Find the thousandth number in the sequence of numbers relatively prime to $105$. 
Suppose that all positive integers which are relatively prime to $105$ are arranged into a increasing sequence: $a_1 ,  a_2 ,  a_3, . . . .$ Evaluate $a_{1000}$

By inclusion exclusion principle I found the following equation:
$n - \left( \left\lfloor \dfrac {n} {3} \right\rfloor + \left\lfloor \dfrac {n} {5} \right\rfloor + \left\lfloor \dfrac {n} {7} \right\rfloor \right) + \left( \left\lfloor \dfrac {n} {15} \right\rfloor + \left\lfloor \dfrac {n} {21} \right\rfloor + \left\lfloor \dfrac {n} {35} \right\rfloor \right) - \left\lfloor \dfrac {n} {105} \right\rfloor=1000$
I tried to solve the equation by setting $n=105k+r ; 0 \leq 0<105$.  But this method is so tiring and lengthy to solve. I thought to invoke inequalities like $a-1 <\lfloor a \rfloor \leq a$, but I couldn't do as the equation contains positive as well as negative terms .Please provide me any other method to solve such equations involving floor function. Any help would be appreciated.

The number of integers $n \in \{1,2,3,\dots, 105\}$ that are relatively prime to $105$ is
$$\phi(105) = \phi(3 \times 5 \times 7) = 2 \times 4 \times 6 = 48$$
We know that $\gcd(105A + n, 105) = \gcd(n, 105)$.
Since $1000 = 20\times 48 + 40$, we know that there are $20 \times 48 = 960$ numbers relatively prime to $105$ in the interval $[1, 2100]$ (where $2100 = 20 \times 105$).
$86$ is the $40^{\text{th}}$ number in that range that is relatively prime to $105$.
So the thousandth number relatively prime to $105$ is $2100 + 86 = 2186$.
