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.

  • 7
    $\begingroup$ Looks good. You should also add a couple of words about why there cannot be more than $2$ prime divisors. $\endgroup$
    – 5xum
    Aug 27, 2015 at 10:17


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