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Positive integers( A short question )

  1. Dec 14, 2004 #1
    How many positive integers less than 500 have exactly 15 positive integer factors?
    I know the answer, but not sure it. Can you give me the answer ?
     
  2. jcsd
  3. Dec 14, 2004 #2

    Tide

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    HINT: [itex]2^{15} = 32768[/itex]
     
  4. Dec 14, 2004 #3

    matt grime

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    I'm not sure I get the hint. 2.3.5.7 has 15 positive integer factors (excluding itself).
     
  5. Dec 14, 2004 #4

    Tide

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    Oh, I took "has 15 postive factors" as referring to prime factorization. (2x3x5x7 has 4 prime factors!)

    Apparently that's not what primarygun was looking for.
     
  6. Dec 14, 2004 #5

    matt grime

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    I'm not sure if we're supposed to include 1 and itself to be honest. If we are then the only numbers are those... actually, I'm sorry, I always get suspicious when people say "i've got the answer" but don't say what it is. Could you explain how you got your answer, Primarygun, as i don't want to give it away too easily?
     
  7. Dec 14, 2004 #6

    shmoe

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    It's more natural to include the number itself and 1 when counting its number of positive divisors (it makes for a multiplicative function that way), and this is how I would interpret the question. Clarification wouldn't hurt of course.

    Some simple hints in any case. If a number has 15 divisors, what does this say about it's prime factorization? Specifically, can you say how many distinct prime factors it has? What can you say about the exponents appearing in the prime factorization?
     
  8. Dec 15, 2004 #7

    CTS

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    Doesn't
    [tex]
    2 * 3 * 5 * 7
    [/tex]
    have
    [tex]
    \tau (2 * 3 * 5 * 7) = \tau (2) * \tau (3) * \tau (5) * \tau (7) = 2 * 2 * 2 * 2 = 16
    [/tex]
    positive factors?

    I think what you need to look at is the factors of 15: 1, 3, 5, and 15
    [tex]
    15 = 1 * 15
    [/tex]
    [tex]
    15 = 3 * 5
    [/tex]
    The first case gives
    [tex]
    \tau (p^{1-1} * q^{15-1}) = \tau (q^{14})
    [/tex]
    Because
    [tex]
    2^{14} > 500
    [/tex]
    Move on to the second case
    [tex]
    \tau (p^{3-1} * q^{5-1}) = \tau (q^4 * p^2) = 15
    [/tex]
    Move through the primes until you surpass 500:
    [tex]
    2^4 * 3^2 = 144 < 500
    [/tex]
    [tex]
    \tau (144) = 15
    [/tex]
     
    Last edited: Dec 15, 2004
  9. Dec 15, 2004 #8

    shmoe

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    No, it's not the only one. Try other values of p, q. Let's not give too much away though?


    ps.when Matt said 2.3.5.7 had 15 positive divisors, note he also said "excluding iself".
     
  10. Dec 16, 2004 #9
    Sorry, haven't been online so long.
    I guess it should be 3.
    As you see, 15 only can be expressed as 1x15 or 3x5 for positive interger factor.
    Let the number have 15 factors be n=pq (p and q : positive integers)
    For 15 factor, p=1, q^14=n (q : prime number) or p^2q^4=n (p , q : prime number)
    But 2^14> 500, so the first expression is rejected.
    p^2 q^4 = n
    Find the limit for both first, start to do it with q.
    So q=2, q^4=16 n=500
    So the possible largest integer is 5.
    Select q from 2,3 Select p from 2,3,5.
    So answer: 400, 144, and 162
     
  11. Dec 16, 2004 #10

    shmoe

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    Method looks good, but check how you ended up with 162.
     
  12. Dec 16, 2004 #11

    dextercioby

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    324 has 17 positive divisors.Excluding 1 and 324,it's left with 15.

    Daniel.
     
  13. Dec 16, 2004 #12

    shmoe

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    Nope. The formula for the number of divisors that's been kicking around the last few posts for a product of two primes, [tex]\tau(p^n q^m)=(n+1)(m+1)[/tex] includes 1 and itself in the count.
     
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