Difference between revisions of "2021 AMC 12B Problems/Problem 3"

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==Problem==
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Suppose<cmath>2+\frac{1}{1+\frac{1}{2+\frac{2}{3+x}}}=\frac{144}{53}.</cmath>What is the value of <math>x?</math>
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<math>\textbf{(A) }\frac34 \qquad \textbf{(B) }\frac78 \qquad \textbf{(C) }\frac{14}{15} \qquad \textbf{(D) }\frac{37}{38} \qquad \textbf{(E) }\frac{52}{53}</math>
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==Solution==
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{{solution}}
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==See Also==
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{{AMC12 box|year=2021|ab=B|num-b=2|num-a=4}}
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{{MAA Notice}}

Revision as of 21:08, 11 February 2021

Problem

Suppose\[2+\frac{1}{1+\frac{1}{2+\frac{2}{3+x}}}=\frac{144}{53}.\]What is the value of $x?$

$\textbf{(A) }\frac34 \qquad \textbf{(B) }\frac78 \qquad \textbf{(C) }\frac{14}{15} \qquad \textbf{(D) }\frac{37}{38} \qquad \textbf{(E) }\frac{52}{53}$

Solution

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See Also

2021 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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 12 Problems and Solutions

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