-1 (\text { mod }11)\]
\[4\equiv 4x( \text { mod }11)\]

shouldn't this be
\[6\equiv 4x( \text { mod }11)\]

Answer 14

oh you're right

Answer 15

then
\[4x\equiv 6 ( mod 9)\]
\[2x\equiv 3( mod 9)\]
\[2x\equiv 12(mod 9)\]
\[x\equiv 6 (mod 9)\] or did i make a mistake there? i think this is right,

Answer 16

but you should be able to proceed with your method once you correct that error right?

Answer 17

I will try, when I was writing it out I started working more of it out

Answer 18

good luck!

Answer 19

thanks for your help!

Answer 20

I think the answer is complete solution is x ≡ 69 (mod 99)

Answer 21

i have what i think may be right but you can correct me if i am wrong. i want to solve
\[x\equiv 3(11)\] and
\[x\equiv 6(9)\] so i write
\[x=3+11k\] and then
\[3+11k\equiv 6(9)\]
\[11k\equiv 3(9)\]
and then i keep adding 9 until i can divide by 11 and get
\[11k\equiv 66(9)\] and so
\[k\equiv 3(9)\] so
\[k=3+9j\] and substitute back to get
\[x = 11(3+9j)+3\]
\[x=36+99j\]
and therefore

\[x\equiv 36(99)\] but don't take my word for this

Answer 22

too bad i got 66 and you got 69, because now i doubt my method. let me check my arithmetic

Answer 23

no mine is wrong

Answer 24

in the second equation if you put int 36 then you get
5 ≡ 4(36)-1 (mod 9)
5 ≡ 143 (mod 9)
5 ≡ 8 (mod 9)
which isn't true I don't think

Answer 25

no you are right and i am wrong. i am not sure why but i will try to figure it out

Answer 26

whatever you did worked when i checked for 69 + 99

Answer 27

ok that's a relief haha
I've been stuck on this one for a while

Answer 28

i don't want you to do more work, but isn't
\[5\equiv 4x-1(9)\] the same as
\[x\equiv 6(9)\]?

Answer 29

never mind you are done and i will figure it out on my own

Answer 30

what did you do with the 4?

Answer 31

i put
\[6\equiv 4x(9)\]
\[4x\equiv 6(9)\] then either divide by 2 and solve or add 9 until i can divide by 4
i get
\[4x\equiv 24(9)\] and so
\[x\equiv 6(9)\]

Answer 32

5 ≡ 4x-1 (mod 9)
6 ≡ 4x (mod 9)
6 - 4x ≡ 0 (mod 9)
-4x ≡ -6 (mod 9)

I don't know if you can multiply the - away

Answer 33

you can

Answer 34

oh I don't think you can do division in the linear congruence or else something like 13x ≡ 1 (mod 60) would be very trivial wouldn't it?

Answer 35

or simply switch them
\[6\equiv 4x(9)\]
\[4x\equiv 6(9)\]

Answer 36

I don't think you can divide by the 4 to get rid of it like that

Answer 37

equivalence is an equivalence relation, so i can switch.

Answer 38

at least I've never seen my prof do that so I never have, but I'm not sure

Answer 39

ok well i will figure out my mistake, i will not take any more of your time. good work though!

Answer 40

thanks, and thank you for always helping me!

I may need some more help later on if I get stuck haha