An integer $n$ is said to be square-free if the only perfect square that divides $n$ is $1^2$. How many positive odd integers greater than 1 and less than $100$ are square-free?
Let's find the total number of odd numbers between 1 and 100. There are 9 perfect squares less than 100. That's right. If a positive odd integer is square-free, then it is not divisible by any of these perfect squares, except for 1. So there are $100/2=50$ positive odd integers less than 100. The total number of positive odd integers less than 100 is 49. Ok. So we just need to find how many of these are divisible by perfect squares. Let's use our list of perfect squares. Ok. First, we can cross out 1, because all positive odd integers are divisible by 1. Right. Let's move to the next perfect square, 4. All positive odd integers that are divisible by 4 have a units digit of 2 or 6. Right. Now we can move on to 9. The only positive odd integers less than 100 that are divisible by 9 are 9, 27, 45, 63, 81. Right. So there are 6 odd integers less than 100 that are divisible by 9. The next perfect square is 16. But 16 is an even number, so no odd integer less than 100 is divisible by 16. That's correct. Let's move to 25 Let's move to the next perfect square, 36 But 36 is an even number, so no odd integer less than 100 is divisible by 36 Right. Now let's move to 49. Only 49 is divisible by 49. Ok, so we've found all the odd integers less than 100 that are divisible by perfect squares. So the total number of odd integers less than 100 that are square-free is 49 - 6 - 2 - 1 = 40. So there are 40 odd integers less than 100 that are square-free.

