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Sets and functions

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  • dwittman@twcny.rr.com
    Suppose that M is an integer with the property that if x is randomly chosen from the set { 1, 2, 3, ..., 999,1000 }, the probability that x is a divisor of M
    Message 1 of 3 , Jun 4, 2006
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      Suppose that M is an integer with the property that if x is randomly
      chosen from the set { 1, 2, 3, ..., 999,1000 }, the probability that x
      is a divisor of M is 1/100, with 1 and M inclusive. If M < 1000,
      determine the maximum possible value of M.
    • adh_math
      ... Or more succinctly: Find the largest integer M
      Message 2 of 3 , Jun 5, 2006
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        --- In mathforfun@yahoogroups.com, dwittman@... wrote:
        >
        > Suppose that M is an integer with the property that if x is randomly
        > chosen from the set { 1, 2, 3, ..., 999,1000 }, the probability that x
        > is a divisor of M is 1/100, with 1 and M inclusive. If M < 1000,
        > determine the maximum possible value of M.
        >
        Or more succinctly: Find the largest integer M < 1000 that has exactly
        ten divisors (including 1 and M).

        If M = (p1^k1)*(p2^k2)*...(pn^kn) is the prime factorization, then M
        has precisely (1+k1)*(1+k2)*...*(1+kn) divisors: the products for
        which the exponent of pj is an integer between 0 and kj inclusive.
        Since 10=2*5, it follows that

        M = p1*p2^4 for some primes p1, p2.

        There are only three primes (2, 3, 5) whose fourth power is smaller
        than 1000. Let PF(x) (prime floor!) be the largest prime smaller than
        x (or 1 if x<=2)

        PF(1000/2^4) = PF(1000/16) = 61 -> M = 61*2^4 = 976 (largest),
        PF(1000/3^4) = PF(1000/81) = 11 -> M = 11*3^4 = 891,
        PF(1000/5^4) = PF(1000/625) = 1 -> M = 5^4, no good.

        Regards,
        adh
      • cooperpuzzles
        ... randomly ... that x ... exactly ... M ... Or M = p1^9. 2^9=512 smaller than 976. 3^9=19683 too big. ... than
        Message 3 of 3 , Jun 7, 2006
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          --- In mathforfun@yahoogroups.com, adh_math <no_reply@...> wrote:
          >
          > --- In mathforfun@yahoogroups.com, dwittman@ wrote:
          > >
          > > Suppose that M is an integer with the property that if x is
          randomly
          > > chosen from the set { 1, 2, 3, ..., 999,1000 }, the probability
          that x
          > > is a divisor of M is 1/100, with 1 and M inclusive. If M < 1000,
          > > determine the maximum possible value of M.
          > >
          > Or more succinctly: Find the largest integer M < 1000 that has
          exactly
          > ten divisors (including 1 and M).
          >
          > If M = (p1^k1)*(p2^k2)*...(pn^kn) is the prime factorization, then
          M
          > has precisely (1+k1)*(1+k2)*...*(1+kn) divisors: the products for
          > which the exponent of pj is an integer between 0 and kj inclusive.
          > Since 10=2*5, it follows that
          >
          > M = p1*p2^4 for some primes p1, p2.
          >
          Or M = p1^9.

          2^9=512 smaller than 976.
          3^9=19683 too big.

          > There are only three primes (2, 3, 5) whose fourth power is smaller
          > than 1000. Let PF(x) (prime floor!) be the largest prime smaller
          than
          > x (or 1 if x<=2)
          >
          > PF(1000/2^4) = PF(1000/16) = 61 -> M = 61*2^4 = 976 (largest),
          > PF(1000/3^4) = PF(1000/81) = 11 -> M = 11*3^4 = 891,
          > PF(1000/5^4) = PF(1000/625) = 1 -> M = 5^4, no good.
          >
          > Regards,
          > adh
          >
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