Question

The number of ways in which the number $10800$ can be resolved as a product of two factors.

Answer 1

Let we have a no. $N$ which we can factorize as

$N={\alpha }_1^{k_1}{\alpha }_2^{k_3}...{\alpha }_n^{k_n}$

Where ${\alpha }_1,{\alpha }_2,...,{\alpha }_n$ are coprime of each other and $k_1,k_2,...,k_n$ are natural numbers

Now, if $N$ is not a perfect square then the no. of ways in which it can be resolved as a product of two factors is given as

$\frac{(k_1+1)(k_2+2)....(k_n+1)}{2}$

Here $N=10800$

It's prime factorization will be

$10800=2^4\times 3^3\times 5^2$

Hence $N$ is not a perfect square

So, $\frac{(4+1)\times (3+1)\times (2+1)}{2}$

$=30$ ways.