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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!

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Three year old competition problem? –  Harald Hanche-Olsen Jun 11 '13 at 20:47
    
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. –  Julian Kuelshammer Jun 11 '13 at 20:58
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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
    
A quick Python program. I don't think that's what the OP is looking for though –  Cocopuffs Jun 11 '13 at 21:32

1 Answer 1

up vote 4 down vote accepted

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 $lcm$.

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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

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