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.

Let $1 \leq a_1, \dots, a_n \leq N$ be number chosen uniformly at random.  What is their least common multiple?
What is the highest power of $2$ dividing any of the $a_1, \dots, a_n$?


*

*All the numbers are odd with probability $(1 - \frac{1}{2})^n$

*All the numbers are odd or even (but not divisible by 4) with probability $(1 - \frac{1}{4})^n$

*All the numbers are odd or even (but not divisible by 8) with probability $(1 - \frac{1}{8})^n$

*...


So the expected power of $2$ dividing all these numbers is (this number might simplify):
$$ \mathbb{E}_2 = \sum_{k \geq 0} \Big(2^k - 2^{k-1} \Big) \left[1- \left( 1- \tfrac{1}{2^k} \right)^n \right]$$
A similar story for $\mathbb{E}_3, \mathbb{E}_5, \mathbb{E}_7$ etc; multiply your answers 
$$ \mathbb{E} = \prod_{p } \mathbb{E}_p$$

See also: Expectation of the maximum of i.i.d. geometric random variables a quick look confirms there is no closed-form answer. 
In your case $n = 21$ and $N = 1,000,000$ so we can hope for an estimate.

One way to state the prime number theorem is that $\mathrm{lcm}(1,2,\dots, n) = e^n$ So the least common multiple is growing exponentially fast in $n$.  In your case, there are $n = 21$ numbers ranging from $1$ to $N = 10^6$.  Perhaps
$$ \mathbb{E} \approx e^{n} \times \left(\frac{N}{n}\right)^n \approx \frac{N^n}{n!} = \frac{10^{21}}{ 21!} \approx \frac{10^{21}}{ 5 \times 10^{19}} = 40 $$  
Still doesn't seem quite right.  See Granville Prime Number Races)
