7x-3-mod-18-

Question Number 92255 by jagoll last updated on 05/May/20

7 x = 3 (\mod 18)

Answered by Rasheed.Sindhi last updated on 05/May/20

7 x \equiv 3 (m o d 18)

7 x \equiv 3 + 18 (m o d 18)

7 x \equiv 21 (m o d 18)

x \equiv 3 (m o d 18)

Commented by Joel578 last updated on 06/May/20

y e s, t h e r e a r e m a n y s o l u t i o n s

Commented by jagoll last updated on 06/May/20

only x = 3 sir ?

Commented by Maclaurin Stickker last updated on 06/May/20

x = 18 n + 3 f o r n \in Z

t h i n k t h e r e a r e m a n y s o l u t i o n s \dots

o r a m I w r o n g ?

Commented by Rasheed.Sindhi last updated on 06/May/20

∙ x \equiv 3 (x i s c o n g r u e n t t o 3

m e s n s x = 3 + 18 n f o r n \in Z

∙ x \equiv 3 (m o d 18) m e a n s a l l v a l u e s

w h i c h l e a v e s r e m a i n d e r 3 w h e n

t h e y^{'} r e d i v i d e d b y 18 .

Answered by Rio Michael last updated on 07/May/20

if \gcd (7, 18) = d

and d ∣ 3 \Rightarrow there is a solution

but relative to 18, 7 is prime

so from my suggestion

7 x \equiv 3 (\mod 18) has 1 solution

since d = \gcd (7, 18) ∣ 3 .

these theorem might help .

if a x \equiv b (\mod n)

and \gcd (a, n) = d

then :

(1) d ∣ b implies a solution

(2) a x \equiv b (\mod n) has d solutions .

solution

7 x \equiv 3 (\mod 18)

\gcd (18, 7)

18 = 7 (2) + 4

7 = 1 (4) + 3

4 = 1 (3) + 1

3 = 3 (1) + 0 by {euclid}^{'} s algorithm

moving backwards through the steps in {euclid}^{'} s algorithm .

1 = 4 - 1 (3)

= 4 - 1 (7 - 4)

= 4 - 1 (7) + 1 (4)

= 2 (18 - 7 (2)) - 1 (7)

1 = 2 (18) - 4 (7) - 7

1 = 2 (18) - 5 (7)

\Rightarrow - 5 (7) = 1 - 2 (18)

multiplicative inverse of 7 is - 5 .

\Rightarrow 7 x \equiv 3 (\mod 18)

- 5 \times 7 x \equiv 3 \times - 5 (\mod 18)

\Rightarrow x \equiv - 15 (\mod 18)

x \equiv 3 (\mod 18) is the only solution .

Commented by Maclaurin Stickker last updated on 09/May/20

t h a n k y o u s i r