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Find the smallest positive integer that satisfies both of the following equations: x≡3(mod4) and x≡5(mod6).

Respuesta :

ChoiSungHyun

Step-by-step explanation:

x≡3(mod4) and x≡5(mod6)

=> x≡-1(mod4) and x≡-1(mod6)

Since LCM of 4 and 6 is 12,

=> x≡-1(mod12)

=> x≡11(mod12)

The smallest positive integer for x is 11.

CSMedrano

Answer:

11

Step-by-step explanation:

Because 4 and 6 aren't co prime we can't start the chinese remainder theorem

so first we check

[tex]x\equiv 3\mod4\\\\\rightarrow x\equiv 3 \mod 2\\\\\rightarrow x \equiv 3 \mod 2[/tex]

because 4 = 2*2

[tex]x\equiv3\equiv1\mod2[/tex]

and the other one

[tex]x\equiv5\mod6\\\\\rightarrow x\equiv5\mod2\\\\\rightarrow x\equiv5\mod3\\\\x\equiv5\equiv1\mod2\\\\x\equiv5\equiv2\mod3[/tex]

so now we have

[tex]x\equiv3\mod4\\\\x\equiv1\mod2\\\\x\equiv2\mod3[/tex]

but the mod 2 we don't need it

so now we have

[tex]x\equiv3\mod 4\\\\x\equiv 2\mod 3\\\\[/tex]

for the first one we can say that

[tex]x=4n+3[/tex]

so we plug in that in the second one

[tex]4n+3\equiv2\mod3\\\\4n\equiv2\mod3\\\\n\equiv2\mod3[/tex]

we can say that

[tex]n=3k+2[/tex]

so for x

[tex]x=4(3k+2)+3\\\\x=12k+8+3\\\\x=12k+11[/tex]

so if k=0

a solution is 11

we check if it works

[tex]11\equiv8+3\equiv0+3\equiv3\mod4[/tex]

[tex]11\equiv6+5\equiv0+5\equiv5\mod6[/tex]

so it works so the smallest solution is 11