Question

If $\frac{37}{13}=2+\frac{1}{x+\frac{1}{y+\frac{1}{z}}}$, where $x,y,z$ are natural numbers, then $x,y,z$ are

• A. $1,2,5$
• B. $1,5,2$
• C. $5,2,11$
• D. $11,2,5$

Answer 1

The correct option is B $1,5,2$

$2+\frac{1}{x+\frac{1}{y+\frac{1}{z}}}=\frac{37}{13}=2\frac{11}{13}=2+\frac{11}{13}$
$\Rightarrow \frac{1}{x+\frac{1}{y+\frac{1}{z}}}=\frac{11}{13}\Rightarrow x+\frac{1}{y+\frac{1}{z}}=\frac{13}{11}$
$\Rightarrow x+\frac{1}{y+\frac{1}{z}}=1+\frac{2}{11}$
$\Rightarrow x=1,y+\frac{1}{z}=\frac{11}{2}=5\frac{1}{2}=5+\frac{1}{2}$
$\Rightarrow x=1,y=5,z=2$