- 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. - --- In mathforfun@yahoogroups.com, dwittman@... wrote:
>

Or more succinctly: Find the largest integer M < 1000 that has exactly

> 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.

>

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 - --- In mathforfun@yahoogroups.com, adh_math <no_reply@...> wrote:
>

randomly

> --- In mathforfun@yahoogroups.com, dwittman@ wrote:

> >

> > Suppose that M is an integer with the property that if x is

> > 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,

exactly

> > determine the maximum possible value of M.

> >

> Or more succinctly: Find the largest integer M < 1000 that has

> ten divisors (including 1 and M).

M

>

> If M = (p1^k1)*(p2^k2)*...(pn^kn) is the prime factorization, then

> has precisely (1+k1)*(1+k2)*...*(1+kn) divisors: the products for

Or M = p1^9.

> 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.

>

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

> than 1000. Let PF(x) (prime floor!) be the largest prime smaller

> 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

>