Multiple of $k$ in $k$ consecutive natural numbers.

Question

If $Q$ denotes the set of $k$ arbitrarily chosen consecutive natural numbers, prove that there exists exactly one natural number in the set $Q$, such that it is of the form "$kn$" and $n,k \in \mathbb{N}$ { $\mathbb{N}$ is the set of natural numbers}?

Answer 1

Hints:

Suppose $n, n+1,\dots, n+k-1$ one such list of $k$ consecutive numbers.

If none of them is a multiple if $k$, there's one multiple of $k$ that's $<n$ and the next multiple is $\ge n+k$. Explain why it's impossible.

If two or more multiples of $k$ are in the list, there are two consecutive multiples. Explain again why it' impossible.

Answer 2

$Q$ contains a minimum value, $m$, and the succeeding $k{-}1$ numbers, $m{+}1, \ldots,m{+}(k{-}1)$. So $Q=\{m{+}0,m{+}1,\ldots, m{+}k{-}1 \}$

Now let $ck$ be the largest multiple of $k$ with $ck<m$. Then $m=ck+r$, with $0< r\le k$.

Now $(c+1)k = ck + k = ck+r+(k-r) = m + (k-r)$ is a multiple of $k$. And $k-r \in \{0,1,\ldots, k{-}1\}$ so $m + (k-r)\in Q$ as required.