# Show that for every set of 18 integers there will be two that are divisible by 17 [closed]

I understand the pigeonhole principle is needed here and I see the solution in the back of the book, but the explanation is week. If anyone could explain step-by-step that would be awesome!

## closed as off-topic by Daniel, Harish Chandra Rajpoot, user147263, N. F. Taussig, yoknapatawphaOct 28 '15 at 13:36

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• As it is stated it's false : the set $\{17k+1 : k\in[|1,18|]\}$ has no number divisible by $17$. Maybe it's "the difference between two integers" instead – Tryss Oct 28 '15 at 4:27
• – user147263 Oct 28 '15 at 5:31

Given a set of $18$ integers $x_1,x_2,\dots,x_{18}$, there will exist two integers $x_i, x_j$ with $1\leq i<j\leq 18$ such that $x_i-x_j$ is divisible by $17$.
By the quotient remainder theorem, each of the integers $x_i$ can be written in a unique way as $17q_i+r_i$ where $q_i$ and $r_i$ are both integers and $0\leq r_i<17$. We treat the values of $r_i$ as the holes and the elements in our set as the pigeons. There are then $17$ holes and $18$ pigeons.
By the pigeon-hole principle, there must be two terms with the same remainder. Without loss of generality, suppose that it was $x_i$ and $x_j$ where $x_i=17q_i+r$ and $x_j=17q_j+r$. Then $x_i-x_j = 17q_i+r-17q_j-r = 17(q_i-q_j)$ is therefore divisible by $17$.
The way it is currently worded (Show that for every set of 18 integers there will be two that are divisible by 17) is false. We can have a set of $18$ integers such that none of them are divisible by $17$, for example the set $\{1,2,3,4,\dots,16,18,19\}$.
Even if we were to add additional constraints that each of the integers is consecutive, we have the counterexample $\{1,2,3,\dots,18\}$ has exactly one element divisible by $17$, not two.