7-digit numbers divisible by 7-digit numbers?

Question

The Seven Dwarfs write down all 7-digit numbers that can be formed using each of the digits 1,2,3,4,5,6,7 exactly once (for example, 3175426 is one of their numbers). The Wicked Witch, using the phrase "divide and conquer" literally, looks for pairs (a,b) of such numbers so that b is a multiple of a but a cannot equal b. How many pairs are there?

Answer 1

There aren't any, which I found by doing a computer search.

I could not come up with a short argument to support this conclusion,

just a few bits and pieces.

Let the two numbers be a and b, a < b.

Since 7654321/1234567 = 6.2, the multiplier m such that m * a = b,

cannot be mroe than 6.

It cannot be 1, therefore it is {2,3,4,5,or 6}

It cannot be 3 or 6, since by a sum of digits argument,

none of the numbers is divisible aby 3 (and therefore 6 as well).

So the multiplier is 2, 4, or 5.

Then the first digit of a cannot be 5 or more (generates 8 digit product),

nor can it be 4 (generates an 8 or 9 or an 8-digit product).

There are 191 products consisting of the digits 1,2,3,4,5,6, and 8

(ranging from 2 * 1256734 = 2513468 to 2 * 4327156 = 8654312),

but that's as close as it gets.

Multiplying by 2 or 4 always generates a 0, 8, or 9 somewhere,

and multiplying by 5 generates repeat digits.

Those are just observations of the results of the computer search.

An analytical proof would be quite interesting.