Brain Teasers
I will most some mighty cool brain teasers here.๐
Replies

raj87verma88Consider an arrow in flight. At any given moment of time, a snapshot could be taken of this arrow. In this snapshot, the arrow would not be moving. Let us now take another snapshot, leaving a very small gap of time between them. Again, the arrow is stationary. We can keep taking snapshots for each moment of time, each of which shows the arrow to be stationary. Therefore the overall effect is that the arrow never moves, however it still hits the target!
Where lies the flaw in the logic? 
raj87verma88You can imagine an arrow in flight, toward a target. For the arrow to reach the target, the arrow must first travel half of the overall distance from the starting point to the target. Next, the arrow must travel half of the remaining distance.
For example, if the starting distance was 10m, the arrow first travels 5m, then 2.5m.
If you extend this concept further, you can imagine the resulting distances getting smaller and smaller. Will the arrow ever reach the target? 
raj87verma88Would you believe that this amazing sentence contains ninety two letters, one comma and a single question mark?

raj87verma88Assume that penguins live with a density of 1,000 penguins per square mile and can run at an average speed of 7 miles per hour on land and swim at 20 miles per hour. Also assume that a polar bear has a territory of 10 square miles, can run at 25 miles per hour and swim at 10 miles per hour, how many penguins will an average polar bear eat in any given month, remembering that a polar bear could, as a maximum, only eat one penguin per hour and 7% of the land is next to the sea.

Kaustubh Katdare
I don't see any flaw in the logic. Its pretty obvious. You're taking snapshots in which the distance between the target and the arrow is constantly decreasing. After some time, it will hit the target. ๐raj87verma88Consider an arrow in flight. At any given moment of time, a snapshot could be taken of this arrow. In this snapshot, the arrow would not be moving. Let us now take another snapshot, leaving a very small gap of time between them. Again, the arrow is stationary. We can keep taking snapshots for each moment of time, each of which shows the arrow to be stationary. Therefore the overall effect is that the arrow never moves, however it still hits the target!
Where lies the flaw in the logic? 
suyash
the arrow wont reach the target ever...raj87verma88You can imagine an arrow in flight, toward a target. For the arrow to reach the target, the arrow must first travel half of the overall distance from the starting point to the target. Next, the arrow must travel half of the remaining distance.
For example, if the starting distance was 10m, the arrow first travels 5m, then 2.5m.
If you extend this concept further, you can imagine the resulting distances getting smaller and smaller. Will the arrow ever reach the target?
since the addition of distance covered is a the sum of terms in geometric progression which is "tending to 10 mts" 
raj87verma88The answer lies in the theory of infinite series. I will give all the answers in one or two days. Till then let me see how many come up with the correct answers for all the problems.

Kaustubh Katdare
Why not? The arrow does travel 'half distances' always, as mentioned in the post.suyashthe arrow wont reach the target ever...
since the addition of distance covered is a the sum of terms in geometric progression which is "tending to 10 mts"
Its like while typing this post, I typed the half of the post, then then half of remaining, then half of remaining  like that. ๐
๐ 
gohmWell a photograph only shows objects as stationary. However the viewer would know the object is really moving. As long as the trajectory is correct and velocity, it will reach the target. Is this like if a tree falls in the woods and nobody is there to hear it, does it make a sound?
672 penguins at least. However we need to know which month to know number of days (672 is based on leap year month) and also how many polar bears as realistically you would not have only one polar bear (it would die out) and also polar bears sleep, eat other animals etc. which would reduce the number. The polar bear would hunt only on land and wait for the penguins which would have to return. OK so I know this isn't supposed to be a realistic problem... laugh. 
Ashraf HZ
Mathematical theories aside, here's another perspective. Even if the arrow ends up embedded in the target, they never really touch. The electrons at the surface of the arrowhead are repelling the ones in the target. However small the distance between them are, they are still not touching (and thus never "reached").suyashthe arrow wont reach the target ever...
since the addition of distance covered is a the sum of terms in geometric progression which is "tending to 10 mts"
*ahem*
carry on.. ๐ 
raj87verma88The "pictures of the arrow" puzzle seems to have aroused the most interest in the members......this teaser is a paradox. It is my second and final hint for this teaser. Try it and the others, while i look for other brain teasers.

raj87verma88Actually all the teasers are a paradoxes. The answer to both the arrow puzzles lie in the theory of infinite numbers.

raj87verma88I suppose no one wants to give it a shot. So why bother when no ones interested:neutral:

Ashraf HZBro, give it another day or two ๐ Its barely been 24 hours since you've posted it! You know how slow things move in CE ๐

raj87verma88Yeah i can see. I will wait for some time

raj87verma88
The answer lies in the theory of infinitesimals, these are numbers that exist between all other numbers. That is, when you keep choosing a smaller and smaller number, no matter how small, there always lies a smaller one. Therefore, with the arrow, we have a infinite series that has a finite sum, thus the arrow reaches the target.raj87verma88Consider an arrow in flight. At any given moment of time, a snapshot could be taken of this arrow. In this snapshot, the arrow would not be moving. Let us now take another snapshot, leaving a very small gap of time between them. Again, the arrow is stationary. We can keep taking snapshots for each moment of time, each of which shows the arrow to be stationary. Therefore the overall effect is that the arrow never moves, however it still hits the target!
Where lies the flaw in the logic? 
raj87verma88raj87verma88You can imagine an arrow in flight, toward a target. For the arrow to reach the target, the arrow must first travel half of the overall distance from the starting point to the target. Next, the arrow must travel half of the remaining distance.
For example, if the starting distance was 10m, the arrow first travels 5m, then 2.5m.
If you extend this concept further, you can imagine the resulting distances getting smaller and smaller. Will the arrow ever reach the target?
Yes. This is because the sum of an infinite series can be a finite number. Thus, 1/2 + 1/4 + 1/8 + ... = 1 and the arrow hits the target. 
gohmvery cool Raj!

suyash
the sum is never completely equal to 1 ... its always barely short...raj87verma88Yes. This is because the sum of an infinite series can be a finite number. Thus, 1/2 + 1/4 + 1/8 + ... = 1 and the arrow hits the target.
come on guys... there is a difference between the value "tending to 1" and "actually equal to 1"... in this case, to hit the target, it needs to be "actually equal to 1"...
at every point of its course of journey, especially at the end, there always remains an infinitesimal distance remaining to be covered !!
now since the arrow always covers only half of the distance which is remaining, it never reaches the target ever !! (do I need to elucidate even more?? )
I thought everyone here learnt Engineering mathematics, I mean this was even before that !! ๐ 
raj87verma88A series can converge or diverge. A series that converges has a finite limit, that is a number that is approached. A series that diverges means either the partial sums have no limit or approach infinity. The difference is in the size of the common ratio. If r < 1, then the series will converge. If r > or = 1 then the series diverges.
SUM OF AN INFINITE GEOMETRIC SERIES
If r < 1, the infinite Geometric Series:
t1 + t1r + t1r2 + . . . + t1rn + . . .
converges to the sum s = t1 / (1  r).
If r > 1, and t1 does not = 0, then the series diverges.
Consider Question 1/2 + 1/4 + 1/8 + 1/16 + . . .
r = 1/2, so this series converges. Apply the sum
sum = .5/(1  .5) = .5/.5 = 1
Any more doubts Mr. Engineer who has studied Engineering Maths:sleeping: 
gohmAh that is true, however ONLY if you are dealing with abstract mathematics. Determining which form of the two types of mathematics being applied here is key. Applying the wrong form would lead to erroneuos results of theory. I would wager, that since we are describing a natural phenomenon explaining numerically the laws of physics, we are using pure mathmetics which agrees with the natural laws of physics in the the arrow does indeed mathematically reach the target. Thought it would be good to mix in a little philosophy & debate along with the science to answer the puzzle. ๐
suyashthe sum is never completely equal to 1 ... its always barely short...
come on guys... there is a difference between the value "tending to 1" and "actually equal to 1"... in this case, to hit the target, it needs to be "actually equal to 1"...
at every point of its course of journey, especially at the end, there always remains an infinitesimal distance remaining to be covered !!
now since the arrow always covers only half of the distance which is remaining, it never reaches the target ever !! (do I need to elucidate even more?? )
I thought everyone here learnt Engineering mathematics, I mean this was even before that !! ๐ 
raj87verma88Imagine a prisoner in a prison. He is sentenced to death and has been told that he will be killed on one day of the following week. He has been assured that the day will be a surprise to him, so he will not be anticipating the hangman on a particular day, thus keeping his stress levels in check.
The prisoner starts to think to himself, if I am still alive on Thursday, then clearly I shall be hanged on Friday, this would mean that I then know the day of my death, therefore I cannot be hanged on Friday. Now then, if I am still alive on Wednesday, then clearly I shall be hanged on Thursday, since I have already ruled out Friday. The prisoner works back with this logic, finally concluding that he cannot after all be hanged, without already knowing which day it was.
Casually, resting on his laurels, sitting in his prison cell on Tuesday, the warden arrives to take him to be hanged, the prisoner was obviously surprised!
Ponder this? 
raj87verma88The four people in this puzzle all competed in different classes of dog agility at a recent competition. The competitions all required the dogs to run over jumps, through tunnels and various other obstacles in as quicker time as possible. Each had a different result  one came first, one third, one fourth and one ninth. All four dogs were each of a different breed.
Can you work out who handled which dog, at what level each competed, the place each finished in and the breed of each dog?
1. If Tiff finished first then Terry finished fourth.Handler's Names: Jane, Mark, Ruth and Terry
2. If Terry finished fourth then Jago is a collie otherwise Jago is not a collie.
3. If Jane competed in the Senior class then she finished third.
4. If Jane competed in Novice then she finished fourth.
5. The dog that finished ninth was an alsatian. This was either Jago, in which case Jago competed in the Elementary class, or this was Kelly, in which case Terry handled Kelly.
6. Mark won Starters.
7. If Mark's dog is called Patti then Patti is a labrador otherwise Patti is a collie.
8. Ruth's dog is called Jago.
9. If Jago finished fourth then she competed in the Novice class otherwise she competed in the Senior class.
10. If Patti finished first then Terry's dog is an alsatian otherwise Terry's dog is a collie.
11. If Jane's dog is a doberman then Jane finished fourth otherwise Jane finished third.
โ
Dog's Names: Tiff, Patti, Jago and Kelly
Breed: Alsatian, Collie, Labrador and Doberman
Level: Starters, Elementary, Novice or Senior 
raj87verma88Try some more. And this time there can be no debate on mathematics

gohmWas this inmate arrested for his faulty logic? laugh.
His error was running on a diferent logic thread than everyone else. Just like if you convince yourself it is 3 PM when your local time is 9AM and switch on your TV at your concieved 4PM time to watch your favorite program, it will not be on.
raj87verma88Imagine a prisoner in a prison. He is sentenced to death and has been told that he will be killed on one day of the following week. He has been assured that the day will be a surprise to him, so he will not be anticipating the hangman on a particular day, thus keeping his stress levels in check.
The prisoner starts to think to himself, if I am still alive on Thursday, then clearly I shall be hanged on Friday, this would mean that I then know the day of my death, therefore I cannot be hanged on Friday. Now then, if I am still alive on Wednesday, then clearly I shall be hanged on Thursday, since I have already ruled out Friday. The prisoner works back with this logic, finally concluding that he cannot after all be hanged, without already knowing which day it was.
Casually, resting on his laurels, sitting in his prison cell on Tuesday, the warden arrives to take him to be hanged, the prisoner was obviously surprised!
Ponder this? 
suyashraj87verma88You can imagine an arrow in flight, toward a target. For the arrow to reach the target, the arrow must first travel half of the overall distance from the starting point to the target. Next, the arrow must travel half of the remaining distance.
For example, if the starting distance was 10m, the arrow first travels 5m, then 2.5m.
If you extend this concept further, you can imagine the resulting distances getting smaller and smaller. Will the arrow ever reach the target?raj87verma88Yes. This is because the sum of an infinite series can be a finite number. Thus, 1/2 + 1/4 + 1/8 + ... = 1 and the arrow hits the target.suyashthe sum is never completely equal to 1 ... its always barely short...
come on guys... there is a difference between the value "tending to 1" and "actually equal to 1"... in this case, to hit the target, it needs to be "actually equal to 1"...
at every point of its course of journey, especially at the end, there always remains an infinitesimal distance remaining to be covered !!
now since the arrow always covers only half of the distance which is remaining, it never reaches the target ever !! (do I need to elucidate even more?? )
I thought everyone here learnt Engineering mathematics, I mean this was even before that !! ๐
Dude... Seriously, after reading that last line of yours, you have really got me talking here.raj87verma88A series can converge or diverge. A series that converges has a finite limit, that is a number that is approached. A series that diverges means either the partial sums have no limit or approach infinity. The difference is in the size of the common ratio. If r < 1, then the series will converge. If r > or = 1 then the series diverges.
SUM OF AN INFINITE GEOMETRIC SERIES
If r < 1, the infinite Geometric Series:
t1 + t1r + t1r2 + . . . + t1rn + . . .
converges to the sum s = t1 / (1  r).
If r > 1, and t1 does not = 0, then the series diverges.
Consider Question 1/2 + 1/4 + 1/8 + 1/16 + . . .
r = 1/2, so this series converges. Apply the sum
sum = .5/(1  .5) = .5/.5 = 1
Any more doubts Mr. Engineer who has studied Engineering Maths:sleeping:
First of all, before explaining my solution for the final time on this thread, let me assure you, with utmost sincerity that I didn't exactly flunk in my engineering mathematics tests and even I am aware of the formula for calculation of sum of infinite terms in geometric progression with a common ratio (r) which is less than 1. I swear by GOD, most of us on this forum know it for sure.
Now coming back to the point stated by me previously: According to the problem statement, the arrow, at the end of each day, covers exactly half of the distance which was remaining at the start of the day. Isn't this what is implied in the problem?
 Now according to the above mentioned statement, at the end of each day, there always remains some distance to be covered at the end of every single day, since the arrow covers only half of the distance remaining on each particular day.
 Before you think further, let me first tell what exactly should be the mathematical condition under which one can say that the arrow reaches the target: one can say that the arrow reaches the target, only when the distance in between the arrow and target becomes zero.
 In this case, however, since the distance between the arrow and the target is always a nonzero entity, we can confirm the fact that the arrow never reaches the target. And that's that. Period.
When we consider a sum of infinite terms (beginning with the first term as "a") in geometric progression (i.e. the common ratio r < 1), it is given by [a/(1r)]. Has anyone considered how this formula is derived? It is based on Newtonian Mathematics. Ever heard of Philosophiรยฆ Naturalis Principia Mathematica?? Following is a short image which depicts how the formula has been derived. Do take a look. It won't take more than a minute.
#LinkSnipped#
Intelligent CEans may not even need to read the remainder of this post since I hope I've made it clear that the formula is derived from the concept of "limits", which itself comprises of certain assumptions so as to make things easier in the world of mathematics (especially, calculus). Hence, it would be ridiculous, if we use such a formula to predict the outcome of real life applications like the one mentioned in this puzzle.
As a result the sum of 1+0.5+0.25+0.125+ ............ + .... up to infinite terms may be equal to 1, based on the abstract mathematical formula, but not in this case. We can't take into account the assumption made by Sir Isaac Newton (for "limits") while dealing with the idealistic scenario of events which are defined strictly by certain constraints. When it comes to real life, we have to be idealistic, where 0.999999999 is simply not equal to 1.
Bingo. That is it. Hence the arrow doesn't hit the target. There always was, and there always is, and there always will be, a certain nonzero distance left to be covered in between the target and the arrow.
OK. Now that we are done with it, let me congratulate you for the superb application of the above formula for the solution of this puzzle. But in fact, on second thought, I need to strip that credit from you too, since the solution of the puzzle seems to be ripped off from another website, word by word (BrainBashers  Paradoxes: See puzzle 3 on paradoxes page, and click on show answer, you will see that it matches surprisingly with the text posted in this thread as the solution). This puts in new questions for me. Did you even bother about giving a serious thought to the puzzle before copy pasting the solution??
In the end, I would only like to mention 2 points:
1) The last line you wrote in this post (#LinkSnipped#). Ouch. That surely hurt me. Because Mathematics has always been my favorite subject. Please don't, never ever, post such a snide remark about someone, especially on a forum like this one which I presume consists of such highly capable and intelligent minds in this world.
2) Don't treat a real life problem/puzzle as an equivalent of a mathematical question. They are different. A good puzzle requires much more higher ingenuity and creative thinking as compared to numerical computations based on mindless application of mathematical formulae. Yes, they (formulae) are good, since they help us to live an easy life when it comes to numbers, however, it is not wise to use them without thinking just because they can be applied almost everywhere. If you are a fan of Isaac Asimov, you must be already drawing parallels of his ideas about the use of technology.
As Andrew Koenig (C++ expert and former AT&T researcher) says, "Abstraction is selective ignorance."
Adios, Amigos. That is all from my side. Sorry for the long post though. ๐ 
raj87verma88Yes i did copy the solution from the site as i was too lazy to type the solution but had to solve it mathematically to prove the answer later. The riddle also came from that site. Actually all riddles right now are from this site. I will rip the exact answers from the site as I myself may solve it incorrectly. But don't worry about me not solving them. I only post those riddles that i found interesting to solve. ๐
And I am sorry if my sarcastic remark hurt you.
And after your first post about the answer i rechecked it through many sources and this is the answer. It will always come out to be 1. Your points are correct that there is difference between a value tending to 1 and actually being 1 and i do not deny them. So now to prevent such an ugly confrontation happening again i won't put any mathematical riddles in this post. Check out the other ones hope you will enjoy them and i assure you they will tax your brains.
I hope there are no hard feelings as i became rather aggressive.

suyash
I'm Cool. Looking forward to more such thought provoking puzzles. By the way, if you really wanna have some serious fun, do visit this site: [ wu :: riddles(intro) ]it was really helpful for me when I prepared for my campus placement last year !!raj87verma88Check out the other ones hope you will enjoy them and i assure you they will tax your brains. I hope there are no hard feelings as i became rather aggressive.

raj87verma88Thanks for the link mate.

suyash
You are welcome,anytime !! ๐raj87verma88Thanks for the link mate. 
raj87verma88
This is a paradox. It has no true answer. You can go on and on thinking various scenarios. ๐raj87verma88Imagine a prisoner in a prison. He is sentenced to death and has been told that he will be killed on one day of the following week. He has been assured that the day will be a surprise to him, so he will not be anticipating the hangman on a particular day, thus keeping his stress levels in check.
The prisoner starts to think to himself, if I am still alive on Thursday, then clearly I shall be hanged on Friday, this would mean that I then know the day of my death, therefore I cannot be hanged on Friday. Now then, if I am still alive on Wednesday, then clearly I shall be hanged on Thursday, since I have already ruled out Friday. The prisoner works back with this logic, finally concluding that he cannot after all be hanged, without already knowing which day it was.
Casually, resting on his laurels, sitting in his prison cell on Tuesday, the warden arrives to take him to be hanged, the prisoner was obviously surprised!
Ponder this?
You are reading an archived discussion.
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