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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?