Compute the number of ordered pairs of positive integers (a,b) such that the lcm of a and b is $3^3*5^7*11^{13}$.

Question

I know that the gcd of a and b is 1, so is it true that I only need to calculate the ways $3^3\cdot 5^7 \cdot 11^{13}$ can be broken up?

Answer 1

Not exactly. By denoting with $\mathcal{P}=\{p_1,p_2,p_3,\ldots\}=\{2,3,5,\ldots\}$ the set of primes, any $n\in\mathbb{N}^+$ can be written in a unique way as $\prod_{k\geq 1} p_k^{\alpha_k}$ with $\alpha_k\in\mathbb{N}$ (this is the fundamental theorem of Arithmetics, i.e. the statement "$\mathbb{Z}$ is a UFD"). If $m\in\mathbb{N}^+$ is given by $\prod_{k\geq 1}p_k^{\beta_k}$, then $$\text{lcm}(m,n) = \prod_{k\geq 1}p_k^{\max(a_k,b_k)}.$$ If $\text{lcm}(m,n)=3^3\cdot 7^7\cdot 11^{13}$, the only primes potentially appearing in the factorizations of $m$ and $n$ are $3,7,11$, and $$ \left|\{(m,n)\in\mathbb{N}^+\times\mathbb{N}^+:\text{lcm}(m,n)=3^3\cdot 7^7\cdot 11^{13}\}\right| $$ equals $$ \left|\{(a,b,c,A,B,C)\in\mathbb{N}^6: \max(a,A)=3, \max(b,B)=7,\max(c,C)=13\}\right| $$ i.e. $$ (2\cdot 3+1)\cdot (2\cdot 7+1)\cdot (2\cdot 13+1) = 7\cdot 15\cdot 27 = \color{blue}{2835}.$$ On the other hand, $$ \left|\{(m,n)\in\mathbb{N}^+\times\mathbb{N}^+:\text{lcm}(m,n)=3^3\cdot 7^7\cdot 11^{13},\color{red}{\gcd(m,n)=1}\}\right| $$ simply equals $2\cdot 2\cdot 2=\color{blue}{8}$.