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

This Question is Closed

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

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

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.0Could 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).

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.0Well, sorry, by the "prime cases" I mean that they only have prime divisors.

RadEn
 2 years ago
Best ResponseYou've already chosen the best response.2looks it is be a square number

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0what should I do now?

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

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0I just showed my work.

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

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.030 has more than 3 divisors

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.030 has how many, within these rules? Are you counting units and unit transformations?

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0@RadEn But 21 is incorrect too!

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.030 has 1,2,3,6,10,30 as its divisors.

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0Oh, a prime number? But why so?

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

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

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.0Because, 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.

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.02,3,5,7,11,13,17,19 are the primes below 22. So should the answer be 8?

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

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0@LolWolf lol, that has 4 divisors

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

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.06 has the divisors 1, 2, 3, 6

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.0Oh, jeez, you're counting improper... BAH. Yes.

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.0Then, yes, that's the case, it would be 8, indeed.

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, it's 8. Thanks @RadEn!
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