Move the top two to the spare peg in 3 moves, move the big disc across in 1, then move the top two back on top of it in 3 more. Every disc you add doubles it and adds one, so for n discs it’s 2 to the power of n, minus 1.
Three discs, each a different size, sit on one of three pegs, the biggest at the bottom. You have to move the whole stack to another peg, a disc at a time, and you can never put a bigger disc on a smaller one. What’s the fewest moves it takes?
Give me a hintThe hint
Solve it for two discs first, then use that twice.