Question

20 pairwise distinct positive integers are all smaller than 70. Prove that among their pairwise differences there are at least 4 equal numbers

Answer 1

Notice that we have selected 20 pairwise distinct positive integers and all are smaller than 70.

As they are positive so greater than 0, and pairwise distinct means no two integers are same. Thus we can put these numbers in order.

So let the numbers be

.

Now observe that if in this list there are 5 consecutive numbers then we see that there consecutive difference would give us four 1s, for example, say 5,6,7,8,9 are in this list, then we have

6-5=7-6=8-7=9-8=1

that is, their pairwise differences are all 1.

So if such a list exists then we don’t have to prove anything.

This is for any number may be 2 or 3 or any other.

On the contrary let the list be such that their pairwise differences have at most 3 equal numbers.

Now let us form the differences

This consists of 19 numbers (differences) (pigeons).

Now if we add all these numbers we find

Since the largest possible number is 69 and the smallest possible number is 1, so

Further observe that at the least if 3 differences are each 1, 2, 3, 4, 5, 6 which counts for 18 differences and one difference being 7 then also we have

That is, in the minimum possible ay also the sum of the differences put in this order gives sum as 70 but this is >68 and hence our assumption is false that their pairwise differences have at most 3 equal numbers.

This proves that the list must contain 4 pairwise differences which are equal.

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