How many different integer values—not necessarily members of the set—can be…

Concept

For an arithmetic progression a + di, choosing k distinct terms is equivalent to choosing k distinct indices. Each term-sum equals ka plus d times the corresponding index-sum.

If the possible index-sums fill every integer from a minimum L to a maximum U, then there are U − L + 1 distinct term-sums. A one-index exchange proves this gap-free property.

Application

1. Write the eight terms as 3 + 7i, where i = 0, 1, …, 7.

2. For three distinct indices i < j < k, the selected-number sum is (3 + 7i) + (3 + 7j) + (3 + 7k) = 9 + 7(i + j + k). Thus distinct number-sums correspond exactly to distinct values of i + j + k.

3. The minimum index-sum is 0 + 1 + 2 = 3, and the maximum is 5 + 6 + 7 = 18.

4. Unless the chosen indices are already {5, 6, 7}, some chosen index has an unused successor. Replacing that index by its successor increases i + j + k by exactly 1. Repeating this exchange reaches every integer from 3 through 18.

5. Therefore the number of distinct sums is 18 − 3 + 1 = 16. The actual sums run from 9 + 7·3 = 30 to 9 + 7·18 = 135 in steps of 7.

Cross-check

The attained sums are 30, 37, 44, 51, 58, 65, 72, 79, 86, 93, 100, 107, 114, 121, 128, and 135—sixteen values.

Result

Hence 16 different integer values can be formed.