Number of positive integers less than or equal to $1000$ and are not divisible by $17,19,23$. Four Options are there .... $854, 153, 160$ and none My answer was $153$, it's long process to get that, notice that $17,19,23$ are primes so is there any particular way to handle these types of problems? Please help me.
 A: $$A=\{k\mid k\in\mathbb Z^+, k\le 1000,17\mid k\}\\B=\{k\mid k\in\mathbb Z^+, k\le 1000,19\mid k\}\\C=\{k\mid k\in\mathbb Z^+, k\le 1000,23\mid k\}$$
$$\left|\overline{A}\cap \overline{B}\cap \overline{C}\right|=1000-\left|A\cup B\cup C\right|$$
$$=1000-(|A|+|B|+|C|)$$
$$+(|A\cap B|+|B\cap C|+|C\cap A|)-|A\cap B\cap C|$$
$$|A|=\lfloor\frac{1000}{17}\rfloor,\, |B|=\lfloor\frac{1000}{19}\rfloor,\, |C|=\lfloor\frac{1000}{23}\rfloor$$
$$|A\cap B|=\lfloor\frac{1000}{17\cdot 19}\rfloor,\, |B\cap C|=\lfloor\frac{1000}{19\cdot 23}\rfloor,$$
$$|C\cap A|=\lfloor\frac{1000}{23\cdot 17}\rfloor,\, |A\cap B\cap C|=\lfloor\frac{1000}{17\cdot 19\cdot 23}\rfloor$$
Edit: I used a De Morgan's Law, Inclusion-Exclusion Principle and the fact that $17,19,23$ are pairwise coprime, i.e. $\gcd(17,19)=\gcd(19,23)=\gcd(23,17)=1$.
A: There are $\lfloor \frac {1000}{17} \rfloor = 58$ that are divisible by $17$  You can compute the other two and subtract each from $1000$, but you have subtracted the multiples of $17 \cdot 19$ (and the other two pairs) twice, so add back in $\lfloor \frac {1000}{17\cdot 19} \rfloor = 3$ and the other two.  As $17 \cdot 19 \cdot 23 \gt 1000$ you don't need to worry about multiples of all three.  It will be close to $854$
