ParthKohli
  • ParthKohli
``` How many positive integers less than or equal to 500 have exactly 3 divisors? ``` How is my answer wrong?
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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ParthKohli
  • 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
  • ParthKohli
There are \(22\) perfect squares \(\le 500\). So my answer turns out to be \(22\)
anonymous
  • 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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ParthKohli
  • ParthKohli
Oh... not prime
anonymous
  • anonymous
Well, sorry, by the "prime cases" I mean that they only have prime divisors.
RadEn
  • RadEn
looks it is be a square number
RadEn
  • RadEn
except 1
ParthKohli
  • ParthKohli
oh...
ParthKohli
  • ParthKohli
what should I do now?
anonymous
  • anonymous
Plus, I don't quite understand what the case is with them necessarily being square numbers? How'd you derive that?
ParthKohli
  • ParthKohli
I just showed my work.
RadEn
  • 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
  • ParthKohli
30 has more than 3 divisors
anonymous
  • anonymous
30 has how many, within these rules? Are you counting units and unit transformations?
ParthKohli
  • ParthKohli
@RadEn But 21 is incorrect too!
ParthKohli
  • ParthKohli
30 has 1,2,3,6,10,30 as its divisors.
ParthKohli
  • ParthKohli
and 5
ParthKohli
  • ParthKohli
Oh, a prime number? But why so?
anonymous
  • 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
  • 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
  • ParthKohli
oh.
anonymous
  • 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
  • anonymous
(Where \(p, q \ne 1\).
ParthKohli
  • ParthKohli
2,3,5,7,11,13,17,19 are the primes below 22. So should the answer be 8?
anonymous
  • anonymous
Also, don't forget numbers of the form: \[ n=pq \]Where \(p, q\) are prime.
ParthKohli
  • ParthKohli
Oh Lord.
anonymous
  • anonymous
Jaja, yes.
RadEn
  • RadEn
yes, the answer is 8
ParthKohli
  • ParthKohli
@LolWolf lol, that has 4 divisors
anonymous
  • anonymous
They have no more than three divisors. No, the answer is not, note that 6=2*3 also has 3 divisors.
ParthKohli
  • ParthKohli
6 has the divisors 1, 2, 3, 6
anonymous
  • anonymous
Oh, jeez, you're counting improper... BAH. Yes.
anonymous
  • anonymous
I forget.
anonymous
  • anonymous
Then, yes, that's the case, it would be 8, indeed.
ParthKohli
  • ParthKohli
Yes, it's 8. Thanks @RadEn!
RadEn
  • RadEn
you're welcome :)

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