Find the smallest integer $n$ with $\tau(n)=a$ with $\tau(n)$ the number of dividers of n.
- $\tau(n)=1$
- $\tau(n)=2$
- $\tau(n)=3$
- $\tau(n)=6$
The soluion is given by:
The book does not give a solution method, I just tried it by trail and error. What is the best method for such an excercice?
This can be done by factoring the number into prime factor.Read this https://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic , SO there is only one unique to factorize a number into prime factors. Suppose Given number is N for unique way to reduce this into prime factor is $p1^{e1}*p2^{e2}......pn^{e^{m}}$ Here $p1 < p2 < p3....$. Now the total number of divisors of this number is $ (e1 + 1)*(e2 + 1) *(e3 + 1).....(em +1)$ So if you are given a number of divisors of Nyou can easily find the number by factoring it. Just take $τ(n)=3$ so prime factor of 3 is 3 and it is = $(e1 + 1)$. Now put this in form o$p1^{e1}*p2^{e2}......pn^{e^{m}}$ Here $p1 < p2 < p3....$. it will be $2^{2} $ whic is equal to 4 and hence 4 is the smallest number with 3 divisors which are $ 1, 2 ,4$ . Hope this helps.