PUZZLES ......courtesy : babai

[madhu kiran]

[gandham mahendranadh]

8 rotations.................

[madhu kiran]

nah..

On 28 March 2011 13:50, gandham mahendranadh <mahendr...@gmail.com> wrote:

8 rotations.................

[siddartha adusumalli]

4

[madhu kiran]

nah..

On 28 March 2011 14:00, siddartha adusumalli <siddugadu....@gmail.com> wrote:

4

[siddartha adusumalli]

FUNDA

       ~~~~~ The sentence "The quick brown fox jumps over the lazy dog." uses every letter in the english language. 

                ~~~~~   Typewriter, is the longest word that can be made using the letters on only one row of the keyboard. 

[madhu kiran]

Active members of our group. :)

[Nikhil]

Anduke e sentence fonts ki example vadataru 

--
♪√ Ê ∩ K▲τ▲ ♥ ∩ ì K ╫ í └ ♥ M µ þ þ å ñ ε ñ ï ♪

[Nikhil]

On Mon, Mar 28, 2011 at 2:07 PM, siddartha adusumalli <siddugadu....@gmail.com> wrote:

-- Anduke e sentence fonts ki example vadataru 

[Nikhil]

 "The quick brown fox jumps over the lazy dog."

 

@siddu 

[aswan kumar]

i'm all time top....\m/...

but really sad to see that most of the ppl are not active

--
Paleti Aswan Kumar,
Undergraduate student,
Civil Engineering Department,
IIT Madras,
Chennai

[venkata harish babu manne]

@ Madhu : whats the answer? I think its 8 rotations

[madhu kiran]

It is not 8 .
Do you really want me to tell the answer?

[venkata harish babu manne]

what i thought was -
one rotation around B will rotate A around itself 2times.      -        (2*pi*10 = 2(2*pi*5))
so 4 rotations => 8 rotations

[gandham mahendranadh]

Same solution........

[raghavendra chary]

@madhu

ans 12...

[maripi chaitanya]

chary got it

On Mon, Mar 28, 2011 at 7:53 PM, raghavendra chary <char...@gmail.com> wrote:

@madhu

ans 12...

--
MARIPI
guy wit guts

[siddartha adusumalli]

how ?

[raghavendra chary]

imagine the contacting point on 'A'.  for one rotation of 'A'(as you think) imagined point again touches 'B' (exactly oposite side of 'B'). Erlier that point was on right hand side of 'A'. But now it is on left hand side of 'A' which means there is additional rotation of 180 deg. So for every full rotation around 'B'  'A' actually makes 3 roations around itself.
               

             for 4 rotations around 'B',         4*3 = 12

[siddartha adusumalli]

తెలుగు లో చెప్పు మామా . అర్ధం కావట్లేదు .  :పి

[Sailesh]

'a' be the centre of A and 'b' be the centre of B. Initially the line ab is horizontal. Let the point of contact be c. acb is the order of points. Imagine the position of A after it makes half revolution around B. Now, ba is again horizontal, but c is not the point of contact. Initially acb was the order of points, but now bac is the order, ie, c has undergone some n+0.5 rotations around b(n is a non -ve integer). So half revolution implies n+0.5 rotations. One revolution implies 2n+1 rotations. We were ignoring this 1 rotation. Hope it is clear.

But, how do you prove that n = 1? Infact I think the order of the points a, b and c after half revolution depends on the radii of the circles. Comments...

[Sailesh]

On Mon, Mar 28, 2011 at 9:58 PM, Sailesh <ramayana...@gmail.com> wrote:

'a' be the centre of A and 'b' be the centre of B. Initially the line ab is horizontal. Let the point of contact be c. acb is the order of points. Imagine the position of A after it makes half revolution around B. Now, ba is again horizontal, but c is not the point of contact. Initially acb was the order of points, but now bac is the order, ie, c has undergone some n+0.5 rotations around b(n is a non -ve integer). So half revolution implies n+0.5 rotations. One revolution implies 2n+1 rotations. We were ignoring this 1 rotation. Hope it is clear.

But, how do you prove that n = 1? Infact I think the order of the points a, b and c after half revolution depends on the radii of the circles. Comments...

In the above explanation "c has undergone some n+0.5 rotations around a (n is a non -ve integer)". The n+0.5 number of rotations are around a and NOT around b.

--
---
Sailesh
3rd year Undergraduate
Computer Science & Engg.
IIT Kanpur.

[raghavendra chary]

@ sailesh :

       For half rotation around 'B'..
 ------> A travels a distance of (2*pi*5). Because of this we get 1 rotation of A.

-------> As A is moving around a circular surface we are getting that 0.5 rotation..

so total 1.5

[madhu kiran]

@chary...... u made it mama..
Explain the answer to the folks.

On 28 March 2011 20:41, siddartha adusumalli <siddugadu....@gmail.com> wrote:

how ?

[madhu kiran]

My mistake in the last post....I have seen that has already explained it.

[Sailesh]

In general if radius of B is a multiple of radius of A, then we should calculate the angular displacement of "c" w.r.t "a" after A moves a distance equal to the perimeter of A. Let it be x. Then the number of rotations made by A in one revolution is radius of B/radius of A(1+x/360), if x is in degrees.

Seems to be correct, but not sure.

[siddartha adusumalli]

@Madhu 

Ee Thread ki Subject  PUZZLES....... Courtesy : Babai ani pettav 

which babai 

[madhu kiran]

@Sailesh...  Mana siddu babai ra..
Inkentamandi babai lu unnaru? :)

[madhu kiran]

Here is another good ques.... may be a little easier.

[Sailesh]

Sorry guys, another mistake in the explanation. "Now, ba is again horizontal, but c is not the point of contact." This is wrong.
     'c' is the point of contact again, and now bca is the order of points from left to right. Initiially the order was acb. c was to the right of a, but now c is to the left of a, which means half a rotation.

      In general I found the following. R - radius of B, r - radius of A. Then, the number of rotations A will have made after one revolution around B is (R+r)/r. This is valid for any combination of R and r, ie, even if R<r and even if one is not a multiple of the other.

[Sailesh]

A - 1 chip, B - 2 chips, C - 3 chips ==> {1,2,3}

IE - 1 chip from B ==> {1,1,3}
CE - has 4 choices. one chip from C or all the chips from any of the piles. ==> {1,1,2} or {1,1,0} or {1,0,3} or {0,1,3}
IE - In the case of {1,1,2}, pick one chip from C ==> {1,1,1}.After this the steps are obvious.
      In any other case, if C is non-empty pick all the chips in C. If C is empty, pick any of the non-empty piles. Both the cases result in exactly one non-empty pile with one chip. IE is done.

[madhu kiran]

Hmm that's done.

[madhu kiran]

And for the  previous question, explaining in words was very uncomfortable
I think it might helps..

[madhu kiran]

Here is an another way of seeing it.

[madhu kiran]

@Aswan and siddu
I am about to beat you guys.... only a few left ... he he  he he

[Sailesh]

@Madhu: why does it not work? I mean it was not a guess. I got it mathematically. May be I was wrong. But, can you explain?

[maripi chaitanya]

@sailesh .....madhu already explained... you see you got mathematically it is making 2 revolutions per round if it is a rotating over perfect straight line ri8. see it is over a circular ring so revolving around circular ring counts one more, which is to be counted.

On Tue, Mar 29, 2011 at 3:12 PM, Sailesh <ramayana...@gmail.com> wrote:

@Madhu: why does it not work? I mean it was not a guess. I got it mathematically. May be I was wrong. But, can you explain?

--
MARIPI CHAITANYA
B.Tech 3rd Year
Electrical Engineering
IIT Roorke.

"Life is an adventure--- Nature is a gift----I am one of the wonderful inventions of GOD "

[Sailesh]

@chaitu: I think you are talking about the case R going to infinity. In that case B will be a plane and A(which is a finite circle) cannot revolve around B which is reflected in (R+r)/r going to infinity. Am I right? See the attached derivation. I hope it is correct.

[riyaz ahmed]

aaaahhhh.....what a handwriting..!!!
opened the pic and did not even see the solution.
reminded me of all your notes i referred for two years.

[madhu kiran]

@sailesh
Sorry mama. (R+r)/r works.
I was caring only about the logic. While checking your formula I have put some random numbers and I didn't get the correct answer.
May be my calculation was ( in hurry!)

That logic can be explained by your formula also.

(R+r)/r  =  (R/r) + 1
For every one revolution around the big circle the small circle has to make one extra rotation other than rotating on circumference.

Anyways ...great math there. :)

[Sailesh]

@riyaz: thanks :)

[Sailesh]

Now I think madhu is the top poster of the month.

[Sailesh]

In some other thread, Harish posted a question about some smart professor, but did not give the solution. Harish????

[aswan kumar]

@riyaz:i agree that we took the notes...but when did we referred it??

On Tue, Mar 29, 2011 at 7:05 PM, Sailesh <ramayana...@gmail.com> wrote:

In some other thread, Harish posted a question about some smart professor, but did not give the solution. Harish????

--
Paleti Aswan Kumar,
Undergraduate student,
Civil Engineering Department,
IIT Madras,
Chennai

[madhu kiran]

@Aswan
Riyaz gadu chadivevadu lera!! nuvuu mareenu vadini insult chestunnav.

@Riyaz
Nuvvu vadi matalem pattinchukovaddu ra.

[aswan kumar]

lol..riyaz gada chadavatama....chary,dheeraj,riyaz and me always used to chitchat abt arbit stuff during entire 2nd year...i do remember those good old days...kaada riyaz...???

[raghavendra chary]

@aswan


nannu madyalogi llagi mari edvanu chestunnav ga !!!!     :P

[aswan kumar]

nijalu chepthe edavalu iporu ra chari....anyway ninnu edavani chesemundu nannu nenu cheskuntunnatlu nuvvu cheppina prakaram...

On Tue, Mar 29, 2011 at 8:48 PM, raghavendra chary <char...@gmail.com> wrote:

@aswan


nannu madyalogi llagi mari edvanu chestunnav ga !!!!     :P

[raghavendra chary]

nuvvo RGV ra... :)

[aswan kumar]

madhyalo...RGV enduku vachadura...

On Tue, Mar 29, 2011 at 8:54 PM, raghavendra chary <char...@gmail.com> wrote:

nuvvo RGV ra... :)

[raghavendra chary]

@aswan : first vaanni vaade dobbukoni, pakkodini dobbutaadu.. :)
                 
                 lyt le..  

[venkata harish babu manne]

Thanks for reminding sailesh..

Small Modification of question- The Students Stared at each other for a while and then one shouts the color of his hat..

The full answer is as follows:

Let's number the Students #1, #2, and #3 where #1 is the one who yells out.(The color of his hat)

#1 thinks and says, if I am wearing White then #2 would have thought that if s/he was wearing White the #3 would know that s/he had on a Black hat (since there are only 2 White hats) and would yell out. Since #3 didn't yell out then #2 should know that s/he is wearing Black. #3 should be thinking the same thing of #2.

Since neither #2 nor #3 yells out, them #1 knows they must all have a Black hat on.

[riyaz ahmed]

i think the real puzzle is to understand this solution.

anyways, someone please elaborate.

[venkata harish babu manne]

@ Riyaz:

Case 1: 2 white and 1 Black
Case 2: 1 White and 2 Black
Case 3 : 3 Black..

Case 1 and 2 are ruled out since no one shouted their colour..
So, Only possibility is Case 3..

[Divya Goel]

all of them were blindfolded ryt? so, even if 2 of them have white caps on, how would the 3rd one know that??? 

[venkata harish babu manne]

After removing those blind folds... And after spending some time staring at each other, smart one can tell the answer...

But the smartest one can tell the answer before removing the folds.. \m/

[riyaz ahmed]

hmmm....nice question.
moving on....

The below is a number puzzle. It should be read left to right, top to bottom.
1
1 1
2 1
1 2 1 1
1 1 1 2 2 1
? ? ? ? ? ?
? ? ? ? ? ? ? ?
Question 1: What is the next two rows of numbers?
Question 2: How was this reached?

[Nikhil]

The below is a number puzzle. It should be read left to right, top to bottom.
      1
     1 1
     2 1
   1 2 1 1
 1 1 1 2 2 1
 ? ? ? ? ? ?
? ? ? ? ? ? ? ?
Question 1: What is the next two rows of numbers?
Question 2: How was this reached?

   3 1 2 2 1 1 - Thrice 1, twice 2, once 1
1 3 1 1 2 2 2 1 - Once 3, once 1, twice 2, twice 1

1 - 1
1 1 - Once 1
2 1 - Twice 1
1 2 1 1 - Once 2, once 1
1 1 1 2 2 1 - Once 1, once 2, twice 1

It's like saying what we wrote in previous line.

[Nikhil]

[Nikhil]

[Nikhil]

[Nikhil]

[Nikhil]

[Nikhil]

[Nikhil]

[Nikhil]

[Nikhil]