There was this thread about logic puzzles a while ago here at Sajha. Needless to say, it was quite popular. The thread was also a confluence of 'great' minds of sajha, who displayed there abilities to approach problems from different view points, not necessarily the correct point of view, however.

One way to keep this thread moving is to come up with puzzles each time a puzzle is solved. And just to avoid any stress to Google Inc., I encourage people to phrase the puzzles as accurately as they can in their own words.
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Here's the one that I would like to start with.
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There are two threads, with DIFFERENT lengths and DIFFERENT diameters. The threads are NOT uniform, each thread has different diameter at different points along the thread. But if you light up the thread at any one end, it takes exactly one hour for EACH thread to finish buring from one end to the other end. You can light up any end first-- it does not matter. Now, how do I measure 45 minuties by using those threads?

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This is not a riddle, so the aim is not trick anyone in small details.

hetterika..tauko khayo..no more puzzle..

more???

haude_ko_bhai I know what I was talking about , if you didnt get , you simply didnt

Haude ko bhai,
You got it!

Hyperthread,
You assumed that the thread are uniform, ie at half hour, the thread would finish burning half--which is not necessarily true.

Others...depends how boring the day will be tomorrow.

im not sure if someone's posted it but here's an easy solution to the first one:

Fold the first longer thread into half and fold the second shorter thread into 4 equal parts (its easy).
Line them up end to end and light up either one of the threads at one of the folded ends so that every part of the thread burns
Once they're both finished burning its exactly 45 minutes.

.you have
10 stacks of coins
an electronic scale

problem:
of the 10 stacks of coins, 9 stacks are good coins
one stack is all counterfeit coins
the countefeit coins are identical to the real ones, except they weigh one tenth of a gram less than the real coins
you are allowed to weigh any combination of coins

tara
ONLY ONE WEIGHING ALLOWED

figure out which stack is counterfiet

and remember, one weighing allowed in total

For the sake of clarity, I am assuming that each real coins weigh 10g and the fake coins weigh 9g (it does not really matter what they weigh though)

Take 1 coin from stack 1; 2 from stack 2; 3 from stack 3; 4 from stack 4; 5 from stack 5; 6 from stack 6; 7 from stack 7; 8 from stack 8; and, 9 from stack 9. This makes a total of 45 coins.

This combination of coins give different weight depending upon which stack the coin is in. For example, if the fake stack is stack 1, the total weight will be 449g.
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Try the same problem, now, using a BEAM BALANCE. Try to find the fake stack by weighing no more than 2 times.

Ye dai..ani ki..ma sanga pani auta question cha...
ani ki.. tyo Kachuwa ra Kharayo ko..
Baba le malai achel english sikaunu huncha....
Ma aja english ma lekchu la..

In the old story of Rabit( R) and Tortoise (T), I say that Rabit will never meet Tortoise if Rabit starts the race after Tortoise. It doesnt matter how fast Rabit runs. He just can't meet Tortoise.
Here is my argument:

Lets say T starts some seconds earlier than R with his very slow speed and R starts to follow T with his extremely fast speed.

After certain time "t", R reaches the place where T was. But in that time, T will cover some extra distance. Again R will reach that new position after some small time "t1" but again T will have travelled some additional distance in that period. Again R will run to cover the extra distance but T will cover some more distance in that time.. it keeps on going..

R will always be busy to cover the extra distance that T has already covered.

Hence R can never meet T becase by the time R reaches the point where T was, T will have covered some additional distance.

Ani ki ...dai..
Kharayo le ni ..Kachuwa lai kahile pani bhetauna sakdaina..
Mero school ko kitab ma galti lekheko cha..

Dai..chadai bhannu na...
Kharayo le Kasari Kachuwa lai jitcha...
Maile Bholi Miss lai bhannu cha..

kapuri kha, you've totally missed the concept of speed.
incorporate that and youll find that the book was right.

timetraveller dai, speed ko concept bujhai dinu na..

Nature should be the same from whichever perspective you observe it.
Truth should be truth.

Please point out any flaw if there is in my argumentation.

Otherwise I dont see any way the Rabit winning the match.

For the puzzle regarding the thread,

take the first thread and burn it on both ends and only one end of the second thread at the same time. This way when the first thread burns out it will be 30 minutes. Then after as soon as the first thread burns out then light the other end of the second thread as well (which would have been half burnt by now). Then the remaining part of the thread should burn out in 15 minutes. Hence you get 30 + 15 = 45 minutes.
Thank you.

Khoi no one to solve my logical problem? Ali overdose bha jasto cha sajha lai.

Just repeat it for new readers

There is a 200 mile distance in which runners are allowed to carry 2 gallons of water in 2 hands, thats it. The 1 gallons of water is enough for the runner to run 25 miles. He can go a certain miles and store gallons as many times as he likes provided he has enough water to get back to the start. He can come back and get maximum of 2 gallons. So he runs a certain distance and stores gallons of water, comes back to start get more water. What is the lowest number of trips he has to make back to make the 200 mile distance?

must be factorial of 16 or something like that....don't want to think too much...tauko dukhcha...

Ooo...didn't see that.

Is it 64? If not, then will have to think after I get off from my boring work!

I'm thinking like this:
Lets divid total distance(-- = 12.5 miles and each digit = 25 miles)
0--|--1--|--2--|--3--|--4--|--5--|--6--|--7--|--8
1. He needs 2 gallons of water to travell 25 miles
2. He needs 8 gallons of water to travell 50 miles
similarly, 32 for 75 miles (I'm not sure... )
If it works then our general formula will be
2(4 pow(n-1))

to reach at position 8
2(4 pow(8-1))
total distance he needs to travell to cover 200 miles and come back
2(4 pow(7)) X 25 miles
total number of round trip
(2(4 pow(7)) X 25)/200
=(4 pow(7))/4

Thanks for all the responses

Buddhaboy was one of the closest to the answer.

Well the number of times he has to return to the start is

N = 2^(n-1)

n= ((2 x Total Distance)/distance per gallon) - 1

so n = 8
Hence N = 128

Rationale: for every 25 miles he has to use half the number of gallons. At 150th mile he has to have 2 gallons of water to complete the distance. so 125 he has to have 4 gallons, similarly 100 : 8, 75 : 16, 50 : 32, 25 : 64 and thus 0 : 128.

Nevermind, I don't quite agree with your answer.

I agree that the runner should have 2 full gallons of water when he is at 150th mile, but I don't agree that the number of gallons at 125th mile should be 4.

Here's why: If I carry two gallons from 125th mile to 150th mile, then I will finish one gallon just going. I will need another gallon to come back. So, if I have that interval of 25 miles, then I will have to use one gallon while going and the other gallon while coming back--this does not save me any water. So, instead, the travelling interval should be 12.5 miles NOT 25 miles. In effect, the number of gallons of water I will require at different intervals now become:
150m: 2 (2^1)
137.5m: 4 (2^2)
125m: 8 (2^3)
-
-
-
25m: 2^11
12.5m: 2^12 = 4096

That means I need 4096 gallons of water at a distance of 12.5 miles from the start. I am just trying to know the lowest number of trips that I have to make to cover 200 miles, so I do NOT need not to know how many gallons I need at 0th mile (as opposed to what you have said above).
And to have 4096 gallons of water at 0th mile, I need to get back to start 4095 (4096-1) times.
So, I think the correct answer should be 4095 times.
Let me know if you do not agree.

Key assumption: I need water to travel, regardless of moving forward or going back to get the gallons of water.

Among 10 coins which look exactly alike, two are fake coins. All real coins weigh the same. Both fake coins weigh the same, but they weigh less than the real coins. You have a balance. How do you find the fake coins using a balance not more than 5 times?