Find consecutive composite numbers How to find 100 consecutive composite numbers? After many attempts I arrived at the conclusion that to find $m$ consecutive composite numbers we can use this
$n!+2, n! +3, ..., n! + n$
where $n! + 2$ is divisible by $2$, $n! + 3$ is divisible by $3$ and so on...
and where $m$ = $n-1$
Thus $n!+2, n! +3, ..., n! + n$ tells that there are $(n-1)$ consecutive numbers. However, there seems to be some gaps or incompetence. For example:
$4!+2, 4! +3, 4! +4$ $→$ $26, 27, 28$.
Although it's right there are for sure smaller numbers such as $8, 9, 10$ and $14, 15 ,16.$ Is there another method for solving such a problem mathematically? Is it a correct method or have I misunderstood it?
 A: In general, finding the smallest such numbers would be not easy. Primes are, at times, tough to deal with. However, it can be made simpler. Rather than taking $n!$, you may take the LCM of the first $n$ numbers.
For example, for $n=5$, take $LCM(1,2,3,4,5)=60$ instead of $5!=120$.
So, instead of $122,123,124,125,126$, you take $62,63,64,65,66$.
Better than this, take the product of the primes not exceeding $n$.
For example, foR $n=5$, take $2\times 3\times 5=30$.
In this case, you take $32,33,34,35,36$.
A: If it was allowed to do 2 primality test, then you could define
$n = f(x)$
where $f$ is either factorial or primorial, and do a primality check for $n - 1$ and $n + 1$. If those are not primes, then you have
$n - x, ..., n - 3, n - 2, n - 1, n, n + 1, n + 2, ..., n + x$ all composite numbers. If $x$ is odd $> 1$, then you also have $n - x - 1$ and $n + x + 1$ composite, as they are even.
The smallest case if I'm not mistaken is $x = 5$ letting $f$ be factorial, with which we can have 13 consecutive composite numbers: $5! = 120$, $7 \mid 119$, $11 \mid 121$, so it means that from $120 - 5 - 1 = 114$ to $120 + 5 + 1 = 126$ all numbers are composite.
