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``` How many positive integers less than or equal to 500 have exactly 3 divisors? ``` How is my answer wrong?

Mathematics
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\[3 = 3 \times 1 \]So the numbers we're looking for are in the form \(a^{3 - 1}b^{1 -1} = a^2\)
There are \(22\) perfect squares \(\le 500\). So my answer turns out to be \(22\)
Could it also include prime cases? In opposite of only non square-free integers. Because, note that 30 is square-free, yet has 3 divisors (that are not units or multiplied by units).

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Other answers:

Oh... not prime
Well, sorry, by the "prime cases" I mean that they only have prime divisors.
looks it is be a square number
except 1
oh...
what should I do now?
Plus, I don't quite understand what the case is with them necessarily being square numbers? How'd you derive that?
I just showed my work.
a^2 always have (2+1) factors, in other words a^2 have exactly 3 divisors with a must be a prime number
30 has more than 3 divisors
30 has how many, within these rules? Are you counting units and unit transformations?
@RadEn But 21 is incorrect too!
30 has 1,2,3,6,10,30 as its divisors.
and 5
Oh, a prime number? But why so?
Oh, okay, so we're counting improper divisors. Then, yes, the answer must be of the form: \(p^2\) for some prime \(p\).
factors of 2^2 = {1,2,4} factors of 3^2 = {1,3,9} factors of 5^2 = {1,5,25} .... so on
oh.
Because, assume that \(a\) is not prime, then: \[ a=pq \]For some \(p, q\in \mathbb{Z}\). So: \[ p|(pq)^2, p^2|(pq)^2, q|(pq)^2, pq|(pq)^2 \]Et al. Which is greater than 3 divisors. Hence, the number must be prime.
(Where \(p, q \ne 1\).
2,3,5,7,11,13,17,19 are the primes below 22. So should the answer be 8?
Also, don't forget numbers of the form: \[ n=pq \]Where \(p, q\) are prime.
Oh Lord.
Jaja, yes.
yes, the answer is 8
@LolWolf lol, that has 4 divisors
They have no more than three divisors. No, the answer is not, note that 6=2*3 also has 3 divisors.
6 has the divisors 1, 2, 3, 6
Oh, jeez, you're counting improper... BAH. Yes.
I forget.
Then, yes, that's the case, it would be 8, indeed.
Yes, it's 8. Thanks @RadEn!
you're welcome :)

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