TITLE: Find the lcm(a,b,c) given two conditions
QUESTION [9 upvotes]: This is a problem in a Italian competion of three years ago. Could you help me to solve this problem? 
Given $a,b,c\in \mathbb{N}$ such that
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{58}  $$
and 
$$a+b+c=2010$$
Find the $\text{lcm}(a,b,c)$
I'm trying to solve this problem without any use of software, and I would like to prove it with an elegant proof. Thanks!

REPLY [4 votes]: Algebraically manipulating:
$$58(ab+bc+ac)-abc=0$$
$$58^3-58^2(a+b+c)+58(ab+bc+ac)-abc=58^3-58^2(a+b+c)$$
$$(58-a)(58-b)(58-c)=58^2(58-2010)$$
$$(a-58)(b-58)(c-58)=58^2(1952)$$
$$(a-58)(b-58)(1952-a-b)=2^7*29^2*61$$
Replacing conviniently:
$$x=a-58,y=b-58$$
$$xy(1836-x-y)=2^7*29^2*61$$
Then you only have to choose $x$ and $y$ as divisors of the number, check if the final equation holds, replace for $a$ and $b$, and calculate the $lcm$.