# $\lfloor (2^{2})/3 \rfloor +… + \lfloor (2^{1000})/3 \rfloor = \frac{2^{A}-B}{C}$, minimum of $A+B+C$?

$$\lfloor (2^{2})/3 \rfloor + \lfloor (2^{3})/3 \rfloor + \lfloor (2^{4})/3 \rfloor + ... + \lfloor (2^{999})/3 \rfloor + \lfloor (2^{1000})/3 \rfloor = \frac{2^{A}-B}{C},$$ $$A,B,C \in \mathbb{Z}^{+}$$

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.

• Which contest is this from please ? – Ewan Delanoy May 27 at 9:03
• @EwanDelanoy winter camp olympiad for senior high school students – Arief Anbiya May 27 at 9:20
• Since $2^{1001}-1502\equiv(-1)^{1001}-2\equiv0\,\,(\mathrm{mod}\,3)$, you in fact can divide the numerator by $3$. – J_P May 27 at 19:09