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

Question

Let us observe a set that attempts to find 4 numbers that vary by more than 19 from each element. Such a set will be 60, 60-20=40, 60-20-20=20, 60-20-20-20=0. Since 0 is out of range we can say that such a set does not exist. Hence it is a proof by contradiction. Is it satisfactory? Is there another way to prove this directly?

Answer 1

Use the pigeon hole principle. Divide the set into three ranges each of length 20 (1 to 20, 21 to 40, 41 to 60. Then two of the four numbers must land in the same bin and will be within 19 of each other.