2021 AMC 12B Problem 3Problem 3 of 25·Easier·AlgebraSuppose 2+11+12+23+x=14453.2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{2}{3+x}}}=\dfrac{144}{53}.2+1+2+3+x211=53144. What is the value of x?x?x?Answer choicesA34\dfrac{3}{4}431B78\dfrac{7}{8}872C1415\dfrac{14}{15}15143D3738\dfrac{37}{38}38374E5253\dfrac{52}{53}53525Submit answerStuck? Show hintsShow solutionSolutionWorking from the outside in, 14453−2=3853,\dfrac{144}{53}-2=\dfrac{38}{53},53144−2=5338, so the inner fraction equals 3853.\dfrac{38}{53}.5338. Its reciprocal gives 1+12+23+x=5338,1+\cfrac{1}{2+\frac{2}{3+x}}=\dfrac{53}{38},1+2+3+x21=3853, so 12+23+x=1538.\cfrac{1}{2+\frac{2}{3+x}}=\dfrac{15}{38}.2+3+x21=3815. Then 2+23+x=3815,2+\dfrac{2}{3+x}=\dfrac{38}{15},2+3+x2=1538, so 23+x=815,\dfrac{2}{3+x}=\dfrac{8}{15},3+x2=158, giving 3+x=154.3+x=\dfrac{15}{4}.3+x=415. Therefore x=154−3=34.x=\dfrac{15}{4}-3=\dfrac{3}{4}.x=415−3=43. Thus, the correct answer is A.