Solution:

Suppose set A denotes the number of integers between 1 to 2000 divisible by 2.

Set B is the number of integers between 1 and 2000 divisible by 3.

Set $\mathrm{C}$ is the number of integers between 1 and 2000 divisible by 5.

Set D is the number of integers between 1 and 2000 divisible by $7 .$

$|A|=\left[\frac{2000}{2}\right]=1000 \\ $

$|B|=\left[\frac{2000}{3}\right]=666 \\ $

$|C|=\left[\frac{2000}{5}\right]=400 \\ $

$|\mathrm{D}|=\left[\frac{2000}{7}\right]=285 \\ $

$|A \cap B|=\left[\frac{2000}{2 \times 3}\right]=333 \\ $

$|A \cap C|=\left[\frac{2000}{2 \times 5}\right]=200 \\ $

$|A \cap D|=\left[\frac{2000}{2 \times 7}\right]=142 \\ $

$|B \cap C|=\left[\frac{2000}{3 \times 5}\right]=133 \\ $

$|B \cap D|=\left[\frac{2000}{3 \times 7}\right]=95 \\ $

$|C \cap D|=\left[\frac{2000}{5 \times 7}\right]=57 \\ $

$|A \cap B \cap C|=\left[\frac{2000}{2 \times 3 \times 5}\right]=66 \\ $

$|A \cap B \cap D|=\left[\frac{2000}{2 \times 3 \times 7}\right]=47 \\ $

$|A \cap C \cap D|=\left[\frac{2090}{2 \times 5 \times 7}\right]=28 \\ $

$|B \cap C \cap D|=\left[\frac{2090}{3 \times 5 \times 7}\right]=19 \\ $

$|A \cap B \cap C \cap D|=\left[\frac{2000}{2 \times 3 \times 5 \times 7}\right]=9 \\ $

Number of elements divisible by 2 or 3 or 5 or 7 are $|A \cup B \cup C \cup D|$.

From inclusion-exclusion principle

$\begin{aligned} |A \cup B \cup C \cup D|=&|A|+|B|+|C|+|D| \\\\ &-|| A \cap B|+| B \cap C|+| A \cap C|+| A \cap D|+| B \cap D|+| C \cap D \mid] \\\\ &+|| A \cap B \cap C|+| A \cap B \cap D|+| A \cap C \cap D|+| B \cap C \cap D \mid] \\\\ &-|| A \cap B \cap C \cap D \mid] \\\\ \therefore|A \cup B \cup C \cup D|=& 1000+666+400+285-[333+200+142+133+95+57] \\\\ &+[66+47+28+19]-9 \\\\ =& 2351-960+160-9 \\\\ =& 1542 \\ \end{aligned}$