1. "Year of the Ox" 2021 has the following property: $2021=$ $43 \times 47$, that is, it can be expressed as the product of two distinct prime numbers less than 100, such numbers are called "Ox numbers". In the set $M=\{1,2, \cdots, 2021\}$ of the first 2021 positive integers, the number of Ox numbers is $\qquad$
$-, 1.201$.
Notice, there are 25 prime numbers within 100, which are $2,3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37,$ $41,43,47,53,59,61,67,71,73,79, 83, 89,$ 97.
Divide them into two subsets $A$ and $B$, where
$$
\begin{aligned}
A= & \{2,3,5,7,11,13,17,19,23,29,31, \\
& 37,41,43,47\}, \\
B= & \{53,59,61,67,71,73,79,83,89, \\
& 97\} .
\end{aligned}
$$

Obviously, $|A|=15,|B|=10$.
Let $k=p q$ be a cow number, where the primes $p, q \in A \cup B$, and $p<q$.

It is easy to see that the product of any two elements in set $A$ belongs to set $M$, resulting in $\mathrm{C}_{15}^{2}=105$ cow numbers $k$.

If $q \in B$, then $p$ must be in $A$. The number of cow numbers $k$ generated by such $q$ is denoted as $S(q)$.
If $q=53$, since
$$
\begin{array}{l}
53 \times 37=1961 \in M, \\
53 \times 41=2173 \notin M,
\end{array}
$$

it follows that $p \leqslant 37$.
Thus, $S(53)=12$.
Similarly,
$$
\begin{array}{l}
S(59)=S(61)=11, S(67)=10, \\
S(71)=S(73)=S(79)=S(83)=9, \\
S(89)=S(97)=8 .
\end{array}
$$

Therefore, the total number of cow numbers is
$$
\begin{array}{l}
105+12+2 \times 11+10+4 \times 9+2 \times 8 \\
=201 .
\end{array}
$$