logic 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)

## Similar riddles

See also best riddles or new riddles.

logicmathshortTake 9 from 6, 10 from 9, 50 from 40 and leave 6. How is it possible?

SIX - 9 (IX) = S
9 (IX) - 10 (X) = I
40 (XL) - 50 (L) = X

logicmathSam has got three daughters. The eldest daughter is the most honest girl in the universe and she always speaks truth. The middle daughter is a modest woman. She speaks truth and lies according to the situations. The youngest one never speaks truth. Not a single word she spoke was true and would never be true.
Sam brought a marriage proposal for one of his girls. It was John. John wanted to marry either the eldest or the youngest daughter of Sam as he can easily identify whether the girl speaks truth or lie!
John told his desire to Sam. However, Sam laid a condition. He told John that he will not say who the eldest, middle or youngest one is. Also, he allowed john to ask only one question to identify the eldest or youngest so he can marry one.
John asked one question and found the right girl. What was the question and whom should he pick?
He asks this question to one of the daughters.
If he asked this question to older daughter pointing at other two, he probably would know the youngest one! NO matter, she always speaks truth.
If he asked the question to middle one, probably he can choose either.
If he asked the youngest one, she always lies and he can find eldest one! No matter, he has to choose the youngest one based on the answer.

The question he asked is, ‘Is she older than her!’
He asks this question to one of the daughters.
If he asked this question to older daughter pointing at other two, he probably would know the youngest one! NO matter, she always speaks truth.
If he asked the question to middle one, probably he can choose either.
If he asked the youngest one, she always lies and he can find eldest one. No matter, he has to choose the youngest one based on the answer.

logicmathThere are n coins in a line. (Assume n is even). Two players take turns to take a coin from one of the ends of the line until there are no more coins left. The player with the larger amount of money wins.
Would you rather go first or second? Does it matter?
Assume that you go first, describe an algorithm to compute the maximum amount of money you can win.
Note that the strategy to pick maximum of two corners may not work. In the following example, first player looses the game when he/she uses strategy to pick maximum of two corners.
Example 18 20 15 30 10 14
First Player picks 18, now row of coins is
20 15 30 10 14
Second player picks 20, now row of coins is
15 30 10 14
First Player picks 15, now row of coins is
30 10 14
Second player picks 30, now row of coins is
10 14
First Player picks 14, now row of coins is
10
Second player picks 10, game over.
The total value collected by second player is more (20 + 30 + 10) compared to first player (18 + 15 + 14). So the second player wins.

Going first will guarantee that you will not lose. By following the strategy below, you will always win the game (or get a possible tie).
(1) Count the sum of all coins that are odd-numbered. (Call this X)
(2) Count the sum of all coins that are even-numbered. (Call this Y)
(3) If X > Y, take the left-most coin first. Choose all odd-numbered coins in subsequent moves.
(4) If X < Y, take the right-most coin first. Choose all even-numbered coins in subsequent moves.
(5) If X == Y, you will guarantee to get a tie if you stick with taking only even-numbered/odd-numbered coins.
You might be wondering how you can always choose odd-numbered/even-numbered coins. Let me illustrate this using an example where you have 6 coins:
Example
18 20 15 30 10 14
Sum of odd coins = 18 + 15 + 10 = 43
Sum of even coins = 20 + 30 + 14 = 64.
Since the sum of even coins is more, the first player decides to collect all even coins. He first picks 14, now the other player can only pick a coin (10 or 18). Whichever is picked the other player, the first player again gets an opportunity to pick an even coin and block all even coins.

cleanlogicshortwhat am IWash it and it isn't clean. Don't wash it and then it's clean. What I Am?

Water.

logicYou 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.

funnylogicSam is talking to his lawyer in jail. They are very upset because the judge has refused to grant bail. At the end of the conversation Sam is allowed to leave the jail. Why?

Sam is visiting his lawyer, who had been arrested and jailed.

logicshortwhat am IA word I know,
Six letters it contains,
Subtract just one,
And twelve is what remains.

Dozens.

logicmysteryscary Dodge was staying with Cousin Jamie in Jamie's lakeside cabin. They were setting up Jamie's will. As Dodge was Jamie's closest living relative, much of Jamie's estate was being left to him. One day, Jamie went to Dr Dodge very disturbed. "Doctor," he began, "I have just found out that a man named Georgio wants to get me. He will be here very soon. Where will I go? Where can I hide? If he finds me in here, he will surely kill me. I do not have time to leave this clearing and go farther into the woods."
Dr Dodge thought for a moment, and then grabbed a 5' long bamboo pole, with a diameter the size of a quarter. "Jamie, follow me out to the lake. This lake is 4' deep. If you lie on the bottom of the lake and breathe through this pole, Georgio will never find you. I will be in the bulrushes with a shotgun, and I will shoot him when he comes. I will swim down to find you when he is gone." Jamie consented, and lay down on the bottom of the lake with the bamboo pole in his mouth. A few hours later, a ranger passed by. He found Jamie's body, dead. Dr Dodge told the police of the circumstance, and that Jamie had probably panicked, and died. Police arrested Dr Dodge, on the charges of murdering Jamie. Why?
The bamboo pole did not have any cracks or holes. Its opening was above water the whole time.

Jamie died of carbon dioxide poisoning. The pole was 5' long, but only the size of a quarter. The first time he breathed in, he breathed oxygen. When he exhaled, the air could not travel 5' before he breathed in again. He was just breathing what he exhaled. Before long, all he was breathing was carbon dioxide. He died of CO2 poisoning. Doctor Doge was the one who told him to use the pole, therefore the cause of his death. Dodge is a DOCTOR, and therefore knows about the CO2. Dodge did murder Jamie. His motive: the money in the will.

animalfunnylogicshortIf 20 blackbirds are on a fence and you shoot one, how many remain?

None, they would all fly away from the sound of the shot.

logicshortYou have to look at me to say what I show you. I offer two different purposes but I am spelled the same and I am pronounced the same.

WATCH and WATCH (Watch the watch).