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

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ParthKohliBest ResponseYou've already chosen the best response.0
\[3 = 3 \times 1 \]So the numbers we're looking for are in the form \(a^{3  1}b^{1 1} = a^2\)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
There are \(22\) perfect squares \(\le 500\). So my answer turns out to be \(22\)
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
Could it also include prime cases? In opposite of only non squarefree integers. Because, note that 30 is squarefree, yet has 3 divisors (that are not units or multiplied by units).
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
Well, sorry, by the "prime cases" I mean that they only have prime divisors.
 one year ago

RadEnBest ResponseYou've already chosen the best response.2
looks it is be a square number
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
what should I do now?
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
Plus, I don't quite understand what the case is with them necessarily being square numbers? How'd you derive that?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
I just showed my work.
 one year ago

RadEnBest ResponseYou've already chosen the best response.2
a^2 always have (2+1) factors, in other words a^2 have exactly 3 divisors with a must be a prime number
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
30 has more than 3 divisors
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
30 has how many, within these rules? Are you counting units and unit transformations?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
@RadEn But 21 is incorrect too!
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
30 has 1,2,3,6,10,30 as its divisors.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Oh, a prime number? But why so?
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
Oh, okay, so we're counting improper divisors. Then, yes, the answer must be of the form: \(p^2\) for some prime \(p\).
 one year ago

RadEnBest ResponseYou've already chosen the best response.2
factors of 2^2 = {1,2,4} factors of 3^2 = {1,3,9} factors of 5^2 = {1,5,25} .... so on
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
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.
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
(Where \(p, q \ne 1\).
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
2,3,5,7,11,13,17,19 are the primes below 22. So should the answer be 8?
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
Also, don't forget numbers of the form: \[ n=pq \]Where \(p, q\) are prime.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
@LolWolf lol, that has 4 divisors
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
They have no more than three divisors. No, the answer is not, note that 6=2*3 also has 3 divisors.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
6 has the divisors 1, 2, 3, 6
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
Oh, jeez, you're counting improper... BAH. Yes.
 one year ago

LolWolfBest ResponseYou've already chosen the best response.0
Then, yes, that's the case, it would be 8, indeed.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Yes, it's 8. Thanks @RadEn!
 one year ago
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