## Sets and functions

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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
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.
• ... 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,
• ... 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, 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,