If (37/13) = 2 + (1/(x + (1/ y +(1/z)))), where x, y, z are natural numbers, then x, y, z are

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

(b) 1, 5, 2

\(= 2+\frac{1}{x+\frac{1}{y+\frac{1}{z}}}=\frac{37}{13} = 2\frac{11}{13}= 2+\frac{11}{13}\)

 \(⇒\frac{1}{x+\frac{1}{y+\frac{1}{z}}}= \frac{11}{13}⇒x+\frac{1}{y+\frac1z}=\frac{13}{11}\)

⇒ \(x+\frac{1}{y+\frac1z}=1+\frac2{11}\)

⇒ x = 1, y = \(\frac1z=\frac{11}{2}=5\frac12=5+\frac12\)

⇒ x = 1, y = 5, z = 2