## Task B-1.2.

Determine the last three digits of the number

$$
2^{2015}-2^{2013}+2^{2010}
$$
## Solution.

After extracting the common factor, we get:

$$
\begin{aligned}
2^{2015}-2^{2013}+2^{2010} & =2^{2010}\left(2^{5}-2^{3}+1\right)=2^{2010} \cdot 25 & & 2 \text { points } \\
& =2^{2008} \cdot 2^{2} \cdot 25=2^{2008} \cdot 100 . & & 1 \text { point }
\end{aligned}
$$

Powers of two (greater than 1) have the last digit in the sequence $2,4,8,6,2,4,8,6, \ldots . \quad 1$ point

Thus, if the exponent is divisible by 4, the last digit is 6. Therefore, the last digit of $2^{2008}$ is 6, and multiplying by 100 gives the last three digits as $6,0,0$.