Diophantine Question

Directly quoting from Wikipedia: "the Chinese remainder theorem states that if one knows the remainders of the Euclidean division of an integer $n$ by several integers, then one can determine uniquely the remainder of the division of $n$ by the product of these integers, under the condition that the divisors are pairwise coprime".

In this case, our divisors are $7,11,13$ (since $1001=7\cdot11\cdot13$ ). We have three brackets in the factorisation, and three divisors. As an example of how we can use this, say we want to find $x$ such that $7|x-3,\;\;11|x-3,\;\;13|x+3$ (assigning divisors to brackets). The CRT states there will be exactly one solution in the range (in this case it's $x=465$ , but the point is we don't have to work out the particular value).

Any one of the divisors can divide at most one of the brackets; so there are $3^3=\boxed{27}$ such integers.

We have that $x$ satisfies the condition if and only if $(x-3)(x+1)(x+3)$ is divisible by $7,11,$ and $13,$ and since these are all prime, this is equivalent to $\begin{aligned} x &\equiv -3, -1, 3 \pmod 7 \\ x &\equiv -3, -1, 3 \pmod{11} \\ x &\equiv -3, -1, 3 \pmod{13} \end{aligned}$ and each choice of one of the three possibilities gives a unique solution mod $1001,$ so there are $3 \cdot 3 \cdot 3 = \fbox{27}$ solutions.