LCM of randomly selected integers What is the expected LCM of 21 randomly selected positive integers under 10000000?
How would someone even approach this problem?
EDIT: The positive integers are chosen with replacement.

Whether you chose the integer with replacement or not is unlikely to make a difference: the probability to have 21 different integers in your set of 21 is
$$\exp\left(\sum_{i=0}^{20}\log\left(1-\frac i{10^7}\right) \right)\simeq 1 - \frac{20⋅21}{2⋅10^7}=1-2.1⋅10^{-5}$$.
A paper by Cilleruelo, Rué, Šarka and Zumalacárregui (arXiv:1112.3013 , J. Num. Th. 144 92) specifically addresses similar questions. They compute $\DeclareMathOperator\lcm{lcm}$ 
$$ψ(A)=\log\lcm\{a:a∈A\} $$ for random subsets $A⊆\{1…n\}$ of size $|A|=k(n)$ when $n→∞$. When $k(n)→∞$, their theorem 1.2 says that, almost surely 
$$ψ(A)=k\frac{\log\frac nk}{1-\frac kn}\left(1+O\left(e^{-C\sqrt{\log k}}\right)\right), $$
for a constant $C$.
In your case, you have $n=10^7$, $k=21$. Since $k$ is not really large, you should look into the paper to see how this is proven, and to which extent it applies to your case. Boldly assuming $k=21→∞$ (I know !), would led to
$$ψ(A)=21\frac{19.16}{1-2.1⋅10^{-6}}\left(1+O\left(e^{-1.74C}\right)\right) ∼ 400 \text{, if $C$ is big enough.} $$
In other words, we would have $\lcm(A)∼e^{400}∼5⋅10^{173}$ if my rough unjustified approximations are correct.
