# Find the lcm(a,b,c) given two conditions

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!

-
Three year old competition problem? –  Harald Hanche-Olsen Jun 11 '13 at 20:47
It will be some permutation of $a = 90$, $b = 180$, $c = 1740$ –  Cocopuffs Jun 11 '13 at 21:01
Hence $\text{lcm}(a,b,c)$= 5220. @Cocopuffs How do you arrive at your answer? –  K. Rmth Jun 11 '13 at 21:31
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 $lcm$.
So what you've essentially done is to take the polynomial that has $a,b,c$ as roots, and then evaluated it at 58. This is very nice and does get the answer, but it's still a bit unsatisfying to have to check so many cases at the end. Also, one is left wondering why the problem setter asked for the lcm, suggesting that s/he believed the lcm of the roots is easier to find than the roots themselves. I wonder if we're still missing something. –  WillO Jun 16 '13 at 4:52
Yes, it is cumbersome to check $(7+1)*(2+1)*(1+1)=48$ possible cases for $x$ and then try to solve for $y$ in the integers. I think that choosing the values for $x$ appropiately (starting from the lowest divisors) and assuming numerical skills are tested in the competition, this would be just in the limit of acceptability. Anyway, I can't find a way to solve the problem to the $lcm$ of $a$, $b$, and $c$ without actually calculating them. Maybe some cutoffs can be made in the choosing of $x$ and $y$, and that would simplify the problem to a smaller bunch of cases. –  chubakueno Jun 17 '13 at 5:30
For example, $AM \ge GM$ doesn't help very much in stablishing useful bounds. Divisibility conditions and simmetry like if $x$ is even and $y$ is odd, then necessarily $x=2^7*a$, do help in cutting some cases. That is the best tools for cutting off some cases that I can think of. –  chubakueno Jun 17 '13 at 5:38