Find all four–digit numbers in which the product of the digits is equal to the sum of the digits, and the number itself is divisible by the sum of its digits.

Forget about the "divisible" part, and let's find numbers with the product of the digits equal to their sum.

We can assume the number has a lot of 1's, and some other digits. We cannot do just with 1's (except if the number is "1" itself), or with just one "non-one" (no solution), or with three or more "non-ones", so we need TWO "non-ones" and two ones, to fill out to four digits. The only possibility is 2, 4, 1, 1, which produces

4112 as the only possible multiple of 8.