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"We assume that n is an integer shows a multiplication of primal numbers factorization:

\begin{array}{ll} n=p_1^{k_1}*\cdots*p_m^{k_m} \\ &\end{array} that \begin{array}{ll} p_1,......,p_m \\ &\end{array} are different primal numbers and that \begin{array}{ll} m,k_1,......,k_m ∈ N \\ &\end{array}

find the number of the integers that divide n (includes n and 1)

I tried to solve it by failed many times

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    You should include what you have attempted towards a solution (even if it is a "failure"), so that we can see where you are struggling, and then we can help with those points. – 2017-01-26
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    You could say that i was completly on the wrong direction and i want to get idea about how such a thing can be solved – 2017-01-27

1 Answers 1

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The number of positive divisors of $n=p_1^{a_1}\cdots p_k^{a_k}$ , including $1$ and $n$ , is $$(a_1+1)\cdots(a_k+1)$$

If you allow negative divisors, the number of divisors must simply be multiplied with $2$

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    To arrive at this result, consider which of the prime powers can occur in a divisor. – 2017-01-27