How many pairs of natural numbers can be formed whose LCM will be $49000$? I'm stuck on this question.
First, I factorize $49000$.
$49000 = 2^3 \cdot 5^3 \cdot 7^2$
I've just assumed the two numbers : $$P=(2^a \cdot 5^b \cdot 7^c)  \text{ and }  Q=(2^x \cdot 5^y \cdot 7^z)$$
Making the following pairs for :-
1) $$(a,x)=(0,3),(1,3),(2,3),(3,3),(3,0),(3,1),(3,2)$$
Number of $(a,x)$ pairs is $7$
Similarly,
2) Number of $(b,y)$ pairs is $7$
3) Number of $(c,z)$ pairs is $5$
so the total number of $(P,Q)$ pairs which can be formed $= 7\cdot7\cdot5 = 245.$
But I have doubt that I have over-counted. 
 A: The answer is correct if you consider ordered pairs, that means if you want to count $(1,49000)$ and $(49000,1)$ as different pairs.
Since the LCM is obviously symmetric, one could naturally ask about the number of unordered pairs. The gut reaction would be to divide the above answer by 2, but that leads to problems as $245$ is odd. 
The reason that approach doesn't work directly is that while the $2$ different ordered pairs $(1,49000)$ and $(49000,1)$ just become one unordered pair, only the $1$ ordered pair $(49000,49000)$ becomes the unordered pair $(49000,49000)$.
That means all the ordered pairs of the form $(x,x)$ become just one unordered pair, while the 2 ordered pairs $(x,y), (y,x)$ with $x\neq y$ become one unordered pair.
Since LCM$(x,x)=x$, the only ordered pair that applies to is actually $(49000,49000)$.
That means we have $1+\frac{245-1}2=123$ unordered pairs.
A: Your solution is correct if P and Q are not necessarily distinct and ordered.
If P and Q are distinct, then there are only $244$ pairs since you counted $49000$ twice. 
If the pair is unordered then clearly you counted everything twice so there are only $1+\frac{244}{2}=123$ pairs.
