This question was uploaded on 12/09/21 on social media accounts.
I got this idea from AIME 1991 Problem number 7. It was uploaded on YouTube. This is a ladder-type equation in which similar terms are going in denominator up to infinite times.
Reference: https://www.youtube.com/watch?v=eV71XhDdrbw
Solution 1:
Given Equation:
\[x=15-\frac{56}{15-\frac{56}{15-\frac{56}{x}}}\]
If we replace x in RHS every time then we get a ladder-like sequence up to infinity.
\[x=15-\frac{56}{15-\frac{56}{15-\frac{56}{15-\frac{56}{15-\frac{56}{15-\frac{56}{15-\frac{56}{15-\frac{56}{15-\frac{56}{15-\frac{56}{15-\frac{56}{.....(upto⠀\infty)}}}}}}}}}}}\]
This can also be written as:
\[x=15-\frac{56}{x}\]
This form is simpler and easy to solve for x.
\[\Rightarrow x^2=15x-56\]
\[\Rightarrow x^2-15x+56=0\]
\[\Rightarrow x^2-7x-8x+56=0\]
\[\Rightarrow x(x-7)-8(x-7)=0\]
\[\Rightarrow (x-8)(x-7)=0\]
\[\Rightarrow x=7⠀or⠀x=8\]
Solution 2: (Lengthy solution )
Here we will simplify the equation directly from the given equation.
Given Equation:
\[x=15-\frac{56}{15-\frac{56}{15-\frac{56}{x}}}\]
\[\Rightarrow x=15-\frac{56}{15-\frac{56x}{15x-56}}\]
\[\Rightarrow x=15-\frac{840x-3136}{225x-840-56x}\]
\[\Rightarrow x=15-\frac{840x-3136}{169x-840}\]
\[\Rightarrow x=\frac{2335x-12600-840x+3136}{169x-840}\]
\[\Rightarrow x=\frac{1695x-9464}{169x-840}\]
\[\Rightarrow 169x^2-840x=1695x-9464\]
\[\Rightarrow 169x^2-2335x+9464=0\]
\[\Rightarrow x^2-15x+56=0\]
Here we get same equation:
\[\Rightarrow x^2-7x-8x+56=0\]
\[\Rightarrow x(x-7)-8(x-7)=0\]
\[\Rightarrow (x-8)(x-7)=0\]
\[\Rightarrow x=7⠀or⠀x=8\]
I have shown 2nd method to show that we will get the same result by the shortcut method (Solution 1).
Question related to Algebra, Equation.
