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### Re: BR Maths Corner-1

Posted: 24 Jul 2008 03:44
lakshmic wrote:
Three people, A,B,C, participate in this game. The host has 2 kind of caps, white and black. He chooses uniformly at random and places a cap on each of the peoples heads. The trick is, each person can see the other two people's cap but cannot see her own cap....

They are strictly prohibited from communicating with each other (ofcourse, before the game, they can agree on a scheme). The host gives them each a piece of paper. Each participant has to guess her own cap. They can write "White", "Black" or "Pass" on their paper. They cannot see each other's papers.

The group wins the game if at least one participant guesses correctly and no other participant guesses it wrong. For example if A guesses it correctly, and B,C write "Pass" they win. But if A guesses correctly and B guesses her own cap wrong, they lose.

What are the best odds of winning this game ?

Posted from above to keep the reference correctly ...
one can easily get 75% of saving by:
Simple strategy, if you see other two of the same color, guess yours opposite, else pass. One looses only when all three caps are the same color.

### Re: BR Maths Corner-1

Posted: 24 Jul 2008 03:50
Amber G. wrote:Any way for your 3 caps solution .. one can easily get 75% of saving by:
Simple strategy, if you see other two of the same color, guess yours opposite, else pass. One looses only when all three caps are the same color.

amber g, this method will be in deep sea if a crooked person(like me) is the host !!
I'm not sure, but the hint about pre scheming is suggestive.

there is a variation of this problem :
there are n prisoners, all condemned to death. the king of the land decides to give them a sporting chance. the jailor will put either a white or a black cap on each prisoner such that none can see his own cap. the prisoners will be made to stand in a line of decreasing order of height so that each can see the hats of the ones in front of him but not of the ones in the back. each prisoner will have to call out white or black and will be pardoned if the answer is correct. they can't communicate with each other after the caps have been put on.

they however have a night to ponder over the situation !
how many can be saved ? and how ?

### Re: BR Maths Corner-1

Posted: 24 Jul 2008 04:46
Rahul M wrote:there is a variation of this problem :
there are n prisoners, all condemned to death. the king of the land decides to give them a sporting chance. the jailor will put either a white or a black cap on each prisoner such that none can see his own cap. the prisoners will be made to stand in a line of decreasing order of height so that each can see the hats of the ones in front of him but not of the ones in the back. each prisoner will have to call out white or black and will be pardoned if the answer is correct. they can't communicate with each other after the caps have been put on.

they however have a night to ponder over the situation !
how many can be saved ? and how ?

After learning madrassa math, none survive.
Trivially half survive (alternate guys call out the hat color of the guy ahead of them)
After learning earth-e-shatter, n-1 survive, one has 50-50 chance.

The last guy totals up the number of white caps. If it is even, he shouts white. He may or may not survive. The guy ahead of him totals up the white hats, if it is odd, he has a white hat on, if it is even, he has a black hat on. He calls out his appropriate hat.

### Re: BR Maths Corner-1

Posted: 24 Jul 2008 04:53
spot on !
lakshmic, what is the solution to the 3 caps problem ? as amber g indicated or are there some hitherto unseen logic ?

### Re: BR Maths Corner-1

Posted: 24 Jul 2008 05:06
Rahul - just look at the tiny solution: Unless all the three have the same color (2/8 chances) (in that case all will get it wrong) in all other cases the strategy will work ...everyone will be safe .(6/8 chances) Just try it out. (Because there is only one person who sees other 2 caps of same color, and guess correctly .. others passes per pre-agreed strategy ). Try it out.

### Re: BR Maths Corner-1

Posted: 24 Jul 2008 05:18
Rahul M wrote:spot on !
lakshmic, what is the solution to the 3 caps problem ? as amber g indicated or are there some hitherto unseen logic ?

Amber-g's solution is correct.

### Re: BR Maths Corner-1

Posted: 24 Jul 2008 06:58
Rahul M wrote:spot on !
lakshmic, what is the solution to the 3 caps problem ? as amber g indicated or are there some hitherto unseen logic ?

I couldn't find any "high school" level solution either.

### Re: BR Maths Corner-1

Posted: 24 Jul 2008 07:11
amber g, I did understand your solution. I had in fact thought about this solution (before seeing yours) and rejected it as it did not give near 100% success rate (was hoping against hope for a near 100% ) !! looks like education has ruined my thinking !
regards.

### Re: BR Maths Corner-1

Posted: 24 Jul 2008 09:09
Ok Solution to Lakshmic... I guess this time i will make it to mullah Nayakuddin's hitlist.

There are only 8 ways/combinations in which 'W' or 'B' caps can be worn by 3 beebuls.
Now assuming topiwallah has n number of caps; of the three beebul the one who gets to wear a cap at last ,in theory can calculate his probability of getting a particular type of CAP W/B.

Say A and B recieve a Black then C can at least hope that probability of him getting a WHITE is high.

Now these three can decide among themselves that person who gets the cap at last will only write the colour on paper while other two will say 'BASS'.

assumptions : host has equal number of B and W caps .

### Re: BR Maths Corner-1

Posted: 24 Jul 2008 09:52
OK! Since I see No other solutions yet, Here is my solution of:
1. Prove that (1/2)(3/4)(5/6)………..(99/100)
is less than (1/10) and greater than (1/15)

First a Maha Hint:

One notes that 2/3>1/2 and 4/5 > 3/4 etc .. 100/101 > 99/100 in general n/(n+1)>(n-1)/n
which is obviously true.

Rest is easy (if you get the right idea

Let x = (1/2)(3/4)(5/6) ..... (99/100)
so x^2 = (1/2)(1/2) (3/4)(3/4) (5/6)(5/6) ........ (99/100)(99/100)
so x^2 < (1/2)(2/3) (3/4)(4/5) (5/6)(6/7) ...(97/98)(98/99) (99/100)(100/101)
= 1/101 <===== Things do cancel out nicely !
< 1/100
so x<1/10

Similarly you can also show..
x^2 = (1/2)(1/2) (3/4)(3/4) (5/6)(5/6) .................(99/100)(99/100)
> (1/2)(1/2) (2/3)(3/4) ..................................(98/99) (99/100)
> (1/200)
> (1/225)
so x > (1/15)

QED

The Fun part which have made many youngster to get interested in Math, are problems like these. The general solution is of the form : value is between (sqrt(2/pi*100) ) and (sqrt(2/pi*101)) .. You get "pi" out of nowhere (and it is not that hard, for bright young people to get introduced to fun stuff.

### Re: BR Maths Corner-1

Posted: 24 Jul 2008 14:00
nice and simple !! just the way I like it !!
thanks amber g !

### Re: BR Maths Corner-1

Posted: 24 Jul 2008 22:50
Amber G. wrote:BTW did you check out the link I gave (best card game)..

Yes, I read the abstract, and it has been bothering me ever since.
There is ln2(24) bits of info in the shuffling order. An extra bit has
to come from which of the five cards the assistant chose to pass on.
I have thought of some schemes that will work for most configurations
of the five cards, but haven't yet come up with something that will
work for any configuration.

Nice solution for the 100c50 problem. The idea of squaring never enetered my mind.

### Re: BR Maths Corner-1

Posted: 25 Jul 2008 00:24
Lets assume that none of the topi walas can see their own caps for argument sake. Now after the randomly putting topi on these chaps, let say, they shuffle their positions with closed eyes, and then open their eyes to check on who is wearing what? How does the probability of prediction change? One of the possibility could include all Ws in sequence, and all Bs in sequence.!?

### Re: BR Maths Corner-1

Posted: 25 Jul 2008 01:57
Nandu wrote:Yes, I read the abstract, and it has been bothering me ever since. There is ln2(24) bits of info in the shuffling order. An extra bit has to come from which of the five cards the assistant chose to pass on.

Nandu-saar,
You have hit the nail on the head. There are two versions of the problem, version (a) that I posted where M1 does not have a choice as to what card to hide and version (b) covered in the paper where M1 does have a choice as to what card to hide.

Now ideally, you need lg(52) bits to convey information about the hidden card. Since the hidden card is not one of the four cards that are passed, lg(48) bits are enough. Can this be cut down still further ? A big hint is that M1 does have a choice as to what card to hide. Another big hint is that there are 4 suites and the game is played by drawing 5 cards and there is this old saying about pigeons and holes...

### Re: BR Maths Corner-1

Posted: 25 Jul 2008 06:52
Should be a simple problem for the gurus

3 office co-workers walk into a Patel Motel in Kentucky. The desk clerk tells them that their 1 night stay will cost \$30. The co-workers each buck up \$10 to pay and then go upto their room. The desk clerk then remembers that the cost should have been \$25. He takes five \$1 notes and runs upstairs to return the money. However, he is greedy and returns \$1 to each guest and pockets \$2 for himself.

Here's the puzzle - the guests gave \$10 each. The desk clerk returned \$1 to each one (so, in effect - each gave \$9). The desk clerk pocketed \$2.
[So - \$9 times 3 = \$27. \$27+\$2 = \$29. What happened to the remaining \$1?]

### Re: BR Maths Corner-1

Posted: 25 Jul 2008 11:19
Babui wrote:Should be a simple problem for the gurus

3 office co-workers walk into a Patel Motel in Kentucky. The desk clerk tells them that their 1 night stay will cost \$30. The co-workers each buck up \$10 to pay and then go upto their room. The desk clerk then remembers that the cost should have been \$25. He takes five \$1 notes and runs upstairs to return the money. However, he is greedy and returns \$1 to each guest and pockets \$2 for himself.

Here's the puzzle - the guests gave \$10 each. The desk clerk returned \$1 to each one (so, in effect - each gave \$9). The desk clerk pocketed \$2.
[So - \$9 times 3 = \$27. \$27+\$2 = \$29. What happened to the remaining \$1?]

See. The more things change,the more they remain the same. The Shaeedized dhaga keeps coming back all the same. This is a classic example of Madrassa math .

Now, the customers should pay \$ 8 1/3 each, since they should be paying only \$25 collectively. So what the Paki clerk should have returned to them is \$ 1 2/3 each. Now he returned \$1 each and pocketed \$2/3 from each of them. So the actual math is like this guests ( \$8 1/3 + \$1 returned) *3 = 28 and Madrassa Paki 2/3 *3 = 2.. So total \$30 . So what is the praablem saar ?.. The praablem is that the Paki with Pee Yecch Dee in Madrassa Math (Madmath) could not figure out that when he returned \$1, the boarders were notionally paying \$ 9 1/3 and not \$ 9.. Now if you put \$9 1/3 madmath, it turns from haraam to halal.

A much simpler halal way of looking at it , if looking it at from cash accounting basis and not "accrual" (a cruel?) accounting basis. You gave \$10 , and got back \$1 . So your cash outflow is \$9 only. Now look at it all on cash flow basis and not on "a cruel" basis. So. The room should have cost \$25. The boarders cumulatively paid \$9*3 = 27 . Paki stole \$2.

So halal math is \$25 (cost of room) = \$27 (actual cash paid by boarders) - \$2 stolen by Paki !.

The problem comes when you mix "a cruel" accounting and cash accounting onlee saar!.

### Re: BR Maths Corner-1

Posted: 25 Jul 2008 12:44
\$25 + \$2 tip = \$27. Rest all maya onlee.

### Re: BR Maths Corner-1

Posted: 25 Jul 2008 12:46
You guys should stick to bartering, baki style.

Makes things more interesting.

[Nayakuddin, So you did come and post there! But next time do psot some math related stuff. Thanks, ramana]

### Re: BR Maths Corner-1

Posted: 25 Jul 2008 13:13
A nice poem

((12 + 144 + 20 + (3 * 4^(1/2))) / 7) + (5 * 11) = 9^2 + 0
.
.
.
Or for those who have trouble with the poem:

A Dozen, a Gross and a Score,
plus three times the square root of four,
divided by seven,
plus five times eleven,
equals nine squared and not a bit more.

### Re: BR Maths Corner-1

Posted: 25 Jul 2008 14:19
Rupesh, you should have taken the chance to infiltrate the shero-shayari thread.

### Re: BR Maths Corner-1

Posted: 25 Jul 2008 19:53
Guys/Gals :

There is a nukkad thread, (also a nuclear thread ) where some of these, goat/wives/patel_motels etc might be even more fun.

Can I ask authors to clean it up a liitle bit (If you can't delete the old posts, let admin help)
(Let us not fill it with jokes/puzzles which every one may heard about a dozen times )

Nandu/RahulM you are welcome.

Nandu - The card ("best card trick") problem - first I heard was from Martin Gardner's piece, (Scientific American) about 30 years ago, 4!=24 is of course, pretty trivial, to get to 48 cards, his magician was "cheating" a little bit (He had 2 assistants - and he will call appropriate assistant etc)

Lakshmic's version, of course, you get 4X information, and thus can reach a card deck not only 52 cards but 100 (96+4) cards.

Choosing your own card out of 5 cards (as the trick mentioned in the reference given before) actually gives much more than 2x information ( it is 5x ) That is you can guess even if the card deck had 124 cards.!

BTW, pigeon hole principle, here is quite irrelevant, it is just one way (and very easy strategy) when you have ordinary deck, but it may just as well NO suits/ just numbers, . One of the easiest strategy is to add all the 5 cards (assume each card is numbered from 0 to 52 (or 124 for that matter) , take modulo 5. If say it is 2, turn back 2nd card (when 5 cards are arranged in numerical order).. etc... details are in the paper.

Sorry for long post

BTW" We still have the third problem (factors of 2^58+1 ... 5 is one factor - already known.

### Re: BR Maths Corner-1

Posted: 26 Jul 2008 02:43
FWIW - Here is a problem from old Nukkad before it was kicked away... Vinaji and GDji taking about Pakis jumping from K2 and hoping to reach west (as the world turns or what they think how the world turns) -

So no tricks - here is an honest physics problem - If an object fall from 500 meters ( A very tall building like CN tower) where will it lend. Will it lend directly below (horizontally - that is joining a line from the initial point and center of the earth )

(Not trying to be tricky in asking) ...assume normal meanings etc.. assume NO air resistance, Initial velocity = 0 (wrt to tower, IOW it was dropped) and the experiment is taking place at the equator. (Normal assumptions - earth is a sphere, has gravity, makes night and day goes around sun etc.. if deviation is less than, say 1 cm, it can be ignored.. that is no big mountains/jinns etc close by to cause subtle effects)

No discussion etc needed , (unless some elegant way to solve your problem) give your best answer.. a) directly below or b) x meters in (which ) direction.

### Re: BR Maths Corner-1

Posted: 26 Jul 2008 05:40
Amber-G-saar,

approx sqrt(1000/g)*500*2*pi/(60*60*24) meters or approx 0.3 meters away in the east ? or am I missing something

### Re: BR Maths Corner-1

Posted: 26 Jul 2008 08:30
Math Gurus.... wondering if anyone can give a crack at solving the listed Math problem. I already spend better half of last two days to solve this without any luck. Hoping better brains out there than mine might solve this puzzle and arrive at the correct answer.

Hint - This problem is based on multinomial coefficients.

Que: Show that the number of four-digit numbers that can be formed using the digits 1,1,2,3,3,4 ?

Ans of this question is 102

### Re: BR Maths Corner-1

Posted: 26 Jul 2008 09:18
Cases are small enough to list all; Just don't worry of order of the numbers (yet) , just what 4 you are choosing.
A. 1 (case 2 identical pairs (1133)
B. 2*3 = 6 cases of ( I identical pair, + 2 distinct digits) = (1123)(1124)(1134) (3 similar with double 3)
(another way to think of it is there are two ways to select identical pair (1 or 3) then 3 ways to select the rest 2 (out of possible 3)
C 1 case of all distinct.

Case C Keeping in mind, permutation of 4 objects, when all 4 are distinct = 4! = 24
Case B also, permutation of 4 objects, with 2 identical = 4!/2! = 12
Case A permutation of 4 objects with 2 pairs of 2 identical pairs = 4!/(2!2!) = 6

So total = 24+6*12+6=102
(Takes more time to type than to do it)

### Re: BR Maths Corner-1

Posted: 26 Jul 2008 11:08
Amber thanks a ton. It will be really helpful if you can provide a little more insight into case B. In my opinion, there are more than 6 cases for (1 identical pair + 2 distinct digits) because 1132, 1142, 1143, 3342, 3341, 3312 are the digits which are also included. So I believe there should be more than 6 choices. Prefer to be corrected if wrong.

### Re: BR Maths Corner-1

Posted: 26 Jul 2008 14:44
lakshmic wrote:Amber-G-saar,

I think it is actually be Amber-G maam!

### Re: BR Maths Corner-1

Posted: 26 Jul 2008 17:19
Oh my bad....Vina thanks for pointing out...

### Re: BR Maths Corner-1

Posted: 26 Jul 2008 20:15
So I believe there should be more than 6 choices. Prefer to be corrected if wrong.

Which one? (Cases are small enough, to list all) easiest way to check. Notice that when we are counting the case of 1 pair, we are not interested in arrangement (permutation) yet.
Case is where you have one "pair" only. That is of the form 'xxab' . (At present, you don't distinguish between say 1134 and 1143 or 1413) There are two choices for x (1 or 3), once x is selected, only 3 ways ( selecting 2 out of rest of 3) to select 'a b' . (And of course, there are 4!/2! ways to arrange 'xxab').

### Re: BR Maths Corner-1

Posted: 26 Jul 2008 23:37
Amber G. wrote:So no tricks - here is an honest physics problem - If an object fall from 500 meters ( A very tall building like CN tower) where will it lend. Will it lend directly below (horizontally - that is joining a line from the initial point and center of the earth )

(Not trying to be tricky in asking) ...assume normal meanings etc.. assume NO air resistance, Initial velocity = 0 (wrt to tower, IOW it was dropped) and the experiment is taking place at the equator. (Normal assumptions - earth is a sphere, has gravity, makes night and day goes around sun etc.. if deviation is less than, say 1 cm, it can be ignored.. that is no big mountains/jinns etc close by to cause subtle effects)

No discussion etc needed , (unless some elegant way to solve your problem) give your best answer.. a) directly below or b) x meters in (which ) direction.

I assume that the angular velocity of the paltform from which the ball is dropped is the same as the angular velocity of the earth. Then however the horizontal linear velocity will be be different from that of the linear velocity of the initial ground point. So the ball will not land directly below.

### Re: BR Maths Corner-1

Posted: 26 Jul 2008 23:55
Difference in horizontal velocity dv ~= wdr ~= (2pi/(24*3600))*500 = 5pi/432
(w = angular velocity of earth, r = radius of earth, dr = 500)
Time taken for 500 meter descent ~= 10 secs.
Therefore difference in position ~= 10*5pi/432 ~= 0.36 meters.

The body will land 36 centimeters away from the initial ground point.

I note that lakshmic gets same answer.

### Re: BR Maths Corner-1

Posted: 27 Jul 2008 00:12
Aha, I think I have it. The previous answer was incorrect. The thing is that the gravitational field is always pointing towards the centre of the earth. That is the field lines are not parallel. Hence the ball must fall to the center.

So correct answer is: No ball will fall exactly to original ground point.

### Re: BR Maths Corner-1

Posted: 27 Jul 2008 00:51
One can either (a) solve the eqn of motion in radial field or (b) argue as follows: Assume that centre of earth is an inertial frame (Newton's laws hold). Move to the rotating frame of the earth. We can pretend that earth is an inertial frame by adding centrifugal force - which is radial so it does not affect the direction of downward force. Then the initial velocity of the ball is zero in this frame and it can can only move along the lines of force. The result then follows. Thus for example a ball thrown upwards will land exactly at the point from which it was thrown.

### Re: BR Maths Corner-1

Posted: 27 Jul 2008 05:44
Hi Amber-G saar, Thanks for the help last time. I am also banging my head on another problem.

Case:
There are five coins and the number of ways we can flip them to get at least 3 heads is 16.

The way i m trying to solve this is by the following approach:
Set A is Number of combinations of atleast 3 heads ,2 tails= 10 combination
Set B is Number of combinations of atleast 4 heads,1 tail = 5 combination
Set C is Number of combinations of atleast 5 heads = 1 combination

if i wish to represent this problem in the form of venn diagram ,using three cases caseA-3 heads ,2 tails

Simple enough, but I am having trouble in getting answer "16" using Inclusion/Exclusion formula listed below.

can anybody help, by venn diagrams using three circles how can we represent this problem.
In inclusion and exclusion principle v using this formula.
N(0)=N-N(A)-N(B)-N(C)+N(AB)+N(AC)+N(BC)-N(ABC)
N(O) = number of objects with neither property
N=total number of objects .
This problem is similar to using venn diagrams ,here v have to use three circles to represent upper listed three cases.

Any help really appreciated.

### Re: BR Maths Corner-1

Posted: 27 Jul 2008 08:25
Hi mdhoat: Before admins through us both out , for solving homework problems ---

No Venn diagrams are needed. The probability that you will get 3 (or more) heads out of 5 is same as, not getting it, (in other words - same as getting 3 (or more) tails)..so the prob is exactly half. (Think of best out of 5 sets in a game of tennis). Half of 2^5 = 32 is indeed 16.

(For more complicated cases, look up binomial distribution in wiki or any text book)

You may be much better off, asking/discussing these in form like mathcounts.org or ask your own teacher/fellow classmate.. they can go in little bit more details than anyone here.
Hope this helps

### Re: BR Maths Corner-1

Posted: 27 Jul 2008 08:39
you are right, these questions were from a home test I was banging my head to solve for my cousin who's doing MS in Applied Maths. But still thanks for all your help...at the end of day Bharat-Rakshak rocks, a one stop solution for all your information needs in any domain...

### Re: BR Maths Corner-1

Posted: 27 Jul 2008 17:10
AoA math mujahids,
mdhoat,
the case of atleast 3 heads out of 5 tosses can easily be solved by using the bernoulli trails method...

### Re: BR Maths Corner-1

Posted: 27 Jul 2008 21:07
There are 9 pearls. 8 weighing equal and 1 pearl weigh slightly more than others.
We have a balance scale but can use it only twice.Find out the odd pearl.
(solvable in a min)

### Re: BR Maths Corner-1

Posted: 27 Jul 2008 21:19
SK Mody wrote:One can either (a) solve the eqn of motion in radial field or (b) argue as follows: Assume that centre of earth is an inertial frame (Newton's laws hold). Move to the rotating frame of the earth. We can pretend that earth is an inertial frame by adding centrifugal force - which is radial so it does not affect the direction of downward force. Then the initial velocity of the ball is zero in this frame and it can can only move along the lines of force. The result then follows. Thus for example a ball thrown upwards will land exactly at the point from which it was thrown.

No, your first approach was correct. You can't just change to a rotating frame of reference except locally.

In 10 seconds, the angular displacement of the points under consideration will be too small to be significant to the solution of the problem.

IMHO, of course.

### Re: BR Maths Corner-1

Posted: 27 Jul 2008 21:41
I'd like to share an amusing mathematical marshmallow*:
We are all familiar with the geometric series x + x^2 + x^3 + .... (aren't we )
Suppose 0 < x < 1. Then the sum of the series to infinity converges. One way of finding the limit of the sum is as follows:
Let S = x + x^2 + x^3 + ... x^n
Then xS = x^2 + x^3 + x^4 + ... + x^(n+1)

Subtract xS from S. We get:
S(1-x) = x(1-x^n)
=> S = x(1-x^n)/(1-x)

Letting n --> inf we get:
S = x/(1-x)

Thus for example, using the above formula, 1/2 + (1/2)^2 + (1/2)^3 + .... = 1

However there is another "physical" way to look at the sum above. Suppose I need to travel from A to B - a distance of 1 km. First I need to travel half the distance (1/2 km). Then I need to travel half the remaining distance (1/4 km). Then half the remaining distance (1/8 km) and so on. It is obvious that I will eventually get to B. The sum of powers of 1/2 starting from (1/2)^1 is therefore equal to 1.

So here is the question:
Find a "physical proof" of the formula:
1/3 + (1/3)^2 + (1/3)^3 + ... = 1/2

Similarly find "physical proof" of say the formula:
2/5 + (2/5)^2 + (2/5)^3 + ... = 2/3

In general find a physical proof of the formula for p/q + (p/q)^2 + (p/q)^3 + ... (where p and q are positive integers with p < q).