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

- ParthKohli

- katieb

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- ParthKohli

\[3 = 3
\times 1 \]So the numbers we're looking for are in the form \(a^{3 - 1}b^{1 -1} = a^2\)

- ParthKohli

There are \(22\) perfect squares \(\le 500\). So my answer turns out to be \(22\)

- anonymous

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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## More answers

- ParthKohli

Oh... not prime

- anonymous

Well, sorry, by the "prime cases" I mean that they only have prime divisors.

- RadEn

looks it is be a square number

- RadEn

except 1

- ParthKohli

oh...

- ParthKohli

what should I do now?

- anonymous

Plus, I don't quite understand what the case is with them necessarily being square numbers? How'd you derive that?

- ParthKohli

I just showed my work.

- RadEn

a^2 always have (2+1) factors, in other words a^2 have exactly 3 divisors
with a must be a prime number

- ParthKohli

30 has more than 3 divisors

- anonymous

30 has how many, within these rules? Are you counting units and unit transformations?

- ParthKohli

@RadEn But 21 is incorrect too!

- ParthKohli

30 has 1,2,3,6,10,30 as its divisors.

- ParthKohli

and 5

- ParthKohli

Oh, a prime number? But why so?

- anonymous

Oh, okay, so we're counting improper divisors. Then, yes, the answer must be of the form: \(p^2\) for some prime \(p\).

- RadEn

factors of 2^2 = {1,2,4}
factors of 3^2 = {1,3,9}
factors of 5^2 = {1,5,25}
.... so on

- ParthKohli

oh.

- anonymous

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.

- anonymous

(Where \(p, q \ne 1\).

- ParthKohli

2,3,5,7,11,13,17,19 are the primes below 22. So should the answer be 8?

- anonymous

Also, don't forget numbers of the form:
\[
n=pq
\]Where \(p, q\) are prime.

- ParthKohli

Oh Lord.

- anonymous

Jaja, yes.

- RadEn

yes, the answer is 8

- ParthKohli

@LolWolf lol, that has 4 divisors

- anonymous

They have no more than three divisors.
No, the answer is not, note that 6=2*3 also has 3 divisors.

- ParthKohli

6 has the divisors 1, 2, 3, 6

- anonymous

Oh, jeez, you're counting improper... BAH. Yes.

- anonymous

I forget.

- anonymous

Then, yes, that's the case, it would be 8, indeed.

- ParthKohli

Yes, it's 8.
Thanks @RadEn!

- RadEn

you're welcome :)

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