Linear congruence $53x \equiv 62 \pmod{105}$

Question

Find all solutions that aren’t congruent to each other in modulo $105$ of

$53x \equiv 62 \pmod{105}$

I’ve know that this equation has $1$ solution, because $(53 , 105) = 1$. But, how can I find that solution without brute-forcing?

http://www-math.ucdenver.edu/~wcherowi/courses/m5410/exeucalg.html — saulspatz, May 12 '20 at 13:50

score 0 · Accepted Answer · edited May 13 '20 at 09:00

0

$x\equiv 106x \equiv 62*2 \equiv 19 \pmod{105}$

edited May 13 '20 at 09:00

Martin Sleziak

answered May 12 '20 at 13:45

Pavel Kozlov

score 0 · Answer 2 · answered May 12 '20 at 14:02

0

$53^{-1}\cong2\bmod{105}\implies x\cong 2\cdot62\cong124\cong19\bmod{105}$.

answered May 12 '20 at 14:02

2 Answers2