Now generalize this

Let the product be $P$ , then we have:

$\begin{aligned} P & = \left \lceil \sqrt{2} \right \rceil \times \left \lceil \sqrt{3} \right \rceil \times \left \lceil \sqrt{\color{#3D99F6}{4}} \right \rceil \times \left \lceil \sqrt{5} \right \rceil \times \left \lceil \sqrt{6} \right \rceil \times \left \lceil \sqrt{7} \right \rceil \times \left \lceil \sqrt{8} \right \rceil \times \left \lceil \sqrt{\color{#D61F06}{9}} \right \rceil \times \left \lceil \sqrt{10} \right \rceil \times ... \times \left \lceil \sqrt{99} \right \rceil \\ & = 2 \times 2 \times \color{#3D99F6}{2} \times 3 \times 3 \times 3 \times 3 \times \color{#D61F06}{3} \times 4 \times ... \times 10 \quad \quad \small \color{#3D99F6}{\text{Note: }\left \lfloor \sqrt{m} \right \rfloor \text{ increases by 1 when } m-1 \text{ is a perfect square.}} \\ & = 2^{4-1}3^{9-4}4^{16-9}...k^{k^2-(k-1)^2}...10^{99-81} \\ & = \color{#3D99F6}{2^3}3^5\color{#D61F06}{4^7}5^{9}\color{#3D99F6}{6^{11}}7^{13} \color{#D61F06}{8^{15}}9^{17}\color{#3D99F6}{10^{18}} \\ & = \color{#3D99F6}{2^3} \color{#D61F06}{(2^2)^7} \color{#3D99F6}{(2^{11}3^{11})} \color{#D61F06}{(2^3)^{15}} \color{#3D99F6}{(2^{18}5^{18})} 3^5 5^{9}7^{13} 9^{17} \\ & = 2^{\color{#3D99F6}{3}+ \color{#D61F06}{14}+ \color{#3D99F6}{11} + \color{#D61F06}{45}+\color{#3D99F6}{18}} 3^{16} 5^{9}7^{13} 9^{17} \\ & = 2^{\color{#3D99F6}{91}} 3^{16} 5^{9}7^{13} 9^{17} \end{aligned}$

$\Rightarrow n = \boxed{\color{#3D99F6}{91}}$ .

Suppose that the expression is equal to A. Then, $A = 2^{3} \times 3^{5} \times 4^{7} \times 5^{9} \times 6^{11} \times 7^{13} \times 8^{15} \times 9^{17} \times 10^{18}$ Asked : The largest non-negative integer n such that $2^{n}$ is a factor of A So, $n = 3 + 2 \times 7 + 11 + 3 \times 15 + 18$ Then, $n = 91$

Is there any way to generalize it not only for 2 but for other prime factors too?