From the question, we have:\(\displaystyle \left [ \begin{array}{l} N= p_1\times p_2^2\times p_3^2\quad (1) \\ N=p_4\times p_5^8\qquad (2) \end{array} \right.\)

with \(p_1, p_2, p_3,p_4,p_5\)are prime numbers.

From\((2)\), we cannot have \(p_5\leq3 \) because \(3^8=6561\). If \(p_5=2\), \(N=256 \times p_4\), thus,\(p_4 \leq3\) and\(p_4=3\) (beacause \(p_4 \not = p_5\)). So, \(N= 2^8\times3=768\).

From \((1)\):

If \(p_2=2\) and \(p_3=3\) (and vice versa), \(3\leq p_1\leq 27\), therefore \(p_1 \in \{5,7,11,13,17,19,23\}\). We have 7 numbers for this case.

If  \(p_2=2\) and \(p_3=5\)  (and vice versa), \(0\leq p_1 \leq 10\), therefore \(p_1 \in \{3, 7\}\) . We have 2 numbers for this case.

If  \(p_2=2\) and \(p_3=7\)  (and vice versa), \(0\leq p_1\leq5\) , therefore\(p_1 = 3\)  . We have 1 number for this case.

If  \(p_2=2\)  and \(p_3 \geq11\)   (and vice versa),  \(n \geq 2^2 \times 11^2 \times 3 =1452\)  , therefore we don't have any numbers for this case.

If  \(p_2=3\)  and \(p_3=5\)  (and vice versa),  \(p_1 \leq 4\) , therefore  \(p_1=2\) . We have 1 number for this case.

If  \(p_2= 3\)  and \(p_3=7\)   (and vice versa),  \(p_1 \leq 2\)  , therefore ​ ​\(p_1=2\)  . We have 1 number for this case.

​ If   \(p_2=3\)  and   \(p_3 > 7\)  (and vice versa),  \(N > 3^2\times11^2= 1089\), therefore we don't have any numbers for this case.

If \(p_2 >3 \) (or \(p_3>3\)), \(N\geq 5^2\times7^2= 1225\), therefore we don't have any numbers for this case.

Thus, we have 13 numbers satisfied the question.

Note that \(p_1 \not =p_2 \not = p_3\) and\(p_4 \not= p_5\).