Counting problem for the integers 
How many numbers $n < 100$ are not divisible by a square of any integer
  greater than $1$?

Working through the above counting problem. I got $48$ using the Inclusion-Exclusion Principle, do you agree?
 A: I don't see how you got to $48$.
I assume we're working in the natural numbers so $99$ numbers less than $100$ all together. Any number divisible by a square number $>1$ is divisible by a prime square.


*

*$24$ are divisible by $4$.

*$11$ are divisible by $9$ - two of which ($36$ and $72$) are already counted above.

*$3$ are divisible by $25$,

*$2$ are divisible by $49$


Total of $24+(11-2)+3+2=38$ are divisible by squares $>1$, so $61$ numbers $<100$  are not divisible by a square number $>1$.
A: In general the inclusion-exclusion principle gives you
$$
N_{n}=n-\sum_{p}\left\lfloor \frac{n}{p^2}\right\rfloor+\sum_{p < q}\left\lfloor\frac{n}{p^2 q^2}\right\rfloor-\sum_{p<q<r}\left\lfloor\frac{n}{p^2 q^2 r^2}\right\rfloor + \ldots,
$$
for the count $\le n$, where the sums are over all primes.  In this case the non-zero terms are:
$$
N_{99}=99-\left(\left\lfloor\frac{99}{2^2}\right\rfloor + \left\lfloor\frac{99}{3^2}\right\rfloor + \left\lfloor\frac{99}{5^2}\right\rfloor + \left\lfloor\frac{99}{7^2}\right\rfloor\right) + \left\lfloor\frac{99}{2^2 3^2}\right\rfloor \\=99-(24+11+3+2)+2=99-40+2\\=61.
$$
