# Number Theory Least Common Multiple Question

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.

-
I suppose you want strictly positive integers, otherwise the answer is $0$. – Raskolnikov Aug 28 '12 at 16:11
$2005=5\cdot 401$ where both $5$ and $401$ are primes. Since $401$ is the larger one, it should be ideal if all summands contain that factor, that is, are of the form $k\cdot 401$. The three $k$-factors have to add to five, which for three positive integers is only possible for $1+2+2$, which gives $2005 = 401 + 802 + 802$ where $\operatorname{lcm}(401,802,802)=802$, therefore I'm pretty sure the correct answer is $802$. However I don't know how to prove it. – celtschk Aug 28 '12 at 16:12
@Raskolnikov: The way I learned it, "positive" means $>0$. – celtschk Aug 28 '12 at 16:14
Small lcm connote large gcd, which means a large prime factor shared in common, and $5$ and $401$ are the only prime factors of $2005$. – Michael Hardy Aug 28 '12 at 16:15
I got a rough lower bound that the LCM is greater than 668, and by getting the three numbers to divide each other, I got an LCM of 802. I see @celtschk went along the same lines. I think this is a pretty good candidate for a computer check :) – rschwieb Aug 28 '12 at 16:16

$2005=4\cdot401$ 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 $$l =a \leq b \leq c \leq l \Rightarrow a=b=c=l \Rightarrow 3l=2005$$

You can write $5\cdot401$ or $5\times401$. There's no need for such crudidities as $5*401$. – Michael Hardy Aug 28 '12 at 16:18
Shouldn't it be "$b,c\le l$" in line 7 (and then of course also for the first inequality sign in the following line)? That of course doesn't invalidate the proof. – celtschk Aug 28 '12 at 16:25