2
$\begingroup$

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

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.