Four people come to an old bridge in the middle of the night. The bridge is rickety and can only support 2 people at a time. The people have one flashlight, which needs to be held by any group crossing the bridge because of how dark it is.
Each person can cross the bridge at a different rate: one person takes 1 minute, one person takes 2 minutes, one takes 5 minutes, and the one person takes 10 minutes. If two people are crossing the bridge together, it will take both of them the time that it takes the slower person to cross.
Unfortunately, there are only 17 minutes worth of batteries left in the flashlight. How can the four travellers cross the bridge before time runs out?
The two keys here are:
You want the two slowest people to cross together to consolidate their slow crossing times.
You want to make sure the faster people are set up in order to bring the flashlight back quickly after the slow people cross.
So the order is:
1-minute and 2-minute cross (2 minute elapsed)
1-minute comes back (3 minutes elapsed)
5-minute and 10-minute cross (13 minutes elapsed)
2-minute comes back (15 minutes elapsed)
1-minute and 2-minute cross (17 minutes elapsed)
You are standing before two doors. One of the path leads to heaven and the other one leads to hell. There are two guardians, one by each door. You know one of them always tells the truth and the other always lies, but you don’t know who is the honest one and who is the liar. You can only ask one question to one of them in order to find the way to heaven. What is the question?
The question you should ask is "If I ask the other guard about which side leads to heaven, what would he answer?"
It should be fairly easy to see that irrespective of whom do you ask this question, you will always get an answer which leads to hell. So you can chose the other path to continue your journey to heaven.
This idea was famously used in the 1986 film Labyrinth.
Here is the explanation if it is yet not clear.
Let us assume that the left door leads to heaven.
If you ask the guard which speaks truth about which path leads to heaven, as he speaks always the truth, he would say "left". Now that the liar , when he is asked what "the other guard (truth teller) " would answer, he would definitely say "right".
Similarly, if you ask the liar about which path leads to heaven, he would say "right". As the truth teller speaks nothing but the truth, he would say "right" when he is asked what "the other guard( liar ) " would answer. So in any case, you would end up having the path to hell as an answer. So you can chose the other path as a way to heaven.
You have twelve balls, identical in every way except that one of them weighs slightly less or more than the balls.
You have a balance scale, and are allowed to do 3 weighings to determine which ball has the different weight, and whether the ball weighs more or less than the other balls.
What process would you use to weigh the balls in order to figure out which ball weighs a different amount, and whether it weighs more or less than the other balls?
Take eight balls, and put four on one side of the scale, and four on the other.
If the scale is balanced, that means the odd ball out is in the other 4 balls.
Let's call these 4 balls O1, O2, O3, and O4.
Take O1, O2, and O3 and put them on one side of the scale, and take 3 balls from the 8 "normal" balls that you originally weighed, and put them on the other side of the scale.
If the O1, O2, and O3 balls are heavier, that means the odd ball out is among these, and is heavier. Weigh O1 and O2 against each other. If one of them is heavier than the other, this is the odd ball out, and it is heavier. Otherwise, O3 is the odd ball out, and it is heavier.
If the O1, O2, and O3 balls are lighter, that means the odd ball out is among these, and is lighter. Weigh O1 and O2 against each other. If one of them is lighter than the other, this is the odd ball out, and it is lighter. Otherwise, O3 is the odd ball out, and it is lighter.
If these two sets of 3 balls weigh the same amount, then O4 is the odd ball out. Weight it against one of the "normal" balls from the first weighing. If O4 is heavier, then it is heavier, if it's lighter, then it's lighter.
If the scale isn't balanced, then the odd ball out is among these 8 balls.
Let's call the four balls on the side of the scale that was heavier H1, H2, H3, and H4 ("H" for "maybe heavier").
Let's call the four balls on the side of the scale that was lighter L1, L2, L3, and L4 ("L" for "maybe lighter").
Let's also call each ball from the 4 in the original weighing that we know aren't the odd balls out "Normal" balls.
So now weigh [H1, H2, L1] against [H3, L2, Normal].
-If the [H1, H2, L1] side is heavier (and thus the [H3, L2, Normal] side is lighter), then this means that either H1 or H2 is the odd ball out and is heavier, or L2 is the odd ball out and is lighter.
-So measure [H1, L2] against 2 of the "Normal" balls.
-If [H1, L2] are heavier, then H1 is the odd ball out, and is heavier.
-If [H1, L2] are lighter, then L2 is the odd ball out, and is lighter.
-If the scale is balanced, then H2 is the odd ball out, and is heavier.
-If the [H1, H2, L1] side is lighter (and thus the [H3, L2, Normal] side is heavier), then this means that either L1 is the odd ball out, and is lighter, or H3 is the odd ball out, and is heavier.
-So measure L1 and H3 against two "normal" balls.
-If the [L1, H3] side is lighter, then L1 is the odd ball out, and is lighter.
-Otherwise, if the [L1, H3] side is heavier, then H3 is the odd ball out, and is heavier.
If the [H1, H2, L1] side and the [H3, L2, Normal] side weigh the same, then we know that either H4 is the odd ball out, and is heavier, or one of L3 or L4 is the odd ball out, and is lighter.
So weight [H4, L3] against two of the "Normal" balls.
If the [H4, L3] side is heavier, then H4 is the odd ball out, and is heavier.
If the [H4, L3] side is lighter, then L3 is the odd ball out, and is lighter.
If the [H4, L3] side weighs the same as the [Normal, Normal] side, then L4 is the odd ball out, and is lighter.
Bruce is an inmate at a large prison, and like most of the other prisoners, he smokes cigarettes. During his time in the prison, Bruce finds that if he has 3 cigarette butts, he can cram them together and turn them into 1 full cigarette. Whenever he smokes a cigarette, it turns into a cigarette butt.
One day, Bruce is in his cell talking to one of his cellmates, Steve.
"I really want to smoke 5 cigarettes today, but all I have are these 10 cigarette butts," Bruce tells Steve. "I'm not sure that will be enough."
"Why don't you borrow some of Tom's cigarette butts?" asks Steve, pointing over to a small pile of cigarette butts on the bed of their third cellmate, Tom, who is out for the day on a community service project.
"I can't," Bruce says. "Tom always counts exactly how many cigarette butts are in his pile, and he'd probably kill me if he noticed that I had taken any."
However, after thinking for a while, Bruce figures out a way that he can smoke 5 cigarettes without angering Tom. What is his plan?
Bruce takes 9 of his 10 cigarette butts and turns them into 3 cigarettes total (remember, 3 cigarette butts can be turned into 1 cigarette). He smokes all three of these, and now he has 4 cigarette butts.
He then turns 3 of the 4 cigarette butts into another cigarette and smokes it. He has now smoked 4 cigarettes and has 2 cigarette butts.
For the final step, he goes and borrows one of Tom's cigarette butts. With this cigarette butt plus the 2 he already has, he is able to make his 5th cigarette to smoke. After smoking it, he is left with 1 cigarette butt, which he puts back in Tom's pile so that Tom won't find anything missing.