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  n disks and 3 towers, the number of steps required to transfer the disks from 1  tower to another is 2^n -1.

Proof of our lemma:

Image
Let Q_n denote the minimum number of moves required for transferring n discs from one peg to another.Suppose that there are k+1 discs on peg 1 . We will first transfer the top k discs to peg 3 . This can be done in Q_k moves.
Now we transfer the k+1th disc to peg 2 . One move will be required for this.
Finally we will transfer k discs which are on peg 3 to peg 2 . This will require another m_k moves. Thus the transfer will be done in Q_k+1+Q_k moves which will be equal to Q_{k+1} movesTherefore we have formed a recurrence relation that
Q_{k+1} = 2Q_k + 1

Now we will obtain a general formula for Q_{k+1}

Q_n=2Q_{n-1}+1
Q_{n-1}=2Q_{n-2}+1
\vdots
Q_2=2Q_1+1

We will multiply both sides of the above relations by 1 , 2 , 2^2 , \cdots 2^{n-2}
we get

Q_n=2Q_{n-1}+1
2Q_{n-1}=4Q_{n-2}+2
\vdots
2^{n-2}Q_2=2^{n-1}Q_1+2^{n-2}
Solving this gives us

Q_n=2^{n-1}m_1+1+2+2^2 \cdots 2^{n-2}
so
Q_n=2^{n-1}m_1+2^{n-1}-1

Since Q_1=1 ( one move will be required to move )
so
Q_n=2^{n-1}+2^{n-1}-1

THIS FINALLY GIVES

Q_n = 2^n - 1

\blacksquare

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

NMTC 2012 ( CLASS 9th and 10th) Question paper


The AMTI is a pioneer organisation in promoting and conducting Maths Talent Tests in India. Last year (43rd TC Data) (in the 43rd National level tests) 54058 students from 332 institutions spread all over India, participated at the screening level; 10% of them insitutionwise were selected for the final test. For the benefit of final level contestants and the chosen few for INMO, special orientation camps were conducted. Merit certificates and prizes were awarded to the deserving students.

DOWNLOAD THE 2012 PAPER :
NMTC Question Paper

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

CGMO – 2012 ( China Girls Math Olympiad 2012 ) Problem 7


Let  \{a_n\} be a sequence of nondecreasing positive integers such that  \frac{r}{a_{r}}= k+1 for some positive integers k and r. Prove that there exists a positive integer s such that  \frac{s}{a_s} = k

CGMO 2012 ( China Girls Math Olympiad 2012) Problem 6


There are n cities, 2 airline companies in a country. Between any two cities, there is exactly one 2-way flight connecting them which is operated by one of the two companies. A female mathematician plans a travel route, so that it starts and ends at the same city, passes through at least two other cities, and each city in the route is visited once. She finds out that wherever she starts and whatever route she chooses, she must take flights of both companies. Find the maximum value of  n

CGMO 2012 ( China Girls Math Olympiad 2012) Problem 5


As shown in the figure below, the in-circle of ABC is tangent to sides AB and AC at D and E respectively, and O is the circumcenter of BCI. Prove that \angle ODB = \angle OEC.
import graph; size(5.55cm); pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-5.76,xma...

CGMO-2012 (China Girls Math Olympiad 2012) Problem 4


There is a stone at each vertex of a given regular 13-gon, and the color of each stone is black or white. Prove that we may exchange the position of two stones such that the coloring of these stones are symmetric with respect to some symmetric axis of the 13-gon

CGMO-2012 (China Girls Math Olympiad 2012) Problem-3


Find all pairs (a,b) of integers satisfying: there exists an integer  d \ge 2 such that a^n + b^n+1  is divisible by d for all positive integers n

CGMO 2012 (China Girls Math Olympiad 2012)- Problem 2


Circles Q_1 and Q_2 are tangent to each other externally at T. Points A and E are on Q_1, lines AB and DE are tangent to Q_2 at B and D, respectively, lines AE and BD meet at point P
Prove that

(1) \frac{AB}{AT}=\frac{ED}{ET};
(2) \angle ATP + \angle ETP = 180^{\circ}.
import graph; size(5.97cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real ...

CGMO 2012 (China Girls Math Olympiad 2012) Problem 1


Let  a_{1}, a_{2},\ldots, a_{n} be non-negative real numbers. Prove that                                                                          \LARGE \frac{1}{1+a_{1}}+\frac{ a_{1}}{(1+a_{1})(1+a_{2})}+\frac{ a_{1}a_{2}}{(1+a_{1})(1+a_{2})(1+a_{3})}\cdots+\frac{ a_{1}a_{2}\cdots a_{n-1}}{(1+a_{1})(1+a_{2})\cdots (1+a_{n})}\le 1.