mscroggs.co.uk
mscroggs.co.uk

subscribe

# Puzzles

## Two

Find $$a$$ such that $$a+(a+A)^{-1}=2$$, where $$A=(a+A)^{-1}$$.
ie. $$a + \frac{1}{a + \frac{1}{a + \frac{1}{a + \frac{1}{...}}}} = 2$$.
Find $$b$$ such that $$b+(b+B)^{\frac{1}{2}}=2$$, where $$B=(b+B)^{\frac{1}{2}}$$.
ie. $$b + \sqrt{b + \sqrt{b + \sqrt{b + \sqrt{...}}}} = 2$$.
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 $$D=(d+D)^{k}$$.
Tags: numbers
If you enjoyed this puzzle, check out Sunday Afternoon Maths VIII,
puzzles about numbers, or a random puzzle.

## Archive

Show me a random puzzle
▼ show ▼