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Divisors problem

mathworker

Active member
May 31, 2013
118
Let \(\displaystyle N\) be a set such that \(\displaystyle K \text{ belongs to }N\) iff \(\displaystyle K=P_1P_2....P_n\) such that \(\displaystyle P_i=P_j \text{ iff }i=j\) find a smallest K in N such that number of divisors of K=32
edit:sorry for using same variable in two places.k is changed to n
 
Last edited:

mathworker

Active member
May 31, 2013
118
Re: divisors problem

Hint:
though backstabing humans is not allowed backstabing problems is allowed.
To find the smallest begin from the smallest
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Re: divisors problem

[JUSTIFY]Just to check, $P_1, P_2, \cdots$ are meant to be prime, right? (I think so because of the $P_i = P_j ~ ~ \iff ~ ~ i = j$ part).

If so, then $N$ basically contains all integers which no prime divides twice, i.e:

$K = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_n^{\alpha_n} \in N ~ ~ \implies ~ ~ \alpha_1 = \alpha_2 = \cdots = \alpha_n = 1$

Then from this prime power factorization it follows that the number of divisors of $K$ is:

$$d(K) = \prod \left ( \alpha_i + 1 \right ) = 2^n ~ ~ (\text{since all} ~ \alpha_i = 1)$$

And $d(K) = 32$ when $n = 5$. So we are looking for the smallest $K$ which has five distinct prime factors. Trivially, the solution is:

$$p_5 \# = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 = 2310$$
[/JUSTIFY]
 

mathworker

Active member
May 31, 2013
118
Re: divisors problem

Bacterius+ (Clapping)you got it(flower)