A tricky elementary number theory question - least common multiple $M=m_1+m_2+...+m_n$. Show that $lcm\{m_1,m_2,...,m_n\}=M$ iff $m_i=M$ for some $i$ and $m_j=0$ for $j \neq i$.
My attempt:
I can only prove this when $n=2$. My strategy is to solve $M=i \cdot m_1$, $M=j 
\cdot m_2$, $m_1+m_2=M$. This requires $i=\frac{j}{j-1} \notin \mathbb{N}$. But I can't prove the general case. Is there a established result here?
 A: $$6=1+2+3$$
is a counterexample.
Note that your problem is equivalent to $\sum \frac{m_i}{M}=1$, which reminded me of the well known relation
$$1=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$$
Joffan showed below how to iterate this.
Another counterexample
Here is another way to construct a counterexample: Pick $n$ to be a perfect number.
Then 
$$n=\sum_{d|n, d<n} d$$
and the lcm of the elements on the right hand side is $n$.
The first perfect number is $6$ which leads to the same example as above, so let us look at the next one:
$$28=1+2+4+7+14$$
A: Building on the nice counterexample result from N.S., I'll just capture the iteration I noticed quickly:
$$\begin{align}
1=\frac 12 + \frac 13 + \color{blue}{\frac 16} & \quad \to \quad 6= 3+2+1 \\ 
1=\frac 12 + \frac 13 + \color{blue}{\frac1{12} +\frac 1{18} +\frac 1{36}} 
   & \quad \to \quad 36 = 18+12+3+2+1
\end{align}$$
Then I went looking for other results on unit fractions and found this paper by Nishiyama, Unit Fractions that sum to 1, and used some of the techniques suggested there to find a set that didn't include $1$:
$$1=\frac 12 + \frac 14 + \frac 17 +\frac 1{12} +\frac 1{42} 
   \quad \to \quad 84 = 42+21+12+7+2$$
