Re: smallest positive integer that has exactly k divisors

On Oct 24, 10:04 am, "mensana...@xxxxxxxxxxx" <mensana...@xxxxxxx>
wrote:
On Oct 24, 2:43 am, Proginoskes <CCHeck...@xxxxxxxxx> wrote:

On Oct 23, 11:11 pm, "mensana...@xxxxxxxxxxx" <mensana...@xxxxxxx>
wrote:
[...]
## Did you really mean 2000 factors? You do know that the
## number of divisors is 2**factors, don't you?

Really? Then how can 12 have 6 divisors?

It doesn't, it has 6 UNIQUE divisors.

(6 is not a power of 2.)

No, but 8 is:

000 12 2 * 2 * 3
001 4 2 * 2 * 1
010 6 2 * 1 * 3
011 2 2 * 1 * 1
100 6 1 * 2 * 3
101 2 1 * 2 * 1
110 3 1 * 1 * 3
111 1 1 * 1 * 1

Note, 6 appears twice as does 2, that's why there's 6.

That's why the real answer is 6, not 8.

My definition (and the math community's) of a factor of N is a number
k which divides into N with no remainder. The positive divisors of 12
are 1, 2, 3, 4, 6, and 12. If I look at the set of the divisors, I get
a set with 6 elements (even if I "stutter" and list a divisor more
than once).

Hence {1, 2, 2, 3, 4, 6, 6, 12} has 6 elements in it, not 8.

This is true, _if_ N is square-free

No, square-free just means the UNIQUE divisors will be the same
as 2**factors, but there will always be 2**factors divisors.
It's like rolling dice, there are 36 possible outcomes, but
only 11 unique sums.

And I'm interested in counting "sums" above.

And you DO know that you have to count
the duplicates when calculating probability?

There is no probability involved here.

To do this algorithmically, you would need to look at all possible
factorizations of 2000, and given such a factorization

a(1) * a(2) * ... * a(k),

where a(1) >= a(2) >= ... >= a(k), you look at the number

2^[a(1)-1] * 3^[a(2)-1] * ... * p(k)^[a(k)-1],

where p(k) is the kth prime. This is the smallest number of the form

q(1)^[a(1)-1] * q(2)^[a(2)-1] * ... * q(k)^[a(k)-1],

where q(1), q(2), ..., q(k) are distinct primes. (I think.) Then you
would take the minimum of all of these numbers. There might be a
faster way, though, which I can't think of, right at the moment.

That looks like too much work. [...]

Only because I'm putting in all the steps.

--- Christopher Heckman

.

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