Puzzles
Two
Find \(a\) such that \(a+(a+A)^{-1}=2\), where \(A=(a+A)^{-1}\).
ie. \(a + \frac{1}{a + \frac{1}{a + \frac{1}{a + \frac{1}{...}}}} = 2\).
ie. \(a + \frac{1}{a + \frac{1}{a + \frac{1}{a + \frac{1}{...}}}} = 2\).
Find \(b\) such that \(b+(b+B)^{\frac{1}{2}}=2\), where \(B=(b+B)^{\frac{1}{2}}\).
ie. \(b + \sqrt{b + \sqrt{b + \sqrt{b + \sqrt{...}}}} = 2\).
ie. \(b + \sqrt{b + \sqrt{b + \sqrt{b + \sqrt{...}}}} = 2\).
Find \(c\) such that \(c+(c+C)^{2}=2\), where \(C=(c+C)^{2}\).
In terms of \(k\), find \(d\) such that \(d+(d+D)^{k}=2\), where \(D=(d+D)^{k}\).
If you enjoyed this puzzle, check out Sunday Afternoon Maths VIII,
puzzles about numbers, or a random puzzle.
puzzles about numbers, or a random puzzle.