The number is
because we can rewrite the three given conditions as follows, where $H$, $T$, and $U$ are the hundreds, tens, and units digits respectively:
Adding the first two gives
Also from the first equation,
So the number must be
To calculate the answer, all we need are some bounds on the order of magnitude of $x$ and $y$. (In particular, we won't be looking at the answer options, nor will we need to resort to any kind of reasoning "from the assumed uniqueness of the answer", which is questionable at best.)
Here's all it takes:
The starting number has 1998 digits. Therefore ...
Assuming we can use the floor function (denoted here with brackets $\lfloor\cdot\rfloor$):
So we can make
Then we could simply do:
Edited to add: If floor function is allowed we can make each of the following using a single four:
and the following using two fours:
then we could do:
Without allowing any tricks, especially not:
Then here are all the solutions:
d1 = lambda x,y: 10*x + y
d2 = lambda x,y: x + y/10
d3 = lambda x,y: x + y/100
for f,g in itertools.permutations([operator.add, operator.sub, operator.mul, operator.truediv, operator.floordiv, operator.pow, d1,d2,d3, operator.xor]*2,...
I tried a computer program to solve this problem.
I got 425 different expressions giving 2016.
So I added restrictions. I used only addition and multiplication. I stil get 26 expressions. You'll find them below.
To answer the question about "no brackets" and "not ending in a product":