For this kind of problems, you typically want to start assuming a solution exists, and then focus on finding all the properties that solution must have. Then, two things can happen: either you reduce the space of potential solutions enough that you can try them all; or you end up with two properties that contradict themselves and you conclude there is no solution. Let's see what this approach gives us here!

Let \(x\) be a solution to the above. The first thing to notice is that \(x\) cannot include \(0\), since \(0\) does not divide any number. So \(x\) includes all digits from \(1\) to \(9\) except one.

Now assume \(x\) includes \(5\). Then, it must end with a \(0\) or a \(5\). But \(x\) does not include \(0\) so \(5\) must be \(x\)'s last digit. But \(x\) also includes one of \(2\) or \(4\) (it can't exclude both since we've said all digits from \(1\) to \(9\) are included except one). So \(x\) must be even. Clearly, we have a contradiction, so \(x\) cannot include 5. So \(x\) must have all digits except \(0\) and \(5\).

From the above, we also conclude that \(x\) includes \(3\) and must therefore be divisible by \(3\). But the sum of \(x\)'s digits is \(1+2+3+4+6+7+8+9 = 40\) which is not divisble by \(3\). Another contradiction!

So there does not exist an \(8\)-digit number \(x\) such that all its digits are different and each digit divides \(x\).