11. a) How many integers less than 1000 are not divisible by 5 or 7?

b) How many of these numbers are not divisible by 3, 5, or 7?
11. a) There are a total of 999 numbers less than 1000 (numbers $1,2,3, \ldots, 999$). Now, let's strike out those that are divisible by 5; there are $\left[\frac{999}{5}\right]=199$ such numbers (numbers $5,10,15,20, \ldots, 995=199.5$). Next, let's strike out all numbers divisible by 7; there are $\left[\frac{999}{7}\right]=142$ such numbers (numbers 7, $14,21,28, \ldots, 994=142 \cdot 7$). But among the numbers divisible by 7, there are $\left[\frac{993}{35}\right]=28$ numbers that are also divisible by 5 (i.e., divisible by $5.7=35$); these numbers are $35,70,105, \ldots, 980=28.35$. These 28 numbers will be struck out twice (both when striking out numbers divisible by 5 and when striking out numbers divisible by 7). Therefore, a total of $199+142-28=313$ numbers will be struck out. Thus, among the numbers less than 1000, $999-313=686$ numbers will remain unstruck; these are the numbers that are not divisible by 5 or 7.

b) Let's strike out all numbers divisible by 5 and all numbers divisible by 7 again; in this case, we will strike out 313 numbers (problem 11a)). Now, let's strike out all numbers divisible by 3; there are $\left[\frac{999}{3}\right]=333$ such numbers. But among these numbers, all numbers that are also divisible by 5 or 7, i.e., all numbers divisible by 15 or 21, have already been struck out earlier. Among the numbers less than 1000, $\left[\frac{999}{15}\right]=66$ numbers are divisible by 15, and $\left[\frac{999}{21}\right]=47$ numbers are divisible by 21. But $\left[\frac{999}{105}\right]=9$ of these numbers are divisible by both 15 and 21 (105 is the least common multiple of 15 and 21) and, therefore, among the 66 numbers divisible by 15 and the 47 numbers divisible by 21, there are 9 common numbers. Therefore, among the 333 numbers divisible by 3, 104 numbers were already struck out earlier, i.e., in the last strike-out, we will strike out another $333-104=229$ numbers. Thus, the total number of unstruck numbers (i.e., not divisible by 3, 5, or 7) is $686-229=457$.

Note. Reasoning similarly to the solution of problem 11, it is not difficult to show that the number of numbers less than some $N$ and not divisible by any of the pairwise coprime numbers $p_{1}, p_{2}, p_{3}, \ldots, \ldots, p_{n-1}, p_{n}$ is

$$
\begin{aligned}
N-\left[\frac{N}{p_{1}}\right]- & {\left[\frac{N}{p_{2}}\right]-\ldots-\left[\frac{N}{p_{n}}\right]+} \\
& +\left[\frac{N}{p_{1} p_{2}}\right]+\left[\frac{N}{p_{1} p_{3}}\right]+\ldots+\left[\frac{N}{p_{n-1} p_{n}}\right]- \\
& -\left[\frac{N}{p_{1} p_{2} p_{3}}\right]-\ldots+\left[\frac{N}{p_{n-2} p_{n-1} p_{n}}\right]+ \\
& \left.\ldots \ldots \ldots \ldots \ldots \cdots \cdots \cdots \cdots+\frac{N}{p_{1} p_{2} \ldots p_{n}}\right]
\end{aligned}
$$

(see the first solution of problem 78a).