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ParthKohli

  • 3 years ago

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

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

  2. ParthKohli
    • 3 years ago
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    There are \(22\) perfect squares \(\le 500\). So my answer turns out to be \(22\)

  3. LolWolf
    • 3 years ago
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    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).

  4. ParthKohli
    • 3 years ago
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    Oh... not prime

  5. LolWolf
    • 3 years ago
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    Well, sorry, by the "prime cases" I mean that they only have prime divisors.

  6. RadEn
    • 3 years ago
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    looks it is be a square number

  7. RadEn
    • 3 years ago
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    except 1

  8. ParthKohli
    • 3 years ago
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    oh...

  9. ParthKohli
    • 3 years ago
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    what should I do now?

  10. LolWolf
    • 3 years ago
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    Plus, I don't quite understand what the case is with them necessarily being square numbers? How'd you derive that?

  11. ParthKohli
    • 3 years ago
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    I just showed my work.

  12. RadEn
    • 3 years ago
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    a^2 always have (2+1) factors, in other words a^2 have exactly 3 divisors with a must be a prime number

  13. ParthKohli
    • 3 years ago
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    30 has more than 3 divisors

  14. LolWolf
    • 3 years ago
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    30 has how many, within these rules? Are you counting units and unit transformations?

  15. ParthKohli
    • 3 years ago
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    @RadEn But 21 is incorrect too!

  16. ParthKohli
    • 3 years ago
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    30 has 1,2,3,6,10,30 as its divisors.

  17. ParthKohli
    • 3 years ago
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    and 5

  18. ParthKohli
    • 3 years ago
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    Oh, a prime number? But why so?

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

  20. RadEn
    • 3 years ago
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    factors of 2^2 = {1,2,4} factors of 3^2 = {1,3,9} factors of 5^2 = {1,5,25} .... so on

  21. ParthKohli
    • 3 years ago
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    oh.

  22. LolWolf
    • 3 years ago
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    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.

  23. LolWolf
    • 3 years ago
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    (Where \(p, q \ne 1\).

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

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

  26. ParthKohli
    • 3 years ago
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    Oh Lord.

  27. LolWolf
    • 3 years ago
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    Jaja, yes.

  28. RadEn
    • 3 years ago
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    yes, the answer is 8

  29. ParthKohli
    • 3 years ago
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    @LolWolf lol, that has 4 divisors

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

  31. ParthKohli
    • 3 years ago
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    6 has the divisors 1, 2, 3, 6

  32. LolWolf
    • 3 years ago
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    Oh, jeez, you're counting improper... BAH. Yes.

  33. LolWolf
    • 3 years ago
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    I forget.

  34. LolWolf
    • 3 years ago
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    Then, yes, that's the case, it would be 8, indeed.

  35. ParthKohli
    • 3 years ago
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    Yes, it's 8. Thanks @RadEn!

  36. RadEn
    • 3 years ago
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    you're welcome :)

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