# Blog Archives

## Towers of Hanoi problem

Introduction :

The Tower of Hanoi (also called the Tower of Brahma or Lucas’ Tower, and sometimes pluralised) is a mathematical game or puzzle. It consists of three rods, and a number of disks of different sizes which can slide onto any rod. The puzzle starts with the disks in a neat stack in ascending order of size on one rod, the smallest at the top, thus making a conical shape.

The objective of the puzzle is to move the entire stack to another rod, obeying the following rules:

• Only one disk may be moved at a time.
• Each move consists of taking the upper disk from one of the rods and sliding it onto another rod, on top of the other disks that may already be present on that rod.
• No disk may be placed on top of a smaller disk.

With three disks, the puzzle can be solved in seven moves.

Our Lemma:

for   disks and  towers, the number of steps required to transfer the disks from   tower to another is .

Proof of our lemma:

Let  denote the minimum number of moves required for transferring  discs from one peg to another.Suppose that there are  discs on peg  . We will first transfer the top  discs to peg  . This can be done in  moves.
Now we transfer the th disc to peg 2 . One move will be required for this.
Finally we will transfer  discs which are on peg  to peg  . This will require another  moves. Thus the transfer will be done in  moves which will be equal to  movesTherefore we have formed a recurrence relation that

Now we will obtain a general formula for

We will multiply both sides of the above relations by
we get

Solving this gives us

so

Since  ( one move will be required to move )
so

THIS FINALLY GIVES

Sources :
1. wikepedia ( for the introduction) : http://en.wikipedia.org/wiki/Tower_of_Hanoi

## CGMO-2012 (China Girls Math Olympiad 2012) Problem 8

Find the number of integers $k$ in the set $\{0, 1, 2,\cdots, 2012\}$   such that  $\binom{2012}{k}$ is a multiple of $2012$

## IMO 1983 – Problem 3

Let $N$ and $k$ be positive integers and  let $S$ be a set of  $n$ points in the plane such that

$(i)$ no three points of $S$ are collinear, and
$(ii)$  for any point $P$ of  $S$, there are at least $k$ points of $S$ equidistant from $P$

Prove that  $k$  $<$  $\frac{1}{2}$  $+$  $\sqrt{2n}$

Try the question …
Solution will be updated soon