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2021 AMC 12B Problem 3

Problem 3 of 25EasierAlgebra

Suppose 2+11+12+23+x=14453.2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{2}{3+x}}}=\dfrac{144}{53}. What is the value of x?x?

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Solution

Working from the outside in, 144532=3853,\dfrac{144}{53}-2=\dfrac{38}{53}, so the inner fraction equals 3853.\dfrac{38}{53}. Its reciprocal gives 1+12+23+x=5338,1+\cfrac{1}{2+\frac{2}{3+x}}=\dfrac{53}{38}, so 12+23+x=1538.\cfrac{1}{2+\frac{2}{3+x}}=\dfrac{15}{38}. Then 2+23+x=3815,2+\dfrac{2}{3+x}=\dfrac{38}{15}, so 23+x=815,\dfrac{2}{3+x}=\dfrac{8}{15}, giving 3+x=154.3+x=\dfrac{15}{4}. Therefore x=1543=34.x=\dfrac{15}{4}-3=\dfrac{3}{4}. Thus, the correct answer is A.

More practice

Concepts: continued fraction · work backwards

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.