its not really a riddle, more of a paradox. But it got me to really think about time travel:

What happens if someone goes back in time and kills their own grandmother, before his grandmother gives birth to his mother? what happens to him?
now, you're going to think that, well he wouldn't exist, right? well, if he doesn't exist, how can he go back in time and kill his grandmother?
Nemesis-301 wroteits not really a riddle, more of a paradox. But it got me to really think about time travel:

What happens if someone goes back in time and kills their own grandmother, before his grandmother gives birth to his mother? what happens to him?
now, you're going to think that, well he wouldn't exist, right? well, if he doesn't exist, how can he go back in time and kill his grandmother?
Time travel would directly imply that there is no causation. This is why physicists have trouble with the concept of time travel.
This one was asked to my brother during his interview with a big financial institution for a trading position.
There is a round table. You are playing against an opponent and are provided with identical glasses. When his turn comes up, a player will place one glass on the table. The players will continue alternating turns until the table is full. The player whose turn comes up and can't place a glass on the table loses.

Now the question: if you are given the choice, would you choose to be the first to put a glass on the table? If so, where would you place it?
@samer:
I'd go first and put the first glass at the center. Next, I'd wait for his move and place a glass at the symmetrical opposite of his each time. This would work because:

- glasses are identical.
- for each A on the table, the symmetrical opposite (with respect to the center) B is unique and sym(A) = B implies sym(B) = A. (does this property have a name?)

Also, now that I think of it, by the above reasoning, you could argue that there's a finite amount of glasses you can fit on the table and this number is odd ((every point + its opposite) + the center). That proves that as long as you go first, no matter where you put the glasses, you'll win.
@ Samer, it has the same basic concept of tic-tac-toe.
You are a secret agent, your mission is to blow up an arms factory. For reasons we won't get into, you plan to do it old school, using fuses and gun powder, not timers and C4 and that stuff. You have 2 fuses, each burn from end to end in exactly 30 minutes. The fuse does not have a uniform distribution of gun powder, in other words, it does not burn in a constant speed, for ex: 99% of the fuse could take 1 minute to burn and the remaining 1% of its length 29 minutes. You estimate it takes you 45 minutes to make it out of the factory and need to use the 2 fuses you got to build a time bomb that will explode after 45 minutes from lighting the fuse(s).

EDIT

To avoid confusion, let me correct my scenario:
All you care about is creating a timer, forget about how the fuse is attached (or not) to the bomb
Georges wrote- How to put an elephant in the fridge in 3 moves ?
- How to put a giraffe in the fridge with 4 moves ?
- How to put 5 elephants in a mini cooper ?
- How to put 10 elephants in a Cadillac ?
- How do you know that 10 elephants went out to a pub ?
-1- open the fridge ; 2- put the elephant ; 3- close the fridge.
-1- open the fridge ; 2-remove the elephant from there :P ; 3-put the giraffe ; 4-close the fridge
-2 in the front seats and 3 in the backseats
-sell the Cadillac,buy 2 mini coopers put 5 in each mini the way i mentioned above (don't know if it's still legit :P)
-you see 2 mini coopers parking outside.

Keep it simple ;)
@MSD: Can you "waste" a third fuse to measure 15 mins by lighting it simultaneously from both sides?
MSD wroteYou are a secret agent, your mission is to blow up an arms factory. For reasons we won't get into, you plan to do it old school, using fuses and gun powder, not timers and C4 and that stuff. You have 2 fuses, each burn from end to end in exactly 30 minutes. The fuse does not have a uniform distribution of gun powder, in other words, it does not burn in a constant speed, for ex: 99% of the fuse could take 1 minute to burn and the remaining 1% of its length 29 minutes. You estimate it takes you 45 minutes to make it out of the factory and need to use the 2 fuses you got to build a time bomb that will explode after 45 minutes from lighting the fuse(s).
ok tie the first fuse by half (put both ends of the fuse together and twist it together so it become a full fuse with half the size of a normal fuse) attach it to the other fuse (keep this one normal) and light it.the unmodified fuse will take 30 minutes, the twisted fuse will start burning both layers that are twisted together so until it finish burning it will take 15 minutes so 45 min in total.
m.sabra wrote
MSD wroteYou are a secret agent, your mission is to blow up an arms factory. For reasons we won't get into, you plan to do it old school, using fuses and gun powder, not timers and C4 and that stuff. You have 2 fuses, each burn from end to end in exactly 30 minutes. The fuse does not have a uniform distribution of gun powder, in other words, it does not burn in a constant speed, for ex: 99% of the fuse could take 1 minute to burn and the remaining 1% of its length 29 minutes. You estimate it takes you 45 minutes to make it out of the factory and need to use the 2 fuses you got to build a time bomb that will explode after 45 minutes from lighting the fuse(s).
ok tie the first fuse by half (put both ends of the fuse together and twist it together so it become a full fuse with half the size of a normal fuse) attach it to the other fuse (keep this one normal) and light it.the unmodified fuse will take 30 minutes, the twisted fuse will start burning both layers that are twisted together so until it finish burning it will take 15 minutes so 45 min in total.
You don't really need to twisted into a fuse, All you need is to make a loop:

-- FUSE 1
-- FUSE 2
become --O
You light it from the left and when the flame reaches the point of intersection the second fuse would light from both ends.
rahmu wrote@MSD: Can you "waste" a third fuse to measure 15 mins by lighting it simultaneously from both sides?
You don't need a third fuse to do that
OK here's another:

You got a box of cigarette packs containing 20 packs where each pack has 20 cigs in it. That is 400 cigs total.
You are told that one of the packs (with the 20 cigs inside it) has a defect, the defect is that the amount of tobacco inside is less than the usual by 0.1 grams (per cig). Usually a cigarette weighs 1 grams, so a pack (excluding the weight of the wrapping) weighs 20 x 1g = 20g, the defective one would weigh 20 x 0.9 = 18g.
In a locked cell, you are butt naked (don't ask me why) and are provided with the box of cigarettes, and a digital scale that will give you only one reading before it is permanently disabled (for some action it might also automatically blow up a few seconds after you use it). All the packs are marked by a serial number so that they can be identified. The people who are testing your skills (and who probably are the ones that stripped you naked) already have the number of the defective pack. You are assigned with the task of identifying the defective pack.

And yeah, the room is being cooled to make you feel cold (since you are ... yes you guessed it, NAKED)

P.S: Your attire _or lack of it_ has nothing to do with this problem
@MSD you take n cigarettes from each pack where n is the index of the pack and weigh everything.
n_defectivepack = (210 - weigh_result) * 10.


Then I throw the scale at the lock on the cell's door, it explodes and I go out, look for a blanket (cause, you know it's so damn cold in this cell) and go back to sleep. Tomorrow I'll look for clothes.
a year later
i remembered this thread today,so why not revive it a bit.
Riddle 1:
(a classic one) you have 8 identical balls in size,but 1 ball weigh a little more than the others,you have a balance scale and 2 tries to determine which ball is the heavy one.

Riddle 2:
4 campers have 1 flashlight that have enough battery to last 17 minutes,they have to cross a rope bridge but it's too dangerous to cross it without light,and the bridge can hold at most 2 people,the campers can cross the bridge at different speeds.
Camper 1: needs 1 minute.
Camper 2: needs 2 minutes.
Camper 3: needs 5 minutes.
Camper 4: needs 10 minutes.
How can they all cross the bridge in 17 minutes with light.

Don't google the answers :P
This thing all things devours:
Birds, beasts, trees, flowers;
Gnaws iron, bites steel;
Grinds hard stones to meal;
Slays king, ruins town,
And beats high mountain down.
m.sabra wrotei remembered this thread today,so why not revive it a bit.
Riddle 1:
(a classic one) you have 8 identical balls in size,but 1 ball weigh a little more than the others,you have a balance scale and 2 tries to determine which ball is the heavy one.
You put 3 balls on the left side and 3 on the right side
If both sides are balanced (equal), then you remove the 6 balls and put 1 of the 2 remaining balls to the left and 1 to the right, one of them should be heavier
If both sides were not equal, you take the 3 balls that are heavier, you put 2 of them on the scale, one to the right and one to the left, if they are equal then the 3rd one is the heaviest
m.sabra wrote Riddle 2:
4 campers have 1 flashlight that have enough battery to last 17 minutes,they have to cross a rope bridge but it's too dangerous to cross it without light,and the bridge can hold at most 2 people,the campers can cross the bridge at different speeds.
Camper 1: needs 1 minute.
Camper 2: needs 2 minutes.
Camper 3: needs 5 minutes.
Camper 4: needs 10 minutes.
How can they all cross the bridge in 17 minutes with light.

Don't google the answers :P
You're going to have to waste 10 mins on one camper, that's inevitable. So the goal is to minimize loss by hiding the 5 mins in this round:
  • camp1 + camp2 cross (2mins)
  • camp1 returns alone (1min)
  • camp4 + camp3 cross (10 mins)
  • camp2 returns alone (2mins)
  • camp2 + camp1 cross (2mins)
Total: 17 minutes

Nice riddle :)
m.sabra wrotei remembered this thread today,so why not revive it a bit.
Riddle 2:
4 campers have 1 flashlight that have enough battery to last 17 minutes,they have to cross a rope bridge but it's too dangerous to cross it without light,and the bridge can hold at most 2 people,the campers can cross the bridge at different speeds.
Camper 1: needs 1 minute.
Camper 2: needs 2 minutes.
Camper 3: needs 5 minutes.
Camper 4: needs 10 minutes.
How can they all cross the bridge in 17 minutes with light.

Don't google the answers :P
Well, I'm not sure if my solution is meant to be acceptable, but since the bridge can hold two people at once:
C1 and C3 cross at once, it will take C1 one minute to arrive, and C3 will still need four minutes.
C1 arrives. C2 goes on, which means that now, C3 has 4 minutes to go and C2 needs two minutes.
Two minutes later, C2 arrives and C3 still needs two minutes.
C4 goes on. After two minutes, C3 arrives, and C4 arrives after eight minutes.

Which means:
In 5 minutes: C1 + C2 + C3 crossed + C4 crossed 2 / 10.
In 8 minutes: C4 crossed the other 8 / 10.

TOTAL TIME: 5 + 8 = 13 minutes.
There's also this solution but it's longer:

Camper 1 + Camper 3 = 5 minutes.
Camper 2 + Camper 4 = 10 minutes.
Total time = 15 minutes.

I didn't realise that a camper needs to come back in order to hold the flash-light. Rahmu did it.
Ayman wrote
This thing all things devours:
Birds, beasts, trees, flowers;
Gnaws iron, bites steel;
Grinds hard stones to meal;
Slays king, ruins town,
And beats high mountain down.
Time. Its from the hobbit :p