Find a positive integer containing all ten digits: $0,1,2,3,4,5,6,7,8,9$ that is a multiple of $126$
I don't really know where to start. I guess I could find the prime factorization of $126$, which is $2*3^2*7$, but I don't know how that helps. The only thing I can think of is just multiplying the number, but that would take forever.
On
You said a positive integer, not the smaller positive integer. You said also that is must contain each integer, not that it must contain each integer once.
Then I have a solution in the pocket
Take $1234567890000$ , compute its modulo $126$ and the difference of the latter with $126$ and you find $108$. Then just add them and deliver your result : $1234567890108$
If it is not a valid answer, I'm sorry for the trick. Let's always be careful with the statements :)
Since a number containing all the digits $\ 0,1,2,\cdots,9\ $ exactly once is divisble by $9$, we only have to make sure that the number is even and divisible by $7$.
Since the numbers $\ 91,56,203,84,7\ $ are divisible by $7$, the number
$$9156203784$$ must be divisible by $7$ and hence must be divisble by $126$.