Number of squarefree positive integers less than $100$ 
An integer is called squarefree if it is not divisible by the square
  of a positive integer greater than $1$. Find the number of squarefree
  positive integers less than $100$.

My attempt:
I apply the inclusion-exclusion principle directly.


*

*Total number of integers = 99

*Number of integers divisible by $2^2$ = $24$

*Number of integers divisible by $3^2$ = $11$

*Number of integers divisible by $4^2$ = $6$

*Number of integers divisible by $5^2$ = $3$

*Number of integers divisible by $6^2$ = $2$

*Number of integers divisible by $7^2$ = $2$

*Number of integers divisible by $8^2$ = $1$

*Number of integers divisible by $9^2$ = $1$

*Number of integers divisible by $2^2$ and $3^2$ = $2$

*Number of integers divisible by $2^2$ and $4^2$ = $1$


Then the required solution would be $99-(24+11+6+3+2+2+1+1)+2+1=52$. But the solution is $61$. Where is my mistake?

An integer is square-free if and only if none of its prime factors appear with an exponent $\ge 2$ in its prime factorization.
The only primes that can appear with an exponent of 2 or more in a number less than $100$ are those that are less than $\sqrt{100}=10$, that is, $2$, $3$, $5$, and $7$.
So the numbers you need to exclude are just those that are multiples of $4$, $9$, $25$ or $49$.
Neither of the multiples of $25$ and $49$ are multiples of any of the other squares, so they can be subtracted separately. All we need to care about is the double-counting of multiples of $36$, so:


*

*Start with $99$ possible numbers.

*Subtract $24$ multiples of $4$.

*Subtract $11$ multiples of $9$.

*Add back $2$ because $36$ and $72$ were each subtracted twice.

*Subtract $3$ multiples of $25$.

*Subtract $2$ multiples of $49$.


$$ 99 - 24 - 11 + 2 - 3 - 2 = 61 $$
