A set of six positive integers contains an arithmetic sequence of four terms, a geometric sequence of four terms, and a harmonic sequence of four terms. What are the numbers in the set when the largest member of the set is a minimum?
Note: A harmonic sequence is a sequence whose reciprocals form an arithmetic sequence. ex: ({10, 12, 15, 20} is harmonic since {1/10, 1/12, 1/15, 1/20} is arithmetic)
(In reply to
solution by SilverKnight)
My thoughts behind this solution are:
Start with the harmonic sequence.
We need to find reciprocals, which when expressed in lowest terms, have a numerator of 1.
To minimize the numbers in the original set, we want to minimize the denominator.... but we need a denominator which is evenly divisible by the numerator (so that it can reduce to one).
The lowest integer that has 4 factors that are adjacent is 12. (1, 2, 3, 4, 6, and 12).
So... let's look at 1/12, 2/12, 3/12, 4/12 (arithmetic sequence)... the reciprocals are 12, 6, 4, and 3 respectively.
Let's assume that those four numbers are in the set of 6.
3, 6, and 12 are already in arithmetic sequence if we add the missing '9'...
3, 6, and 12 are already in geometric sequence if we add the missing '24'...
So... that's how I came up with these numbers.
re: solution
