What is the smallest natural number such that it has $ 40 $ distinct positive (integer) divisors (inclusive of $ 1 $ and itself?
At first I was stunned of seeing the problem.It's not possible to find all the divisors of all the numbers. I think it's a huge calculation. How can I solve this type of problem easily? help me.How can I proceed?
Elementary Number Theory – Find Smallest Number with 40 Distinct Positive Divisors
algebra-precalculuselementary-number-theory
Best Answer
Letting $$N={p_1}^{q_1}\cdot {p_2}^{q_2}\cdots \cdot {p_k}^{q_k}\ \ (p_i\ \text{are primes}, q_i\ge 1\in\mathbb N,k\in\mathbb N)$$ be your number, the following has to be satisfied (see here for details): $$(q_1+1)(q_2+1)\cdots(q_k+1)=40=2^3\cdot 5.$$ (Here, LHS represents the number of the positive divisors of $N$.)
So, separate it into cases as the followings :
(1) Since $40=40$, $N=p^{39}$. Hence, we have $2^{39}.$ (Note this is the smallest number in this case)
(2) Since $40=2\times 20$, $N={p_1}^{1}\cdot {p_2}^{19}$. Hence, we have $3^1\cdot 2^{19}.$
(3) Since $40=4\times 10$, $N={p_1}^{3}\cdot {p_2}^{9}$. Hence we have $3^3\cdot 2^9$.
Can you take it from here? Note that there are still several cases.