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For any prime number $p$, let $A_p$ be the set of integers $d\in \{1,2,\dots, 999\}$ such that the power of $p$ in the prime factorization of $d$ is odd. Then what is the cardinality of $A_p$?

I have used to the following result, but failed to show the required result.
The largest exponent $e$ of a prime $p$ such that $p^e$ is a divisor of $n!$ is given by $$ e=\left\lfloor \dfrac{n}{p} \right\rfloor + \left\lfloor \dfrac{n}{p^2} \right\rfloor + \left\lfloor \dfrac{n}{p^3} \right\rfloor $$
How can I do next?

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  • $\begingroup$ If I've understood correctly, we have $|A_p|=0$ for $p>999$, right? $\endgroup$ – Sambo Aug 16 '18 at 20:29
  • $\begingroup$ Please concoct a relevant title. $\endgroup$ – Did Aug 16 '18 at 21:55
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The number of integers less than or equal to $n$ and divisible by $d$ is $\lfloor n/d \rfloor $. You need to count the number of integers divisible by $p$, minus the number of primes divisible by $p^2$, and so on...

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