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
$\displaystyle x\equiv{2(mod3)}$......[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.
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