To solve these problems, we can use the pigeonhole principle, which states that if $n$ items are put into $m$ containers, and if $n > m$, then at least one container must contain more than one item.

Problem 15

Set: $\{1, 2, 3, 4, 5, 6\}$

The pairs that add up to 7 are:

• $(1, 6)$
• $(2, 5)$
• $(3, 4)$

These form three distinct pairs. Thus, we have three "containers."

To ensure that at least one pair adds up to 7, we need to select numbers from different pairs. If we pick one number from each of the three pairs, we could pick $1, 2, 3$ without any pairs summing to 7.

Therefore, selecting one more number (the fourth number) will guarantee that at least one pair sums to 7.

Conclusion for Problem 15: You must select 4 numbers.

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Problem 16

Set: $\{1, 3, 5, 7, 9, 11, 13, 15\}$

The pairs that add up to 16 are:

• $(1, 15)$
• $(3, 13)$
• $(5, 11)$
• $(7, 9)$

This gives us four distinct pairs.

Using the same logic as before, we can select one number from each pair. For example, we could select $1, 3, 5, 7$. In this case, we wouldn’t have any pairs that sum to 16.

If we select one more number (the fifth number), we will ensure that at least one pair sums to 16.

Conclusion for Problem 16: You must select 5 numbers.

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Summary:

• For the first problem, select 4 numbers.
• For the second problem, select 5 numbers.

Would you like any details or further explanations on these problems? Here are some related questions you might consider:

1. What are the pairs that sum to other numbers in different sets?
2. How does the pigeonhole principle apply to different types of problems?
3. Can this principle be used in combinatorics beyond simple addition?
4. How can we apply this concept in real-world scenarios?
5. What are some similar problems that involve pairs and sums?

Tip: When dealing with sets and sums, always identify pairs or groups first to simplify the problem.