Common LCM

First factorize $337500=2^2 \times 3^3 \times 5^5$

Now consider two numbers $a,b$ in the following way

$lcm(5^{a_1} \times 3^{b_1} \times 2^{c_1} , 5^{a_2} \times 3^{b_2} \times 2^{c_2})=337500$

To maintain the lcm these two numbers must contain the full powers of each $2,3,5$ .Therefore, any three of $a_1,b_1.c_!,a_2,b_2,c_2$ will equal to $2,3,5$ .So , performing this categorization there can be three possibilities as follows

$\large lcm(5^{5} \times 3^{3} \times 2^{c_1} , 5^{a_2} \times 3^{b_2} \times 2^{2})=337500$

$\large lcm(5^{5} \times 3^{3} \times 2^{2} , 5^{a_2} \times 3^{3} \times 2^{2})=337500$

$\large lcm(5^{5} \times 3^{3} \times 2^{2} , 5^{5} \times 3^{b_2} \times 2^{2})=337500$

In each case there are three variables which can vary in $(2+1) \times (3+1) \times (5+1)=72$ ways.But there are some common numbers are there Consider first two cases from above.Put $a_2=5$ and consider $b_1=b_2$ and $c_1=c_2$ at the same time.Then we can find (3 \times 4=12) same numbers just they are in different orders.As it is in the question that only unordered pairs are allowed so $12$ numbers should be subtracted.In this way fixing other two variables we can get $3 \times 6=18$ and $4 \times 6=24$ .So, total extra numbers are $18+12+24=54$ .So, allowed cases are $3 \times 72 -54=162$

Now another case is $a_1=5,b_1=3,c_1=2$ and other three will vary.Then there will be $2 \times 3 \times 5=30$ pairs.

Again last pair is $(1,337500)$ .

So, total number of pairs are $=162+30+1=193$