2021 AMC 12B Spring Problem 3

Below is the professionally curated solution for Problem 3 of the 2021 AMC 12B Spring, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12B Spring solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:continued fractionwork backwards

Difficulty rating: 1170

3.

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?

34\dfrac{3}{4}

78\dfrac{7}{8}

1415\dfrac{14}{15}

3738\dfrac{37}{38}

5253\dfrac{52}{53}

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.

Problem 3 in Other Years