A problem on chinese remainder theorem (CSIR NET DEC 2015)

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Which of the following intervals contains an integer satisfying following three congruences $$x=2\pmod5\\ x=3\pmod7\\ x=4\pmod{11}$$ $a) [401,600] \\ b)[601, 800] \\ c)[801,1000] \\ d)[1001,1200]$

(CSIR NET 2015 Dec)

I tried this question and I got answer but it is not in the option. I applied Chinese remainder theorem.

$$x=2\pmod5\\ x=3\pmod7\\ x=4\pmod {11}$$

$$N_1=7\times11=77\\ N_2=5\times11=55\\ N_3=7\times5=35$$

$77x=1\pmod5\implies b_1=3\\ 55x=1\pmod7\implies b_2=6\\ 35x=1\pmod {11} \implies b_3=6$

then, $x=2\times77\times3+3\times55\times6+6\times35\times4=2292$

This answer is not in the option.

If my work is wrong please correct it.

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There are 2 best solutions below

3
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You know that the answer you get applying the CRT is not a unique integer, right?

It is only unique modulo $5\cdot 7\cdot 11$.

In particular, $752$ and $1137$ are solutions.

I'll leave you with the tasks of deciding if this is complete, and how you will use it to answer your question.

0
On

Here $~5,~7,~11~$ are pairwise prime to each other.
Let $~m=5\times 7\times 11=385.~$
Let $~M_1=\frac m5=77,~~M_2=\frac m7=55,~~M_3=\frac m{11}=35.~$
Then $~~\gcd(M_1,5)=1,~~\gcd(M_2,7)=1,~~\gcd(M_3,11)=1~.$

Now since $~\gcd(M_1,5)=1,~$ the linear congruence $77x\equiv 1~\pmod5~$ has a unique solution and the solution is $~x\equiv 3~\pmod5~.$

Again since $~\gcd(M_2,7)=1,~$ the linear congruence $55x\equiv 1~\pmod7~$ has a unique solution and the solution is $~x\equiv 6~\pmod7~.$

Also since $~\gcd(M_3,11)=1,~$ the linear congruence $35x\equiv 1~\pmod{11}~$ has a unique solution and the solution is $~x\equiv 6~\pmod{11}~.$

Therefore $~x_0=2\cdot(77\cdot3)+3\cdot(55\cdot6)+4\cdot(35\cdot6)=2292~$ is a solution and the solution is unique modulo $~385~.$

Hence the solution of the given system is $~x\equiv 2292~\pmod{385}\equiv 367~\pmod{385}~.$

i.e., the solutions are $$367+385k~,~~~~\text{where}~k=0,~1,~2,\cdots$$ $$=367,~752,~1137,~1422,~\cdots~$$

Answer of the given question:

The intervals contains an integer satisfying the given three congruences are $b)[601, 800]~~~~\text{and}~~~~ d)[1001,1200]$