A natural number has exactly 10 divisors including 1 and itself.How many distinct prime factors can this natural number have?
options given:
(a) either $1$ or $2$ (b)$1$ or $3$ (c)either $2$ or $3$ (d) either $1$, $2$ or $3$
My Approach:
Any natural number is represented by its prime factors (eg. $2^a$ . $3^b$ . $5^c$ * ..... and so on).
The number of factors or divisors will be the product of all the powers after incremented by $1$ like $(a+1)$.$(b+1)$.$(c+1)$..and so on.
So, in the question we got 10, now 10 can be a result if either r a single power or product of 2 powers only and not more than that. So, here
Case 1: $(a+1)$=$10$ Case 2: $(a+1)$$(b+1)$ = $10$.
so the answer is either $1$ or $2$.
Is there Any better approach to solve this problem?Correct me if i am wrong.
