MHB Can You Solve This Sum of Floor Functions Challenge?

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The challenge involves evaluating the sum of floor functions from 1/3 to 2^1000/3 without a calculator. Participants are encouraged to follow the provided guidelines for submitting their solutions. Members who successfully solved the problem include castor28, kaliprasad, and MegaMoh. The discussion highlights the importance of understanding floor functions in mathematical evaluations. Engaging with such challenges can enhance problem-solving skills in mathematics.
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Here is next week's (23th December 2019) POTW:

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Evaluate $$\left\lfloor{\frac{1}{3}}\right\rfloor+\left\lfloor{\frac{2}{3}}\right\rfloor+\left\lfloor{\frac{2^2}{3}}\right\rfloor+\left\lfloor{\frac{2^3}{3}}\right\rfloor+\cdots+\left\lfloor{\frac{2^{1000}}{3}}\right\rfloor$$, without the help of a calculator.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!
 
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Congratulations to the following members for their correct solution!

1. castor28
2. kaliprasad
3. MegaMoh

Solution from castor28:
Let us define:

  • $\displaystyle S = \sum_{k=0}^{1000}{\left\lfloor\frac{2^k}{3}\right\rfloor}$. This is the expression to be evaluated.
  • $\displaystyle T = \sum_{k=0}^{999}{\left\lfloor\frac{2^k}{3}\right\rfloor}$. This is the same as $S$, without the last term.
  • $\displaystyle U = \sum_{k=0}^{999}{\frac{2^k}{3}}$. This is the same as $T$ without the floor function.

$U$ is a geometric progression, with sum $\dfrac{2^{1000}-1}{3}$.

Since $2^{2k}\equiv1\pmod{3}$ and $2^{2k+1}\equiv2\pmod3$, we have:
$$
\left(\frac{2^{2k}}{3} + \frac{2^{2k+1}}{3}\right)-\left(\left\lfloor\frac{2^{2k}}{3}\right\rfloor+\left\lfloor\frac{2^{2k+1}}{3}\right\rfloor\right)=\frac13+\frac23=1
$$
As $T$ and $U$ contain $500$ such pairs of consecutive terms, we have $T = U - 500 = \dfrac{2^{1000}-1}{3}-500$.

As $ S = T + \left\lfloor\dfrac{2^{1000}}{3}\right\rfloor = T + \dfrac{2^{1000}-1}{3}$, we have $S = \dfrac{2(2^{1000}-1)}{3} - 500$.
 
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