Prove that one of $n$ consecutive integers must be divisible by $n$

Question

If we have $n$ consecutive integers, then one of these integers is divisible by $n$. Prove the above statement.

Answer 1

HINTS: Let the integers be $m,m+1,\dots,m+n-1$. If $m$ is a multiple of $n$, you’re done. If not, write $m=qn+r$, where $q$ and $r$ are integers, and $0<r<n$. Here $q$ and $r$ are just the quotient and remainder when you divide $m$ by $n$. Let $c=n-r$.

1. Is $m+c$ one of the $n$ consecutive integers? In other words, is it true that $0\le c<n$?

2. Is $m+c$ divisible by $n$? Why?

Answer 2

There are $n$ residue classes mod $n$. If you have $n$ consecutive integers, they will fill up all $n$ residue classes. One of them must be in the residue class of things divisible by $n$; i.e. things $=0\pmod{n}$.

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Clarification: Let $C_k=\{k+j:0\le j\lt n\}$ be a set of $n$ consecutive integers.

Proposition 1: No two distinct elements of $C_k$ can be in the same residue class mod $n$.

Suppose otherwise: $j_1\lt j_2$ and $$ k+j_1\equiv k+j_2\pmod{n}\tag1 $$ Then $$ j_2-j_1\equiv0\pmod{n}\tag2 $$ which is impossible since $\color{#C00}{j_1\lt j_2}$ and $\color{#090}{k+j_1,k+j_2\in C_k}$ imply $$ 0\color{#C00}{\lt}j_2-j_1\color{#090}{\lt}n\tag3 $$ Proposition 2: For any $m$ some element of $C_k$ is $m\pmod{n}$.

Suppose otherwise: $C_k$ is a subset of the $n-1$ residue classes mod $n$ that are not $m\pmod{n}$. The Pigeonhole Principle says that two elements of $C_k$ must be in the same residue class mod $n$, which Proposition 1 says is impossible.

Answer 3

Proof by induction.

For all $i \in \mathbb{N}_{\ge 1},$ let $P(i)$ be the proposition: $$ \exists q \in \mathbb{N}_{\ge 1} \text{ such that } qn \in \{i, i+1, \ldots, i+n-1\}. $$

1. Base case: $P(1)$ is trivial since $n \in \{1, 2, \ldots, n\}.$

2. Induction hypothesis: Assume $P(i)$ holds for some $i > 1.$ In other words, there exists some $q$ such that $qn \in \{i, i+1, \ldots, i+n-1 \} \tag{1}$

3. Induction step: Consider the set $\{i+1, i+2, \ldots, i+n \}.$ The induction hypothesis in equation $(1)$ implies two cases:

   Case I: $i \neq qn,$ i.e. $qn \in \{i+1, \ldots, i+n-1\}.$ Hence $qn \in \{i+1, i+2, \ldots, i+n \}.$ So $P(i+1)$ holds.

   Case II: $i = qn,$ and hence $i+n = n(q+1).$ Since $i+n \in \{i+1, i+2, \ldots, i+n \},$ we have that $P(i+1)$ holds.

   We have just shown that $P(i) \implies P(i+1).$

QED.

Answer 4

Hint $\ $ Let $\rm\ a\ $ be the largest in the sequence. $\rm\: mod\ n\!:\ a\equiv r\,\in\, [\:\!0,\:\!n)\:$ so $\rm\:n\ |\ a\!-\!r\,\in\, [\!\:a,\:\!a\!-\!n)$

This has widespread applications, e.g. this proof that $\rm\:n\:|\:(n\!-\!1)!\:$ for composite $\rm\:n.$

Note $\ $ This is just a shifted equivalent of the Division Algorithm, i.e. the following equivalence is true in any euclidean domain $\rm Z\:$ (domain with an algorithm for division with 'smaller' remainder)

$$\begin{array}{rcrlrl} &\rm \exists\!\!\!&\rm q,r \in Z, &\rm\ a\: =\: q\:n + r, \!\!\!\!&\rm |r| \!\!\!&\rm <\ |n| \\ \displaystyle\rm{b\: =\: a\!-\!r\ \atop{\huge\iff\atop\phantom{M}}} &\rm \exists\!\!\! &\rm q,b\in Z, &\rm\ b\: =\: q\:n, \!\!\!\!&\rm |a\!-\!b| \!\!\!&\rm <\ |n| \end{array}\qquad$$

Answer 5

$\ let \ k,k-1,k-2,....,(k-(n-1)) \ are \ 'n' \ consecutive \ integers.$

$\begin{equation} \begin{aligned} k(k-1)(k-2)(k-3).....(k-(n-1))=& \frac{k(k-1)(k-2)(k-3).....(k-(n-1))(k-n)....1}{(k-n).....1}\\ =&{\frac{k!}{(k-n)!n!}.n!}\\ \end{aligned} \end{equation}$ $$\implies n!|k(k-1)(k-2)....(k-(n-1))$$ $$\implies n|k(k-1)(k-2)....(k-(n-1))$$