<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] $126000 = 2^4\cdot 3^2\cdot 5^3\cdot 7^1$
For two numbers to have an LCM of $126000$, either $a$ or $b$ has to have $2^4$ in its prime factorization, and similarly with $3^2, 5^3, $ and $7$.
We need either $a$ and/or $b$ to have $2^4$, so we assign $2^4$ to $a$ first. Then, $b$ can have a factor from $2^0$ to $2^4$, for a total of $5$ total possibilities. We do the same for if $b$ as $2^4$, so $5\cdot 2 = 10$. However, we double-counted the case where both $a$ and $b$ have $2^4$ in it, so we subtract one to get $9$.
Similarly for $3^2$, set $a$ to have $3^2$ in its factorization. There are three choices for $b$, then swap $a$ and $b$ to get $3\cdot 2$, subtract one at the end to get $5$.
Same for $5^3$, we get $7$ and for $7^1$, we get $3$. $9\cdot 5\cdot 7\cdot 3 = 945$. However, we have a special case of $(126000, 126000)$ because this can be ordered in just one way. Therefore, we add one to $945$ to get $946$, then divide by $2$ because the pairs are unordered.
Hence, the answer is $$\fbox{473}.$$
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