$$ \lfloor (2^{2})/3 \rfloor + \lfloor (2^{3})/3 \rfloor + \lfloor (2^{4})/3 \rfloor + ... + \lfloor (2^{999})/3 \rfloor + \lfloor (2^{1000})/3 \rfloor = \frac{2^{A}-B}{C},$$ $$ A,B,C \in \mathbb{Z}^{+}$$
What is the minimum possible of $A+B+C$?
Attempt:
We can easily show that if $k>1$ is even, then remainder of $(2^{k})/3$ is 1, and if $k>2$ is odd then remainder is 2. So the problematic summation will become:
$$ (2^{2}-1)/3 + (2^{3}-2)/3 + (2^{4}-1)/3 + ... + (2^{999}-2)/3 + (2^{1000}-1)/3 $$ there are exactly 499 of $-(1/3) -(2/3) = -1$, so we get $$ \frac{2^{2}+...+2^{1000}}{3} - 499 - 1/3 $$ $$ \frac{2^{1001}-4}{3} - \frac{1498}{3} = \frac{2^{1001}-1502}{3}$$ So one possible set of values is $A=1001,B=1502,C=3$.
Notice that the minimum value for $C$ is clearly 3 (because we cannot divide the numerator with 3). If we increase $A$ or $B$, then $A+B+C$ will also increased. So my argument is that $A=1001,B=1502,C=3$ make $A+B+C$ minimum. But.. notice that if we increase $A$ and decrease $B$, this makes more possibilites.