Can I find a $2p_{k}$ consecutive numbers such that all of them are multiples of some $p_{1},p_{2},...,p_{k}$ Suppose that $p_{1},p_{2},...$ is the list of the prime numbers ($p_k$ represents the $k$-th prime, $p_{1}=2, p_{2}=3,...$ etc).  
Can I find $2p_{k}$ consecutive numbers such that all of them are multiples of some  $p_{1},p_{2},...,p_{k}$?
That is, given a number $k$, there is $a$ such that $\gcd(a+i,p_{1}p_{2}\cdots p_{k})\neq 1$ for all $i\leq 2p_{k}$?
 A: No.  Take $2,3,5$.  Any $10$ consecutive numbers have common factors with all three, actually.
$\pmod {p_1}$ there are only $p_1$ numbers, so some of the $2p_k\gt p_1$ would be zero.
As to the edit: if $\pi(a+2p_k)-\pi(a)\ge1$, then it's no.  Arbitrarily large prime gaps exist, though, so this doesn't rule it out.
A: No for $p_k\le 13$, and it is getting hard to find such ranges for $p_k\ge17$ (i.e., we can tell that $a$ must be large-ish if it exists at all).

For $k=1$, we can't: Among $4$ consecutive integers, one (in fact two)will be odd.
For $k=2$, among $6$ consecutive integers, $3$ are even and $2$ are multiples of $3$, hence at least $1$ is neither.
For $k=3$, among $10$ consecutive integers, $5$ are even, at most $2$ are odd multiples of $3$, and exactly one is an odd multiples of $5$, which gives us at most $8$ out of $10$ numbers with the desired property.
For $k=4$, among $14$ consecutive integers, there are $7$ even, at most $3$ odd multiples of $3$, at most $2$ odd multiples of $5$, and one odd multiple of $7$. Still $7+3+2+1<14$.
For $k=5$ and $22$ consecutive integers, the corresponding count gives $11+4+3+2+1<22$.
For $k=6$ and $26$ consecutive integers, the corresponding count gives $13+5+3+2+2+1=26$, but if there are indeed $5$ odd multiples of $3$ in the range, then one of them is in fact a multiple of $15$ that we overcount.
While we seem to be getting closer, it is meanwhile helpful to look for large prime gaps instead. For $p_k\ge17$ we need a gap $p_{m+1}-p_m>34$ and by inspection find that we therefore need $a\ge 9551$. It should be noted that now products of large primes such as $41\cdot 233=9553$ start taking the role of notable additional obstacles. 
A result by (Dusart 1998) tells us that for $x\ge 3275$, there is a prime between $x$ and $\left(1+\frac1{2\ln^2x}\right)x$. Hence we need
$$\frac a{2\ln^2a}>2p_k.$$
If $a\le 89693$, this means $p_k\le 167$, and it  can be checked by computer if a solution with such small $a$ and $p_k$ exists (I didn't do it, but suspect there is none).
For $a\ge 89693$, another result (Dusart 2016) says that there is a prime between $a$ and $\left(1+\frac1{\ln^3a}\right)a$, so that 
$$ \frac a{\ln^3a}>2p_k\quad\text{if }a\ge 89693.$$
As mentioned above, we meanwhile made $a$ so big that it is worthwhile looking for primes near $\sqrt a$ that (when multiplied) produce an obstacle. This might at least help us improve lower bounds for $a$.
