how would i solve
im getting confused around the mod 13 and finding its inverse would appreciate answer..exam revision!thanks
......[1]
......[2]
........[3]
From [1], we have
Sub into [2]:
Sub into x=3t+2, 3(8s+1)+2=24s+5
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