a) If $k$ is odd . Sequence ($k!+2,k!+3,...,k!+k+1$) .

According to OEIS, there exist $89$ consecutive integers each divisible by a prime in $2,3,\cdots 43$, although I have no Idea how one would attempt to show this on the test... Edit: And according to this OEIS page, the minimal such sequence is $79622514581574,\cdots,79622514581659$, which satisfies $k=86$

RagvaloD wrote: a) If $k$ is odd . Sequence ($k!+2,k!+3,...,k!+k+1$) . To be precise, it works if k is non-prime, e.g., k=9.

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Edit: Turns out the above statement is incorrect. But I think this is true:

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3 is not divisible by any prime number less than 3

Part B (laegolas): Take $x=87890 = 2\cdot 5\cdot 11\cdot 17\cdot 47$. Then the sequence $x, x+1, \dots, x+17$ disproves the hypothesis.

For part a) Consider the simplest case at first. Take $k = 3$. There are only prime which is less than $3$, is $2$. Now think of such sequence that works and this is simple. I'm taking $(4,5,6)$. Notice that $5$ is the only member which isn't divisible by $2$(as here). Particularly, the sequences are just like to be $(2a, 2a+1, 2a+2)$. And $2$ doesn't divide the term $2a+1$. We want to make our hand dirty. Take $k = 5$. So the primes that are less than $5$ are $2$ & $3$. And there will be one such number in the sequence which is neither divisible by $2$ nor $3$. So the form of all such sequence will be $(6a, 6a+1, 6a+2, 6a+3, 6a+4)$. We don't care about whether any term of the sequence is divisible by $4$, because we are dealing with primes. See the case for k = $9, 13$ (just for clearing things). Now we have seen that for $k= 3,4,5$ it works. We may assume the statement is true. To prove this, we will just go through the induction. Proof:- We have checked the base case. By induction hypothesis, the statement is true for $k = m\geq 4$. That is, there exist sequences of $m$ consecutive positive integer $(F_a, F_{a+1}, F_{a+2},\cdots, F_{a+(m-1)})$ such that one number of the sequence is not divisible by any prime less than $m$. Let say those primes are $p_1, p_2,\cdots p_j$. And let $F = p_1.p_2\cdots p_j$ and $F$ doesn't divide the term $F_{a+1}$ Now, for $k = m+1$, we have two short cases to consider. Case $1$:- $m$ is prime. Then the primes less than $m+1$ are $p_1, p_2,\cdots, p_j,m$ Then the sequence be like $(F'_a, F'_{a+1},\cdots, F'_{a+(m-1)}, F'_{a+m})$. Where $F' = p_1.p_2\cdots p_j.m$ and last term that is $F'_{a+m}$ is divisible by $m$ and all other term except $F'_{a+1}$ is not divisible by $F'$ (by induction hypothesis). Case $2$:- $m$ is not prime. The sequence be in the form of $(F_a, F_{a+1}, F_{a+2},\cdots, F_{a+(m-1)}, F_{a+m})$ and m is either divisible by $p1$ or $p2$ or,$\cdots$,or $pj$ and all of the term except $F_{a+1}$ is divisible by $F$ ( also by induction) This completes the proof.