Call a positive integer $n$ extra-distinct if the remainders when $n$ is divided by $2, 3, 4, 5,$ and $6$ are distinct. Find the number of extra-distinct positive integers less than $1000$.
$n$ can either be $0$ or $1$ (mod $2$).
Case 1: $n \equiv 0 \pmod{2}$
Then, $n \equiv 2 \pmod{4}$, which implies $n \equiv 1 \pmod{3}$ and $n \equiv 4 \pmod{6}$, and therefore $n \equiv 3 \pmod{5}$. Using [CRT](https://artofproblemsolving.com/wiki/index.php/Chinese_Remainder_Theorem), we obtain $n \equiv 58 \pmod{60}$, which gives $16$ values for $n$.
Case 2: $n \equiv 1 \pmod{2}$
$n$ is then $3 \pmod{4}$. If $n \equiv 0 \pmod{3}$, $n \equiv 3 \pmod{6}$, a contradiction. Thus, $n \equiv 2 \pmod{3}$, which implies $n \equiv 5 \pmod{6}$. $n$ can either be $0 \pmod{5}$, which implies that $n \equiv 35 \pmod{60}$ by CRT, giving $17$ cases; or $4 \pmod{5}$, which implies that $n \equiv 59 \pmod{60}$ by CRT, giving $16$ cases.
The total number of extra-distinct numbers is thus $16 + 16 + 17 = \boxed{049}$.
~mathboy100