1999 AMC 12 Problem 28
Attempt Problem 28 of the 1999 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1999 AMC 12 solutions, or check the answer key.
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28.
Let be a sequence of integers such that
(i) for ;
(ii) and
(iii)
Let and be the minimal and maximal possible values of respectively. What is
Answer: E
Solution:
Let be the numbers of s, s, and s. Then and giving and with
The sum of cubes is The minimum is at (value ) and the maximum at (value ), so
Thus, the correct answer is E.