25. Between 1 and 8000 inclusive, find the number of integers which are divisible by neither 14 nor 21 but divisible by either 4 or 6 .
25. Answer: 2287 .
$$
\begin{array}{l} 
\text { Let } S=\{1, \ldots, 8000\}, A=\{x \in S: 4 \mid x\}, B=\{x \in S: 6 \mid x\}, C=\{x \in S: 14 \mid x\}, \\
D=\{x \in S: 21 \mid x\} . \\
\left|(A \cup B) \cap(C \cup D)^{\prime}\right| \\
=|A \cup B|-|(A \cup B) \cap(C \cup D)| \\
=|A|+|B|-|A \cap B|-\mid A \cap C) \cup(B \cap C) \cup(A \cap D) \cup(B \cap D) \mid \\
=|A|+|B|-|A \cap B|-|A \cap C|-|B \cap C|-|A \cap D|-|B \cap D| \\
+|A \cap B \cap C|+|A \cap C \cap D|+|A \cap B \cap C \cap D|+|A \cap B \cap C \cap D| \\
+|B \cap C \cap D|+|A \cap B \cap D|-|A \cap B \cap C \cap D|-|A \cap B \cap C \cap D| \\
-|A \cap B \cap C \cap D|-|A \cap B \cap C \cap D|+|A \cap B \cap C \cap D| \\
=\lfloor 8000 / 4\rfloor+\lfloor 8000 / 6\rfloor-\lfloor 8000 / 12\rfloor-\lfloor 8000 / 28\rfloor-\lfloor 8000 / 42\rfloor-\lfloor 8000 / 84\rfloor \\
-\lfloor 8000 / 42\rfloor+3 \times\lfloor 8000 / 84\rfloor+\lfloor 8000 / 42\rfloor-\lfloor 8000 / 84\rfloor \\
= 2287 .
\end{array}
$$