Problem solving competition - CSU Fresno

Problem Solving Competition

Fourth Problem

The original Fibonacci sequence is 1, 1, 2, 3, 5, 8, 13, 21, 34, ...

A Fibonacci type sequence is a sequence such that
1. all terms are positive integers
2. the sequence is nondecreasing
3. every term is the sum of the 2 previous terms.
For example, 10, 12, 22, 34, 56, 90,... is a Fibonacci type sequence.

Question: what two starting numbers would give 100 as the sixth term? If there are any other solutions find them.

Solution

If the first two terms of the sequence are a and b, then the sequence is
a, b, a+b, a+2b, 2a+3b, 3a+5b.
We want 3a+5b=100. Since 3a=100-5b=5(20-b), a must be divisible by 5. Now consider the following cases:

a=5, then 15+5b=100, so b=17. The sequence is then 5, 17, 22, 39, 61, 100.
a=10, then 30+5b=100, so b=12. The sequence is 10, 12, 22, 34, 56, 100.
a=15, then 45+5b=100, so b=11. Then a>b, but the sequence must be nondecreasing.
If a>15, then 3a+5b>45+5b, and we have 100>45+5b. Then 55>5b, so 11>b, adn therefore a>b.

The only pairs giving a sequence satisfying the above conditions are (5, 17) and (10,12).

Questions? Email Rachel Mayo at rachelq@csufresno.edu