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

Question

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

Answer 1

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 $$N=3785942160$$