Bound on lcm of denominators of rational numbers that sum to 1. This is related to the question If a finite set of rational numbers sums to one, does one of the rationals have a denominator equal to the LCM of all the denominators?
Suppose $1 = \sum\limits_{i=1}^n \frac{p_i}{q_i}$, where $p_i,q_i$ are positive integers, and $\gcd(p_i,q_i)=1$. 
Let $k = \max_{i=1}^n \{ q_i\}$. $m=\operatorname{lcm}(q_1,\ldots,q_n)$.
Can we bound $m$ by $k$ and $n$? The obvious bound is $m\leq k^n$; however, I feel that there exists better bounds.
 A: Let $\Pi$ be the product of the $q_i$'s.  If $r$ is a prime dividing $m$, $r$ must divide some $q_i$, so let $e>0$ be maximal such that $r^e$ divides $q_i$, for some $i$, and fix such an $i$.  $r^e$ is then the largest power of $r$ dividing $m$.  Let $\nu_r$ be the $r$-adic valuation on $\Bbb Q$.   Then, since $\gcd(p_i,q_i)=1$, $\nu_r(p_i/q_i)=-e<0$, which is the minimum value $\nu_r(p_j/q_j)$ can take.  However, $\nu_r(\sum_{j=1}^n p_j/q_j)=\nu_r(1)=0$, so there must be some $i'\ne i$ with $\nu_r(p_{i'}/q_{i'})=-e$.  Therefore, $r^e\mid q_{i'}$ and so $r^{2e}\mid \Pi$.  Repeating this step for all primes $r\mid m$ gives $m^2\mid \Pi$.  Therefore, $m\le \sqrt{\Pi}\le k^{n/2}$.
A: Supposing that all the $q_i$ are distinct (otherwise if all $q_i$ are equal and all $p_i=1$ the bound $m=k^{n}$ is met): 
If $t|q_i$ and $t$ is prime then there exists some other $q_j$ where $t|q_j$ since $\sum_{i=1}^n \frac{p_i}{q_i}=\frac{p_1q_2...q_n+q_1p_2q_3...q_n+...+q_1q_2...p_n}{q_1q_2...q_n}$ and $gcd(p_i,q_i)=1$.
If the list $k, k-1,...,k-n+1$ contains x primes, remove those x primes, multiply together the remaining members of that list, and multiply by $(k-n+1)(k-n)...(k-n-x+1)$ to get a marginally improved bound.
