Deep Thought for 2015

First divide out the common factor: $\text{gcd}(n,63000) = 42\ \text{if and only if}\ \text{gcd}\left(\frac n{42},1500\right) = 1.$ So let $n = 42m$ , with $m$ the 2015th positive integer that is coprime to 1500.

Second, we must find values of $m$ that have no factors in common with $1500 = 2^2\cdot 3\cdot 5^3$ . This means that $m$ is not a multiple of 2, 3, or 5; in other words, $\text{gcd}(m,30) = 1$ .

Third, since $\text{gcd}(m + 30k,30) = \text{gcd}(m, 30)$ , the gcd values repeat themselves with period 30. Therefore we only need to find the values $0 \leq m < 30$ for which $\text{gcd}(m, 30) = 1$ . These are easily found: $1, 7, 11, 13, 17, 19, 23, 29.$ (There are only eight, since $\phi(30) = (2-1)(3-1)(5-1) = 8$ . We also save work by noticing that the sequence is its own mirror image around 30/2 = 15.)

Thus for every 30 values of $m$ there are 8 that satisfy the condition $\text{gcd}(m, 30) = 1$ . We need the 2015th value of $m$ ; this means that we skip $\left\lfloor\frac {2015}{8}\right\rfloor = 251$ groups of 30, then take the 7th acceptable number after that (see the list above):

$m = 251\cdot 30 + 23 = 7553.$

Finally, $n = 42\ m = \boxed{317\:226}.$