No the answer is 12.
Let me give you the general rule:
Write the number in factored form, like \(\displaystyle N=(p_1)^{n_1}\cdot(p_2)^{n_2}\cdot(p_3)^{n_3}\cdots(p_k)^{n_k}\) where each \(\displaystyle p_j\) is a prime factor and each \(\displaystyle n_j\in\mathbb{Z}^+~\&~n_j\ge 1\)
then the number of divisors of \(\displaystyle N\) is \(\displaystyle (n_1+1)\cdot(n_2+1)\cdot(n_3+1)\cdots(n_k+1)\)
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