How many natural numbers $n<10^6$ are divisible by $7$ but not with $10,12$ and $25$?
Theorem: Let $n,k\in \mathbb{N}$ and $k\leq n$, then in the set $\{1,2,...,n\}$ we have exactly $\left \lfloor \dfrac{n}{k} \right \rfloor$ numbers divisible by $k$.
$S=\{1,2,...,10^6\}$
$A_7$ - set of all numbers from $S$ divisible by $7$
$A_{10}, A_{12}, A_{25}$ - sets of all numbers from $S$ divisible by $10,12$ and $25$
The solution of our problem is the cardinal number of the following set: $| A_7 \cap \overline{A_{10}} \cap \overline{A_{12}} \cap \overline{A_{25}}|$
Let $T=\overline{A_{10}} \cap \overline{A_{12}} \cap \overline{A_{25}}$
Using the following formula: $|\overline{A}\cap \overline{B} \cap \overline{C}|=|S|-|A|-|B|-|C|+|A\cap B|+|A\cap C|+|B\cap C|-|A\cap B\cap C|$
and the fact that
$|A_7|=\left \lfloor \dfrac{10^6}{7}\right \rfloor=142857 $
$|A_{10}|=\left \lfloor \dfrac{10^6}{10}\right \rfloor=100000 $
$|A_{12}|=\left \lfloor \dfrac{10^6}{12}\right \rfloor=83333 $
$|A_{25}|=\left \lfloor \dfrac{10^6}{25}\right \rfloor=40000 $
$|A_{10}\cap A_{12}|=\left \lfloor \dfrac{10^6}{60}\right \rfloor=16666 $
$|A_{10}\cap A_{25}|=\left \lfloor \dfrac{10^6}{50}\right \rfloor=20000 $
$|A_{12}\cap A_{25}|=\left \lfloor \dfrac{10^6}{300}\right \rfloor=3333 $
$|A_{10}\cap A_{12} \cap A_{25}|=\left \lfloor \dfrac{10^6}{300}\right \rfloor=3333 $
we get that $|T|=81333$.
We have to calculate $|A_7 \cap T|=|S|- |\overline{A_7}\cup \overline{T}|$
This is how far I got!
Hint:The cardinal you want to find is the cardinal of the set $A_7\backslash(A_{10}\cup A_{12}\cup A_{25})$ for which you can use the following formula:
$$|T|=|\overline{A}\cap \overline{B} \cap \overline{C}|=|A_7|-|A|-|B|-|C|+|A\cap B|+|A\cap C|+|B\cap C|-|A\cap B\cap C|$$
with $A=A_7\cap A_{10}=A_{70}$, $B=A_7\cap A_{12}$ and $C=A_7\cap A_{25}$.