Finding an integer (if one exists) $n$ such that $n$, $n+1$, $n+2$, $n+3$, $n+4$ are all composite I started off by thinking I would have to work $\bmod 24$ (as $24=1\cdot2\cdot3\cdot4$)
But I then decided to multiply all of the terms together, and have ended up with a rather large expression. I'm not really sure how to approach this question, if anyone could give me some guidance?
[I didn't know if maybe there was a theorem to do with primes being a certain distance apart that I may have overlooked?]
 A: As pointed out in these many solutions, for $n=m!$, $n+k$ are all composites for $2\le k\le m$. I just want to add another point. If we choose $n$ to be of the form $p-1$ where $p$ is a prime, then by Wilson's lemma, $n!+1$ is also composite with $n+1$ as a divisor.
A: The other answers have already remarked that $m! +k $ is composite for $2 \leq k \leq m,$ whenever $2 \leq k \leq m$ and $m \geq 2.$ The reason for this, is that $m!$ is clearly by each of $2,3,\ldots,m-1,m$ by definition of the factorial. Hence $m!+k$ is divisible by $k$ for $2 \leq k \leq m.$ On the other hand, $m!+k$ is clearly strictly greater than $k,$ so we must conclude that $m!+k$ is not prime, hence is composite, for each such $k.$ This is a fairly well-known method for showing that there are arbitrarily long sequences of consecutive composite numbers. In the case of this problem, we can see that $6! + 2, 6!+3, 6!+4, 6!+5$ and $6!+6$ are all composite, so taking $n = 6!+2$ gives one possible choice of $n$ (though not the smallest choice possible). While this general method does not usually give the smallest possible choice of $n,$ it is systematic, and does produce as long a sequence of consecutive composite numbers as needed.
A: Hint: Choose $n$ such that $n = m!$. Then we only need to verify $m!+1$ is positive. This way, we can say $m!+2, \cdots, m!+m$ are all composite. 
A: For a much smaller number
than $n!$,
choose the product of
all the primes $n$ or less.
By the prime number theorem,
this is about
$e^n$
(from Chebychev's
$\theta$ and $\psi$
functions).
A: Most of the answers so far have relied on the factorial function, so I thought I'd try something a little different.  The goal is the same:  show that there are infinitely many cases of five consecutive composite numbers.
Suppose the opposite, i.e., that every stretch of five consecutive "large" numbers contains at least one prime.  Then starting from any large prime $p$, the next prime is either $p+2$ or $p+4$, since $p+1$, $p+3$ and $p+5$ are all even.  Hence the sequence of differences between prime numbers eventually settles into a string of $2$'s and $4$'s (which we know, of course, not to be the case, but let's set that aside and prove it can't be the case).  This eventual sequence cannot have two consecutive $2$'s, since one of $p$, $p+2$, and $p+4$ must be divisible by $3$, and likewise (for the same reason) it does not have two consecutive $4$'s.  Thus the sequence of differences eventually alternates $...,2,4,2,4,2,4,....$  This implies $p$, $p+6$, $p+12$, $p+18$, and $p+24$ are all primes.  But one of these is divisible by $5$, since they are equivalent to $p$, $p+1$, $p+2$, $p+3$, and $p+4$ mod $5$.  And there's our contradiction.
Added later:  It occurs to me the argument above proves a little theorem:  Any interval of the form $[p,p+24]$ where $p$ is a prime greater than $3$ necessarily contains a string of five consecutive composite numbers.  In fact, a more careful analysis leads to a stronger statement:

Any interval of the form $[n,n+25]$ with $n\ge3$ contains a string of
  five consecutive composite numbers.

On the other hand, we also have:

An interval of the form $[p-4,p+20]$ contains no string of five
  consecutive composite numbers if $p$, $p+4$, $p+6$, $p+10$, $p+12$,
  and $p+16$ are all prime numbers.

The OEIS sequence $7, 97, 16057, 19417,\ldots$ gives the first few values for such primes $p$.  The general prime $k$-tuple conjecture implies there are infinitely many such primes.
A: All $a-1$ integers in the interval $$[a!+2,a!+a]$$
are composite.  because you need to find 5 consecutive numbers, Put $a-1=5$, then $n=6!+2$ is a solution.
