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Mathematics 13 Online
Parth (parthkohli):

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

Parth (parthkohli):

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

Parth (parthkohli):

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

OpenStudy (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).

Parth (parthkohli):

Oh... not prime

OpenStudy (anonymous):

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

OpenStudy (raden):

looks it is be a square number

OpenStudy (raden):

except 1

Parth (parthkohli):

oh...

Parth (parthkohli):

what should I do now?

OpenStudy (anonymous):

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

Parth (parthkohli):

I just showed my work.

OpenStudy (raden):

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

Parth (parthkohli):

30 has more than 3 divisors

OpenStudy (anonymous):

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

Parth (parthkohli):

@RadEn But 21 is incorrect too!

Parth (parthkohli):

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

Parth (parthkohli):

and 5

Parth (parthkohli):

Oh, a prime number? But why so?

OpenStudy (anonymous):

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

OpenStudy (raden):

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

Parth (parthkohli):

oh.

OpenStudy (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.

OpenStudy (anonymous):

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

Parth (parthkohli):

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

OpenStudy (anonymous):

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

Parth (parthkohli):

Oh Lord.

OpenStudy (anonymous):

Jaja, yes.

OpenStudy (raden):

yes, the answer is 8

Parth (parthkohli):

@LolWolf lol, that has 4 divisors

OpenStudy (anonymous):

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

Parth (parthkohli):

6 has the divisors 1, 2, 3, 6

OpenStudy (anonymous):

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

OpenStudy (anonymous):

I forget.

OpenStudy (anonymous):

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

Parth (parthkohli):

Yes, it's 8. Thanks @RadEn!

OpenStudy (raden):

you're welcome :)

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