Answer to the April 19, 2004 Problem of the Week

Turk-o-nacci Sequence!

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Like the Fibonacci sequence 1, 1, 2, 3, 5, 8, . . . a certain turkey flock has as many turkeys on a given day as the sum of the number of turkeys on the previous two days. If there were 79 turkeys on April 7th, and 542 turkeys on April 11th, how many turkeys were there on April 18th?

This problem is for our friends in Harrisonburg, Virginia -- the "Turkey Capital of the World."

 

Solution to the Problem:

There were 15,737 turkeys on the 18th.

Try to work with two unknowns, say A = the 5th number and B = the 6th number.
Then A + B = 79.
B + 79 = B + A + B = A + 2B = 8th,
2A +3B = 9th,
3A + 5B = 10th, and
5A + 8B = 11th = 542.

So solving the system {A + B = 79 , 5 A + 8 B = 542}
gives 5 A + 5 B = 395 , 3 B = 147, B = 49 , then A = 30 ;
we get 30, 49, 79, 128, 207, 335, 542, 877, 1419, 2296, 3715, 6011, 9726, 15737.
My son and my granddaughter were both born on April 18th, so I knew the answer had to be at least 2 turkeys... Just kidding, Mike and Carrie.

Larry Schwartz sent in the following: "This problem is simple if you use the Golden Ratio.
542 * phi^7 = 15737 where phi=(1+sqrt(5))/2."
I didn't think of that, and yet, every year, I assign the following program to my C++ Programming students: Find the first 20 Fibonacci numbers and write the ratio of each number to its successor. The answer approaches the number phi, which is approximately 1.618. Good thinking, Larry!