2024 AMC 10B Problems/Problem 23

Problem

Solution 1

Brute forcing gets you B) 319

Solution 2

Plug in a few numbers to see if there is a pattern. List out a few Fibonacci numbers, and then try them on the equation. You'll find that ${\frac{F_2}{F_1}} = {\frac{1}{1}} = 1, {\frac{F_4}{F_2}} = {\frac{3}{1}} = 3, {\frac{F_6}{F_3}} = {\frac{8}{2}} = 4,$ and ${\frac{F_8}{F_4}} = {\frac{21}{3}} = 7.$ The pattern is that then ten fractions are in their own Fibonacci sequence with the starting two terms being $1$ and $3$ , which can be written as $G_1 = 1, G_2 = 3, G_n = G_{n-1} + G_{n-2}$ for $n \geq 3.$ Summing the first ten terms, you arrive at the answer $textbf{(B) } 319.$ ~Cattycute

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2024 AMC 10B Problems/Problem 23

Contents

Problem

Solution 1

Solution 2

See also