# Membership in a Math Circle

A Math Circle has 31 participant. Their ages are all different and sum up to 434 years. Prove that it is possible to find 20 participants whose total age is at least 280.

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Copyright © 1996-2018 Alexander BogomolnyA Math Circle has 31 participant. Their ages are all different and sum up to 434 years. Prove that it is possible to find 20 participants whose total age is at least 280.

Consider the oldest 20 participants. If the youngest among them is at least 14 then so are the others. Their total ages is then at least

If the youngest among the selected 20 is not yet 14, then so are the younger 11 whose total age is then at most

The problem is simple enough but may be criticized on several counts.

- If 31 kids are of different ages then at least one of them is not younger than 31 - a strange age for a school student.
- If there are 11 participants younger than 14 and all their ages are different, then at least one of them is at most 4 - also a strange age for a school student.
- The requirement that that the ages are different is thus a fluke.

Without the requirement that all ages are different, and assuming that all ages are integers, we have an accurate problem:

The total ages of a group of 31 people is 423. Prove that it is possible to find 20 memebers of the group whose total age is at least 280.

If the total age of the oldest 20 wass less than 280, their average age would be less than 14. Then the age of each of the remaining 11 would be at most 13 and their total

However, if the ages are allowed to be real numbers, 434 is the best (i.e., minimal) possible date for the total age.

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