Question

What is wrong? Solve
\(13x \equiv 6 (mod 15)
Please, help

Answer 1

That is 13x -15y = 6
gcd (13, 15) =1
and 1 | 6, hence there is a solution
15 = 13 +2 , then 2 = 15 -13
13 = (7) 2 + 1, then 1 = 13 -(7) 2 = 13-(7) (15-13) = (8) 13 -(7) 15
multiple both sides by 6
13(48) - 15(42) =6
Hence x = 48 mod 15, or \(x \equiv 3 (mod 15)\)

Answer 2

Now, crazy thing comes.
if I take x = 18, then \(x\equiv 3 (mod 15)\)
then x =18 is one of the solution, right?
Plug back to original one 13*18 = 234 \(\equiv 9 (mod 15)\) not 6 as required. :(

Answer 3

@superdavesuper

Answer 4

@IrishBoy123

Answer 5

But if I stop at 2 = 15 -13, I am ok, since
6 = 15 (3) + 13( -3)
Hence \(x \equiv -3\equiv 12 (mod 15)\)
Test back , take x =12, then 13 * 12 = 156 \(\equiv 6 (mod 15)\)
I don't know what is wrong with the first one.

Answer 6

how about simple typo...
"
15 = 13 +2 , then 2 = 15 -13
13 = (7) 2 + 1, then..."

second line should be 13 = (6)2 + 1
going down dat path, u will get 12 (mod 15) as the right ans :D

Answer 7

damn!! silly mistake. Thank you so much