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I solved it algebraically, one year at a time. Let n_k be the number of people with life span of k years (measured as of 2012). Let a_k be the average life span of all people still alive after k years (measured from 2012).From the problem definition, we havea_0 = 1.6 (average life span of all people)a_1 = 3 (because in one year, the remaining people have life span of 2 years, add 1 year to reference it back to starting year)a_2 = 5a_3 = 7a_4 = 9Now we can solve for the n_k's in order:n_1 + a_1*(100-n_1) = 100*a_0 ---> n_1 = 70, so 30 people remain after 1 year.2*n_2 + a_2*(30-n_2) = 30*a_1 ---> n_2 = 20, so 10 people remain after 2 years.3*n_3 + a_3*(10-n_3) = 10*a_2 ---> n_3 = 5, so 5 people remain after 3 years.4*n_4 + a_4*(5-n_4) = 5*a_3 ---> n_4 = 2, so 3 people remain after 4 years.The 3 people remaining after 4 years have an average life span of 9 years (measured from starting year), so simplest solution is to just make them all the same life span of 9 years.

For the last 3 remaining people, we require that the total life span is 27 people-years and each must have life span of at least 5 years (since they all survive the first four years). So we can have life spans of (9,9,9), (8,9,10), or many other cases. The most extreme would be (5,5,17). So we know for sure that nobody will live longer than 17 years.

yegulalp: Well done.Another motivation for the puzzle was to try to show that ignorance of algebra can lead to bad decisions.

I just posted an article that explains how this kind of process gives U.S. Treasury debt a tendency to lengthen even when the Treasury sells short-term securities.