# Sums in an Arithmetic Progression

Let A be any set of 20 distinct integers chosen from the arithmetic progression 1, 4, 7,..., 100. Prove that there must be two distinct integers in A whose sum is 104.

Solution

Let A be any set of 19 distinct integers chosen from the arithmetic progression 1, 4, 7,..., 100. Prove that there must be two distinct integers in A whose sum is 104.

### Solution

The general term of the progression is in the form 3n + 1. 1 = 3×0 + 1, 100 = 3×33 + 1. There are 34 terms in all. We may split all the terms of the arithmetic progression into 18 subsets: two singletons {1} and {52} and 16 pairs (3n + 4, (33-n)3 + 1}. Of the chosen 19 numbers two are bound to fall into one of the eighteen sets. Since all the numbers are distinct, the singletons are out. Thus two of the chosen 19 numbers form a pair (3n + 4, (33-n)3 + 1}, for some n∈{0, 1, ..., 15}. The terms in a pair sum up to 104.

### Reference

1. 18.S34 Problem Solving Seminar, Fall 2007, MIT Opencourseware

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