Hint for 7-11
We have A + B + C + D = 7.11 and (A)(B)(C)(D) = 7.11.
I find it easier to think about the problem if I convert things to whole numbers.
Let x = 100A, y = 100B, z = 100C, w = 100D.
Then x + y + z + w = 711
xyzw = (100A)(100B)(100C)(100D) = 100000000(ABCD) =100000000(7.11) = 711000000
Consider the factorization of 711000000. Can you express 711000000 as the product of four 3-digit numbers?
Clearly, one of the four 3-digit numbers is a small multiple of 79.
Another must end in 5. (Why?)