Let $k \geq 2 \in \mathbb{N}$ Show that each of the numbers $k! +2, k! +3, ... , k! +k$ is composite.

So, firstly I'm not sure I fully understand the question. Surely let $k = 2$ then $2!+3 = 5$, which is prime. This means the statement fails surely?

Is this reasoning correct and the question is incorrect (or am I missing something obvious?)

Thanks in advance.

  • $\begingroup$ If $k=2$, you're only looking at $2!+2$. For $k=3$, you'll look at $3!+2$, $3!+3$. For $k=4$, you'll look at $4!+2, 4!+3, 4!+4$. ... $\endgroup$ – David Mitra Jan 31 '15 at 16:48

Note that $k!+2,k!+3,\ldots,k!+k$ ends with $k!+k$ which suggests that the sum stops once you reach $k$. So, for $k=2$, this means that the $2!+3$ is not included. I do think that the question could have been better written as, "Let $k\geq2$ with $k\in\mathbb{N}$. Show that each $k!+i$, with $2\leq i\leq k$, is composite."

In order to answer this question, note that $n\mid k!$ for all $1\leq n\leq k$ so that we may write $k!+n = n\left(1+\frac{k!}{n}\right)$ where $\frac{k!}{n}$ must necessarily be an integer.


Notice we are talking about $k!+n$ with $n\leq k$ which was not the case in the example you gave. Notice that in $k!+n$ with $n\leq k$ we can factor $n$ and therefore is composite


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