Remainders of 1, 2, 3, 4

Let the integer be $n$ . Then the question is as follows:

$x=5a+1$

$x=7b+2$

$x=9c+3$

$x=11d+4$

Multiplying these equations by $2$ so that the remainders would also count by 2's, just as the divisors are:

$2x=5(2a)+2$

$2x=7(2b)+4$

$2x=9(2c)+6$

$2x=11(2d)+8$

Now add $3$ to each equation so that the value of $n$ would be divisible by all the known divisors in the problem:

$2x+3=5(2a)+5=5(2a+1)$

$2x+3=7(2b)+7=7(2b+1)$

$2x+3=9(2c)+9=9(2c+1)$

$2x+3=11(2d)+11=11(2d+1)$

This means that $2x+3$ is the LCM of $5,7,9,11$ , which is $3465$

So $2x+3=3465$ and $\boxed{x=1731}$

@ABHISHEK Sahoo this is the question you are asking for