Smallest number with 100 divisors

[Flofloflo]

Hello,

I've been trying to find the smallest number with the most divisors. (i.e. 15 has 3 dividers: 1,3,5,15)
Now, I had this idea to find a number that would at least approach it; by simply taking a number with a maximum amount of divisors, and doubling it enough times.
Basically I used my Ti to find these numbers:P The best I got was 315, having 12 divisors, I kept looking for better ones, but aafter a bit (I searched untill 715) it all got way too slow. So basically, I figured 80640 would be at least pretty close to the smallest number with 100 divisors.

I was wondering, is there some way of finding a number between say, 700 and 2000 with at least 13 divisors, or maybe between 700 and something like 10000 with more divisors then that??

[calc84maniac]

100=5*5*2*2

Thus 2^5-1*3^5-1*5^2-1*7^2-1=45,360 has 100 divisors.

This seems like the smallest which satisfies the conditions.

Edit:
How is 1,3,5,15 three numbers? I count four.

[simplethinker]

If a number's prime factorization is n=p₁^a₁p₂^a₂...p_m^a_m, then the number of divisors d(n) is the product of all the (1 + a_m)'s. For example, 15=3¹ * 5¹, so d(15)=(1+1)(1+1)=4. This means that the number of divisors depends on the exponents in its prime factorization rather than the factors, so if we find combinations of exponents that would equal 100, then we can permute them around the primes 2, 3, 5... until we find the lowest value (since 4=2^2 it can't be in the prime factorization)

As an example, let's find the smallest number with 12 divisors. 12 = 3*4 = (2+1)(3+1) = 2*2*3 = (1+1)(1+1)(2+1), so we can choose either (2, 3) or (1, 1, 2). To get the lowest number, the 3rd power should go to 2 (the smallest divisor other than one) and the second power goes to the next lowest prime, which is 3, which gives n=2³ * 3²=72. We could also try (1, 1, 2), and by a similar argument we get 2²3¹5¹=60. Since 60<72, this means that the smallest number with 12 divisors is 60, with divisors 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.

The smallest number with 13 divisors is 4096, since 13 is prime it can only be written as 13=(12+1), so 2^12=4096.

[edit]Dang, I was beaten.

[thornahawk]

You might wish to look at this...

thornahawk

[Weregoose]

Here's the extended list, in case it's missed.

[Flofloflo]

A, writing there are only three was pretty dumb -.-
But anyway, yeah, I now used my program to find something around 40.000 too (probably the same as you found), but now I really gotta see how you found it so fast

So, just to see if I get this:
15 = 3*5 = (2+1)(4+1) , I thought i'd have to rewrite the four but that seems impossible
2^4*3^2=144. Okay that seems right... And it is right according to the list (altough I understand nearly nothing from the whole On-Line Encyclopedia of Integer Sequences).
Thanks:P

[simplethinker]

[Flofloflo]

Wait! I just realised the answer doesn't satisfy me.
Let's do the same thing as last time:
15 = 3^1 * 5^1, So it has (1+1)(1+1) divisors.
The reason this works is because there are four combinations to make:
3^0*5^0
3^0*5^1
3^1*5^0
3^1*5^1
That's why I think there are four divisors, that's correct right?
So how can I be 100% sure that for example 3^1*5^0 isn't the same as one of the others that I consider divisors? Maybe there are doubles!
Can that be proven too?

[simplethinker]

[Flofloflo]

Okay, I read it all, took some time to figure out what d' meant
I gotta gotta check it out some more to see if this actually proofs that there cant be any doubles in there; it's not really obvious yet to me -.- But I understood everything untill the geometric mean part, I dunno if that's important, but I don't think it has anything to do with my question, so I guess I'm not gonna bother with it (yet).

[simplethinker]

Every number has a distinct prime factorization, and any product of powers of primes corresponds to a single number. If you have a number n=a²b¹, then the factors are a⁰b⁰ a¹b⁰, a²b⁰, a⁰b¹, a¹b¹, a²b¹ which are all distinct. The fact that the number are prime guarantees that this is so (if you write something as 2^a4^b then it's a different story)