Let \[ d_1\triangleq {\rm gcd}\left(a_i\frac{m}{b_i}:1\le i\le k\right)\quad\text{and}\quad d_2\triangleq {\rm gcd}\left(a_i:1\le i\le k\right). \]We will establish that $d_1=d_2$. \paragraph{ Step 1: $d_2\mid d_1$.} We first claim $d_2\mid d_1$: note that $d_2\mid a_i$ for $1\le i\le k$. Hence, $d_2\mid a_i\cdot m\cdot b_i^{-1}$ for $1\le i\le k$. Consequently, $d_2\mid d_1$. \paragraph{ Step 2: $d_1\mid d_2$.} This is the less obvious step. To that end let $p$ be a prime with $v_p(d_1)=\alpha\ge 1$. It follows \[ p^\alpha \mid a_i\frac{m}{b_i},\quad 1\le i\le k. \]We claim $p\mid a_i$ for every $1\le i\le k$. To that end, assume the contrary, and define the sets $I_1=\{i:1\le i\le k,p\nmid a_i\}$ and $I_2=\{i:1\le i\le k,p\mid a_i\}$. By the assumption, $I_1\ne\varnothing$. Note now that for any $i\in I_2$, $p\nmid b_i$ as $a_i$ and $b_i$ are pairwise coprime. Hence, \[ v_p\left(\mathrm{lcm}\left(b_i:1\le i\le k\right)\right) = \max_{i\in I_1}v_p(b_i). \]Assume now that this maximum is attained at $i_0\in I_1$. We then have \[ v_p\left(a_{i_0}\frac{m}{b_{i_0}}\right) = 0. \]But this is a contradiction with the fact $p\mid a_{i_0}\cdot m\cdot b_{i_0}^{-1}$. From here, it follows $I_1=\varnothing$, that is, $p\mid a_i$ for $1\le i\le k$. With this, $p\nmid b_i$, and therefore $v_p({\rm gcd}(a_i\cdot m\cdot b_i^{-1}:1\le i\le k)) = v_p({\rm gcd}(a_i:1\le i\le k))$, thus $d_2\mid d_1$. Combining the findings of Steps 1 and 2 above, we establish the desired conclusion.

So, we use $v_p(\gcd(a_1,a_2,\cdots,a_k))=\min(v_p(a_i))$ and $v_p(\mathrm{lcm}(b_1,b_2,\cdots,b_k))=\max(v_p(b_i))$, we need $$\min(v_p(a_i)+\max(v_p(b_i))-v_p(b_i))=\min(v_p(a_i)).$$Choose $i$ so that $\max(v_p(b_i))=v_p(b_i)$, then $v_p(b_i)>0$, we get that $v_p(a_i)=0$ and thus $$\min(v_p(a_i)+\max(v_p(b_i))-v_p(b_i))=0=\min(v_p(a_i)).$$If $v_p(b_i)=0$, then $v_p(b_i)=0$ for all $i$, hence $$\min(v_p(a_i)+\max(v_p(b_i))-v_p(b_i))=\min(v_p(a_i)+0-0)=\min(v_p(a_i)),$$we are done. I apologize for my vague usage of $i$.

@Amir Hossein AoPS User How did you get a space in your username?

Pi31 wrote:

@Amir Hossein AoPS User How did you get a space in your username? his account was made in the early days of aops where there probably wasn't any restrictions

We wish to prove that $\gcd\left(\frac{a_i m}{b_i}\right)=\gcd(a_i)$ where $a_i=a_1,a_2,\dots,a_k,$ $b_i=b_1,b_2,\dots,b_k,$ and $\frac{a_i m}{b_i}=\frac{a_1 m}{b_1},\frac{a_2 m}{b_2},\dots,\frac{a_k m}{b_k}.$ Notice that \begin{align*}\nu_p\left(\gcd\left(\frac{a_im}{b_i}\right)\right)&=\min\left(\nu_p\left(\frac{a_im}{b_i}\right)\right)\\&=\min(\nu_p(a_i)+\nu_p(m)-\nu_p(b_i))\\&=\min[(\nu_p(a_i)+\max(\nu_p(b_i))-\nu_p(b_i)]\\&=\min(\nu_p(a_i))+\min(\max(\nu_p(b_i))-\nu_p(b_i))\\&=\min(\nu_p(a_i))\\&=\nu_p(\gcd(a_i)).\end{align*}$\square$