<s>[INST] How many unordered pairs of positive integers $(a,b)$ are there such that $\operatorname{lcm}(a,b) = 126000$? 
How many unordered pairs of positive integers $(a,b)$ are there such that $\operatorname{lcm}(a,b) = 126000$?

Attempt:
Let $h= \gcd(A,B)$ so $A=hr$ and $B=hp$, and $$phr=\operatorname{lcm}(A,B)=3^2\cdot 7\cdot 5^3 \cdot 2^4\,.$$ Let $p = 3^a5^b7^c2^d$ and $r = 3^e 5^f 7^g 2^s$. Notice, that given $p$ and $r$, $h$ is determined, so we can count $p$ and $r$. Multiplying $p$ and $r$ we get $$pr = 3^{(a+e)} 5^{(b+f)} 7^{(c+g)} 2^{(d+s)}\,,$$ and so $a+e = 0,1,2$.
For the first case we have $0+1 = 1$ possibility, similarly $2$ and $3$ for the other cases, so the total number is $6$. For $b+f$ we have $b + f = 0,1,2,3$ giving $10$ options. Similarly for $c + g$ we have $3$ choices and for $d + h$ we have $$1+2+3+4+5 = 15$$ choices. Multiplying these together, we get $$15\cdot 3\cdot 6\cdot 10 = 60\cdot 45 = 2700\,,$$ which is not equal to the given answer of $473$.
Edit: sorry for the weird variables. I think I've fixed everything, if not, please do point it out
 [/INST] Think it through:  to find the lowest common divisor of $a$ and $b$ you take the prime factorization of $a$ and $b$.  And find the number whose prime factorization consists of the prime factors of  $a$ and of $b$ and raising them to the higher power they are raised to in $a$ and $b$.  In other words:
If $\{p_i\}$ are the prime factors of $a$ and/or $b$ and $a = \prod p_i^{k_i}$ ($k_i$ may be equal to $0$ if $p_i|b$ but $p_i\not \mid a$) and $b = \prod p_i^{j_i}$ (similarly $j_i$ may be $0$ if $p_i\not \mid b$ but $p_i|a$) then $\operatorname{lcm}(a,b) = \prod p_i^{\max (k_i,j_i)}$.
so if $\operatorname{lcm}(a,b)=126000 = 2^4*3^2*5^3*7$ then $a= 2^{k_1}*3^{k_2}5^{k_3}7^{k_4}$ and $b=2^{j_1}*3^{j_2}5^{j_3}7^{j_4}$ where $\max(k_i,j_i) = 4,2,3,1$
So it becomes a combinatorial matter of counting these.
This is easier if we consider ordered pairs.
One of $a$ or $b$ must have $2^4$ divide it.  There is $1$ way that both $a$ and $b$ have $2^4$ divide it.  There are $2$ ways that $2^4$ divides $a$ or $b$ but not the other. And the $2^0,2^1,2^2,2^3$ are four ways that $2$ may divide then others.  So there are $2*4 + 1=9$ we can factor up $2^4$ between $a$ and $b$.
We can do the same for the factors $3^2,5^3, 7^1$ to get $2*2+1=5;2*3+1=7;2*1+1=3$ ways, respectively to distribute those factors.
So there are $9*5*7*3 = 945$ *ordered pairs $(a,b)$ where $\operatorname{lcm}(a,b) =126000$.
But we need unordered pairs.  So we take the number of pairs where $(a,b)\ne (b,a)$ (ie; $a\ne b$) and divide by $2$.  And we take the pairs where $(a,b)=(b,a)$ (ie; $a=b$) and add those.
But if $a=b$ and $\operatorname{lcm}(a,b)=\operatorname{lcm}(a,a) = 126000$ then $a=b=126000$ and there is only one such pair.
So there are $945-1 = 944$ pairs with $a\ne b$.  So that is $\frac {944}2=472$ unordered pairs where $a\ne b$ and one pair where $a=b= 126000$.  So  there are $473$ pairs of unordered $(a,b)$ where $\operatorname{lcm}(a,b)=126000$.
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