Solve-the-equation-x-3-mod-5-x-4-mod-7-x-2-mod-3-

Question Number 86791 by TawaTawa1 last updated on 31/Mar/20

Solve the equation :

x \equiv 3 (\mod 5)

x \equiv 4 (\mod 7)

x \equiv 2 (\mod 3)

Answered by mr W last updated on 31/Mar/20

x = 5 k + 3

x = 7 h + 4

x = 3 i + 2

5 k + 3 = 7 h + 4

5 k - 7 h = 1

\Rightarrow k = 7 m - 4 \Rightarrow x = 5 (7 m - 4) + 3 = 35 m - 17

\Rightarrow h = 5 m - 3

35 m - 17 = 3 i + 2

35 m - 3 i = 19

\Rightarrow m = 3 n + 2

\Rightarrow i = 35 n + 17

\Rightarrow x = 35 (3 n + 2) - 17 = 105 n + 53

Commented by TawaTawa1 last updated on 31/Mar/20

How sir please .

Please sir . It is the geimetry question i want you to help me .

Thanks for your time sir .

Commented by TawaTawa1 last updated on 31/Mar/20

I got x = 53 for the congruence sir .

Commented by Joel578 last updated on 31/Mar/20

Sorry sir, it should be

2, 5, 8, 11, 14, 17, \dots

Commented by mr W last updated on 31/Mar/20

t h a n k s! n o w f i x e d .

Answered by Joel578 last updated on 31/Mar/20

with Chinese Remainder Theorem

a_{1} = 3, m_{1} = 5

a_{2} = 4, m_{2} = 7

a_{3} = 2, m_{3} = 3

\Rightarrow m = (m_{1}) (m_{2}) (m_{3}) = 105

\Rightarrow M_{1} = \frac{m}{m_{1}} = 21, M_{2} = \frac{m}{m_{2}} = 15, M_{3} = \frac{m}{m_{3}} = 35

Now find invers modulus of M_{k} \mod m_{k}, or in other words,

find y_{k} such that M_{k} y_{k} \equiv 1 (\mod m_{k})

\Rightarrow y_{1} = 1, y_{2} = 1, y_{3} = 2

\Rightarrow y = a_{1} M_{1} y_{1} + a_{2} M_{2} y_{2} + a_{3} M_{3} y_{3}

= 63 + 60 + 140 = 263

\Rightarrow x \equiv y \mod 105 \Rightarrow x = 53 \to smallest positive value

Commented by TawaTawa1 last updated on 31/Mar/20

God bless you sir

Commented by Ar Brandon last updated on 08/Apr/20

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Answered by Rio Michael last updated on 31/Mar/20

another method (algebraic)

x \equiv 3 (\mod 5) \dots \dots (i)

x \equiv 4 (\mod 7) \dots \dots (ii)

x \equiv 2 (\mod 3) \dots \dots . (iii)

from (i) x = 5 k + 3, k \in Z \dots . . (i) in (ii)

5 k + 3 \equiv 4 (\mod 7)

5 k \equiv 1 (\mod 7) by trial and error

k \equiv 3 (\mod 7) \Rightarrow k = 7 s + 3, s \in Z \Rightarrow x = 5 (7 s + 3) + 3

x = 35 s + 18

\Rightarrow 35 s + 18 \equiv 2 (\mod 3)

35 s \equiv - 16 (\mod 3)

but - 16 \equiv 2 (\mod 3) \Rightarrow 35 s \equiv 2 (\mod 3)

so s \equiv 1 (\mod 3) \Rightarrow s = 3 t + 1

x = 35 (3 t + 1) + 18 \Rightarrow x = 105 t + 53

there for x \equiv 53 (\mod 105)

Commented by TawaTawa1 last updated on 31/Mar/20

God bless you sir

Commented by Rio Michael last updated on 31/Mar/20

you too sir

Commented by TawaTawa1 last updated on 31/Mar/20

I am a girl sir

Commented by Rio Michael last updated on 31/Mar/20

oh welcome madam

Answered by mind is power last updated on 31/Mar/20

2 x = 1 (5)

2 x = 1 (3)

2 x = 1 (7)

\Rightarrow 3, 5, 7 ∣ (2 x - 1)

d i n c e (3, 5, 7) a r e c o p r i m e s

\Rightarrow 3.5.7 ∣ 2 x - 1 \Rightarrow 2 x - 1 = 0 (15.7) = 0 (105)

\Rightarrow 2 x - 1 = 0 (105)

\Rightarrow 2 x = 1 (105)

2 . x . 53 = 53 (105) \Rightarrow x = 53 (105)

Commented by TawaTawa1 last updated on 31/Mar/20

God bless you sir