Ordered pairs for a quarter hundred thousand

Since $25000 = 2^35^5$ , then, the number of ordered pairs of $(a,b)$ that satisfy $a\times b = 25000$ , $\quad n=(3+1) \times (5+1) = 4 \times 6 = \boxed{24}$

The 24 ordered pairs are as follows:

$\begin{matrix} 1\times 25000 & 25000\times 1 \\ 2\times 12500 & 12500\times 2 \\ 4\times 6250 & 6250\times 4 \\ 5\times 5000 & 5000\times 5 \\ 8\times 3125 & 3125\times 8 \\ 10\times 2500 & 2500\times 10 \\ 20\times 1250 & 1250\times 20 \\ 25\times 1000 & 1000\times 25 \\ 40\times 625 & 625\times 40 \\ 50\times 500 & 500\times 50 \\ 100\times 250 & 250\times 100 \\ 125\times 200 & 200\times 125 \end{matrix}$

It can be seen that the number of unordered pairs of $(a,b)$ that satisfy the equation is $\frac {1}{2} \times 24 = 12$ .

25000 = 2^3 x 5^5

We can find all positive factor by (3+1) (5+1) which is equal to 24.

2^3x 5^5

No. Of factors of this number =(3+1) * (5+1)= 24. A factor always needs a complimentary factor to form the product. Actually this number cab be expressed in 12 different ways as product of 2 numbers. But since ordered pairs are asked the answer will be 12x2= 24

25000 = 5^5. 2^3 , so have 24 different roots . then many pairs of a,b is (24/2). 2! = 24