Question

Find the product of all the divisors of the number 21600

Answer 1

Dear student
$$\begin{gathered} 21600=2^5\times 3^3\times 5^2 \\ \mathrm{So},\mathrm{number}\mathrm{of}\mathrm{divisors}=\left(2+1\right)\times \left(3+1\right)\times \left(5+1\right) \\ =3\times 4\times 6 \\ =72 \\ \mathrm{So},\mathrm{product}\mathrm{of}\mathrm{all}\mathrm{divisors}=\sqrt{{\mathrm{n}}^{\mathrm{\tau }\left(\mathrm{n}\right)}}=\sqrt{{21600}^{72}}={\left(21600\right)}^{\frac{72}{2}}={21600}^{36} \end{gathered}$$