Show that if four distinct integers are chosen between $1$ and $60$ inclusive, some two of them must differ by at most $19$

Solution 1:

In fact, since $\ 2-1=3-2=1\ $, and $\ 3-1=2\ $ there are three pairs of your numbers that differ by much less than $19$, and hence also by at most $19$.

Let $\ a,b,c,d\ $ be four distinct numbers between $1$ and $60$ arranged in increasing order:$\ a<b<c<d\ $. Then, since $\ d\le60\ $ and $\ a\ge1\ $, we must have $\ d-a\le59\ $. Let $\ m=\min(b-a,c-b,d-c)\ $. Then \begin{align} b-a&\ge m\ ,\\ c-b&\ge m\ \text{, and}\\ d-c&\ge m\ . \end{align} If we add these inequalities, we get $\ 59\ge d-a\ge3m\ $, which gives $\ m\le\frac{59}{3}<20\ $, and since $\ m\ $ must be an integer, it can be at most $\ 19\ $. It follows that at least one of the pairs $\ \{a,b\}, \{b,c\},\ $ and $\ \{c,d\}\ $ of numbers differs by at most $\ 19\ $.