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I don't know how I could find, WITH proof, the smallest possible least common multiple of three positive integers which sum to $2005$.

If someone can provide a proof I would be very happy.

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    @celtschk thank you for your help!2012-08-28

1 Answers 1

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$2005=4\cdot401$ with $401$ prime.

Then

$2005=2\cdot401+2\cdot401+401 \,.$

Thus we found three numbers with lcm 802. We claim that this is the smallest possible value.

Suppose by contradiction that we can find

$a+b+c =2005$ and $l=\operatorname{lcm}(a,b,c) < 802$. We can assume that $a \leq b \leq c$.

Since $b, c \leq l$ it follows that

$ 2005=a+b+c \leq a+2l < a+2\cdot802$

Thus $a > 401> \frac{l}{2}$. Since $a> \frac{l}{2}$ and $a\mid l$, it follows that $a=l$.

Then $l =a \leq b \leq c \leq l \Rightarrow a=b=c=l \Rightarrow 3l=2005$

Contradiction.

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    @celtschk Ty fixed. Nitesh You were welcome. The proof is pretty natural once you realize that the only way some numbers with fixed sum have low lmc is if they have a big common part. If 2005 would had have different prime factorization the problem could had been terrible....2012-08-28