find-the-value-of-x-such-that-x-2-mod-5-x-3-mod-8-x-2-mod-3-

Question Number 127111 by benjo_mathlover last updated on 26/Dec/20

f i n d t h e v a l u e o f x s u c h t h a t

{\begin{cases} x = 2 (m o d 5) \\ x = 3 (m o d 8) \\ x = 2 (m o d 3) \end{cases}

Answered by liberty last updated on 04/Jan/21

g i v e n {\begin{cases} x = 2 (m o d 5) \dots (i) \\ x = 3 (m o d 8) \dots (i i) \\ x = 2 (m o d 3) \dots (i i i) \end{cases}

f o r (i) \Rightarrow 24 a \equiv 2 (m o d 5)

- a \equiv 2 (m o d 5); a \equiv - 2 (m o d 5)

f o r (i i) \Rightarrow 15 b \equiv 3 (m o d 8)

- b \equiv 3 (m o d 8); b \equiv - 3 (m o d 8)

f o r (i i i) \Rightarrow 40 c \equiv 2 (m o d 3)

c \equiv 2 (m o d 3)

n o w w e h a v e t h e g e n e r a l s o l u t i o n

∴ 24 (- 2) + 15 (- 3) + 40 (2) + 120 k; k \in Z

i . e : 120 k - 13; k \in Z or 120 k + 107; k \in Z

Answered by physicstutes last updated on 27/Dec/20

x \equiv 2 (\mod 3) \dots . . (i) \Rightarrow x = 3 t + 2, t \in Z \dots \dots (i)

x \equiv 3 (\mod 8) \dots \dots (i i)

(i) in (i) \Rightarrow 3 t + 2 \equiv 3 (\mod 8)

3 t \equiv - 1 (\mod 8)

but - 1 \equiv 7 (\mod 8) therefore

3 t \equiv 7 (\mod 8) by trial and error we find a number t which when

multiplied by 3 and divided by 8 gives a remainder of 7 .

so i find t = 3 after some tries .

\Rightarrow t \equiv 3 (\mod 8) \Rightarrow t = 8 s + 3, s \in Z \dots \dots (i i i)

putting (i i i) in (i) \Rightarrow x = 3 (8 s + 3) + 2 = 24 s + 11

so : x = 24 s + 11 \dots . . (i v)

x \equiv 2 (\mod 5) \dots \dots (v)

(i v) in (v) \Rightarrow 24 s + 11 \equiv 2 (\mod 5)

24 s \equiv - 9 (\mod 5)

but - 9 \equiv 1 (\mod 5) therefore 24 s \equiv 1 (\mod 5) by trial and error

s \equiv 4 (\mod 5) \Rightarrow s = 5 u + 4 \dots . . (v i i) u \in Z

\Rightarrow x = 24 (5 u + 4) + 11

x = 120 u + 107

x \equiv 107 (\mod 120)

Answered by floor(10²Eta[1]) last updated on 27/Dec/20

x \equiv 3 (\mod 8) \Rightarrow x = 8 a + 3, a \in Z

8 a + 3 \equiv 3 a + 3 \equiv 2 (\mod 5)

3 a \equiv 4 (\mod 5) \Rightarrow a \equiv 3 (\mod 5)

\Rightarrow a = 5 b + 3

\Rightarrow x = 8 (5 b + 3) + 3 = 40 b + 27, b \in Z

40 b + 27 \equiv b \equiv 2 (\mod 3) \Rightarrow b = 3 c + 2

\Rightarrow x = 40 (3 c + 2) + 27

\Rightarrow x = 120 c + 107, c \in Z