Sum $\pmod{1000}$ 
Let $$N= \sum_{k=1}^{1000}k(\lceil \log_{\sqrt{2}}k\rceil-\lfloor \log_{\sqrt{2}}k \rfloor).$$ Find $N \pmod{1000}$.

Let $\lceil x \rceil$ be represented by $(x)$ and $\lfloor x \rfloor$ be represented by $[x]$. 
Consider $0 < x < 1$ then:
$$(x) - [x] = 1 - 0 = 1$$ 
Consider $x=0$ then:
$$(x) - [x] = 0$$
Consider $x=1$ then:
$$(x) - [x] = 1-1 = 0$$
So we must find $k$ such that:
$$\log_{\sqrt{2}}k \in \mathbb{Z}$$
$$\log_{\sqrt{2}}k = \frac{2\log k}{\log 2} = y \in \mathbb{Z}$$
$$\log k = \log(\sqrt{2}^y)\implies k = \sqrt{2}^y$$
So for a few integer $y$ values,
$$k = \sqrt{2}, 2^1, \sqrt{2}^3, 2^2,...,  2^3$$...
$k$ is an integer for powers of $2$. 
$$\log_{\sqrt{2}} 2^x = x\log_{\sqrt{2}} 2 = x \frac{2\log_2 2}{\log_2 2} = 2x \in \mathbb{Z}$$
The sum is $1$ for $k=1$ but then $0$ for $k=2$. 
$2^9 = 512 < 1000$ but $2^{10} = 1024 > 1000$.
but how can I compute the sum?

Note that $(\lceil \log_{\sqrt{2}}k\rceil-\lfloor \log_{\sqrt{2}}k \rfloor)=1$ unless $k$ is an integer power of $\sqrt 2$, which is when $k$ is a power of $2$.  In that case it is zero.   So your sum is $$N= \sum_{k=1}^{1000}k-\sum_{m=0}^92^m$$
