<s>[INST] Show that, for large N, the proportion of positive integers n≤N... 
Show that, for large $N$, the proportion of positive integers $n ≤ N$
(a) not divisible by $2^7$is $1 − \frac{1}{2^7}$
(b) not divisible by any of $2^7, 3^4, 5^3, 7^2, 11^2, 13^2, 17^2, 19^2, 23^2$ is $(1-\frac{1}{3^4})(1-\frac{1}{2^7})\cdots(1-\frac{1}{23^2}) ≈ .931.$

For part a, the number of naturals $i\leq n$ that are divisible by $p$ (call this count $c$) satisfies $\frac{n}{p}-1\lt c\lt \frac{n}{p}+1$, so the density $d = \lim_{n\rightarrow\infty}\frac{c}{n}$ satisfies $\frac{1}{p}-\frac{1}{n}\lt d\lt \frac{1}{p}+\frac{1}{n}$ for all $n$; therefore we must have $d=\frac{1}{p}$
How do I apply this to part (b)? 
 [/INST] It is pure probability at that point. not divisible by an of a,b,c,d, or e is the same as (not divisible by a and not divisible by b and ...) which are independent for large N so you have P(not divisible by a)*P(not divisible by b)P(not divisble by c)... 
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