Hi CoolStuffYT
I did change it a bit, I assumed that a,b,c and d were all POSITIVE integers.
I've spent considerable time playing with it and the smallest i have come up with is 28.
That is with a=1, b=3, c=9 and d=3
It is easy to determine that C=3D
so C must be a multiple of 3
Then I did determine that
\(d^2=\frac{18(b-a)}{4}\)
the smallest solution for d would be if b-a=2
so b=2+a
So then I said, maybe a=1, b=3, d=3 which makes c=9
\(abc+d-2\\ =1*3*9+3-2\\ =28\)
I am not claiming this is the smallest solution though.
You say the smallest is 16.
Do you have the values of a b c or d to get this solution? Maybe some of them are negative?