1. Calculate the sum $\left[\log _{2} 1\right]+\left[\log _{2} 2\right]+\left[\log _{2} 3\right]+\cdots+\left[\log _{2} 256\right]$. (The expression $[x]$ means the greatest integer not greater than $x$.)
III/1. We see that

$$
\left[\log _{2} n\right]= \begin{cases}0, & n=1 \\ 1, & 2 \leq n<2^{2} \\ 2, & 2^{2} \leq n<2^{3} \\ & \vdots \\ 7, & 2^{7} \leq n<2^{8} \\ 8, & n=2^{8}=256\end{cases}
$$

Therefore, the sought sum is equal to

$$
\begin{aligned}
& 1 \cdot\left(2^{2}-2\right)+2 \cdot\left(2^{3}-2^{2}\right)+\cdots+7 \cdot\left(2^{8}-2^{7}\right)+8= \\
= & 8+7 \cdot 2^{8}-\left(2+2^{2}+\cdots+2^{6}+2^{7}\right)= \\
= & 8+7 \cdot 256-(2+4+8+16+32+64+128)= \\
= & 8+1792-254= \\
= & 1546
\end{aligned}
$$