Difference between revisions of "2024 AMC 10B Problems/Problem 23"
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==Solution 1== | ==Solution 1== | ||
Brute forcing gets you B) 319 | 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 <math>{\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,</math> and <math>{\frac{F_8}{F_4}} = {\frac{21}{3}} = 7.</math> The pattern is that then ten fractions are in their own Fibonacci sequence with the starting two terms being <math>1</math> and <math>3</math>, which can be written as <math>G_1 = 1, G_2 = 3, G_n = G_{n-1} + G_{n-2}</math> for <math>n \geq 3.</math> Summing the first ten terms, you arrive at the answer <math>\qquad\textbf{(B) } 319.</math> | ||
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==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=B|num-b=22|num-a=24}} | {{AMC10 box|year=2024|ab=B|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 01:46, 14 November 2024
Contents
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 
and 
The pattern is that then ten fractions are in their own Fibonacci sequence with the starting two terms being 
and 
, which can be written as 
for 
Summing the first ten terms, you arrive at the answer 
See also
| 2024 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
