Find Smallest Number For Number of Factors

I would like to know how you would fine the **smallest ordinary number** with **exactly** a certain number of factors given, for example:

The number of factors is 6

The **smallest number** with 6 factors is 12

12 =

__So the question is really:__

The number of factors is **n**

The smallest number with **n** factors is **X**

How do you find **X**?

Please feel free to use an example because I think doing it all in algebra will be unreasonably difficult

Re: Smallest number with given number of factors

Let N be any natural number and $\displaystyle N={p_1}^{a_1}{p_2}^{a_2}...{p_n}^{a_n}$ , where $\displaystyle p_1,p_2,...,p_n$ are prime numbers and $\displaystyle a_1,a_2,a_n$ are natural numbers.

Then number of factors of N is given by $\displaystyle (a_1+1)(a_2+1)...(a_n+1)$

You are given number of factors say k.First step would be to factorize k.That is write k as $\displaystyle (a_1+1)(a_2+1)...(a_n+1)$.

Once you have found out $\displaystyle a_1,a_2,..,a_n$ such that $\displaystyle a_1 \ge a_2 ...\ge a_n $ , your x would be $\displaystyle 2^{a_1} 3^{a_2} 5^{a_3} .... $

Main thing here is that you can multiple possibilities for $\displaystyle a_i 's $ and you have to optimize for that.

Example:Given k=12,find x.

k=3*2*2 . Corresponding x = $\displaystyle 2^{(3-1)} 3^{(2-1)} 5^{(2-1)} $=60

k=6*2 . Corresponding x = $\displaystyle 2^{(6-1)} 3^{(2-1)} $ = 96

k=4*3 . Corresponding x = $\displaystyle 2^{(4-1)} 3^{(3-1)} $ = 72

So minimum x would be 60.