Find a 10 digit number that contains all digits from 0 to 9 once, starts with a 3, and is also divisible by all whole numbers from 2 to 18 Trial and error is always an option, but this question is from a timed math competition sheet, so it shouldn't take that long. Where do I start?
Edit: As far as the competition aspect is involved, a graphing calculator is allowed

The number starts with $3$ and must end in $0$ as it is divisible by $10$. So we have
$$N=3abcdefgh0$$
As $N$ is divisible by $4$ we know that
$$10h+0\equiv0 \mod{4}$$
$$2h\equiv0 \mod{4}$$
$$\therefore h=6$$
As $N$ is divisible by $8$ we know that
$$100g+10(6)+0\equiv0 \mod{8}$$
$$4g\equiv4 \mod{8}$$
$$\therefore g=1$$
So we have
$$N=3abcdef160$$
As $N$ is divisible by $11$ we have that
$$-3+a-b+c-d+e-f+1-6+0\equiv0 \mod{11}$$
$$a-b+c-d+e-f\equiv8 \mod{11}$$
We also know that the remaining numbers are $\{2,4,5,7,8,9\}$, so the maximum value that this alternating sum can take is $13=7+8+9-2-4-5$ and the minimum  value is $-13=2+4+5-7-8-9$. As there are three even and three odd numbers remaining, their alternating sum must be odd so we must have that
$$a-b+c-d+e-f=-3$$
A quick study of the remaining digits gives that
$$7-8+5-9+4-2=-3$$
So
$$N=3785942160$$
