how would i solve

$\displaystyle x\equiv 2mod3 , x\equiv 5mod8 , x\equiv 11mod13 $ im getting confused around the mod 13 and finding its inverse would appreciate answer..exam revision!thanks

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- May 29th 2008, 12:29 PMskystarchinese remainder theorem
how would i solve

$\displaystyle x\equiv 2mod3 , x\equiv 5mod8 , x\equiv 11mod13 $ im getting confused around the mod 13 and finding its inverse would appreciate answer..exam revision!thanks - May 29th 2008, 12:35 PMMoo
Hello,

The inverse of 11 mod 13 ?

Euclidian algorithm :

$\displaystyle 13=11+2 \implies 2=13-11$

$\displaystyle 11=2*5+1 \implies 1=11-2*5=11-(13-11)*5=11*6-13*5$

Therefore, $\displaystyle 11*6=13*5+1$

--> $\displaystyle 11*6=1 \mod 13$

The inverse of 11 is... ? :) - May 29th 2008, 01:08 PMgalactus
$\displaystyle x\equiv{2(mod3)}$......[1]

$\displaystyle x\equiv{5(mod8)}$......[2]

$\displaystyle x\equiv{11(mod13)}$........[3]

From [1], we have $\displaystyle x=3t+2$

Sub into [2]:

$\displaystyle 3t+2\equiv{5(mod8)}$

$\displaystyle 3t\equiv{3(mod8)}$

$\displaystyle t\equiv{1(mod8)}$

$\displaystyle t=8s+1$

Sub into x=3t+2, 3(8s+1)+2=24s+5

$\displaystyle 24s+4\equiv{11(mod13)}$

$\displaystyle 24s-6\equiv{(mod13)}$

$\displaystyle \frac{24s-6}{13}$

This must be an integer. That occurs first at**s=10**

By resubbing, we get 8(10)+1=82, 3(81)+2=245

**x=245**

We can check this and see it checks on our congruencies

(245-2)/3=81

(245-5)/8=30

(245-11)/13=18

Make sure I didn't go astray somewhere. Make sure this is the smallest. - May 29th 2008, 01:34 PMskystar
where did 3t+2 come from,,typo on [1]?

- May 29th 2008, 01:38 PMgalactus
I had 13 instead of 3 in the first one. Should be mod 3. I had mod 13

- May 29th 2008, 02:20 PMgalactus
Hey, check what I just found. I ran our problem through it and it checks

That reminds me, the solutions should be written as 245 mod 312. I just posted the smallest.

Section 7.3: Solving Lots of Congruences