A Problem From The Russian Math Olympiad Involving Number Theory Concepts

image containing math problem statement
This is a problem taken from the Russian Math Olympiad 2000. This problem is good test of your skill with number theory.
Problem contributed by anonymous user
We will solve this problem using the concept of   .
Observe that $2^n$ can be written as $(3 - 1)^n$.

Taking $modulo\ 3$ we get:
$2^n \equiv (3 - 1)^n \equiv (-1)^n\ (mod\ 3)$

Therefore, $2^n$ is congruent to $1$ for even $n$ and $-1$ that is $3 - 1 = 2$ for odd $n$.

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Therefore, $\left \lfloor \dfrac{2^n}{3} \right \rfloor = \dfrac{2^n - 1}{3}$ for even $n$ and

$\left \lfloor \dfrac{2^n}{3} \right \rfloor = \dfrac{2^n - 2}{3}$ for odd $n$.

The first term of the series, $\left \lfloor \dfrac{2^0}{3} \right \rfloor$ is $0$ so we can ignore it to give the series an even number of terms.

Now, we can write the given series as odd and even $n$ as follows:

$\displaystyle \sum \limits_{n = 1}^{1000} \left \lfloor \dfrac{2^{n}}{3} \right \rfloor$

$= \displaystyle \sum \limits_{k = 1}^{500} \left( \left \lfloor \dfrac{2^{2k-1}}{3} \right \rfloor + \left \lfloor \dfrac{2^{2k}}{3} \right \rfloor \right)$

$= \displaystyle \sum \limits_{k = 1}^{500} \left(  \dfrac{2^{2k-1} - 2}{3} + \dfrac{2^{2k} - 1}{3}  \right)$

$= \displaystyle \sum \limits_{k = 1}^{500} \left( \dfrac{2^{2k-1} - 2 + 2^{2k} - 1}{3} \right)$

$= \displaystyle \sum \limits_{k = 1}^{500} \left( \dfrac{ 2^{2k-1} + 2^{2k} - 3}{3} \right)$

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$= \displaystyle \sum \limits_{k = 1}^{500} \left( \dfrac{ 2^{2k-1} + 2^{2k}}{3} \right) - \sum \limits_{k = 1}^{500} 1$

$= \displaystyle \sum \limits_{n = 1}^{1000} \left( \dfrac{ 2^{n}}{3} \right) - 500$

$= \displaystyle  \dfrac{\dfrac{2}{3} (2^{1000} - 1)}{(2 - 1)} - 500$

$= \displaystyle \dfrac{\dfrac{2}{3} (2^{1000} - 1)}{(2 - 1)} - 500$

$= \displaystyle \dfrac{2}{3} (2^{1000} - 1) - 500$

$= \displaystyle \dfrac{ (2^{1001} - 2 - 1500)}{3}$

$= \displaystyle \dfrac{ 2^{1001} - 1502}{3}$

Therefore,
$a = 1001, b = 1502, c = 3$

$\Rightarrow a + b + c = 2506$