Long clean riddles

logicclean

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.
73.10 %
93 votes
cleanfunnytricky

A boy was at a carnival and went to a booth where a man said to the boy, "If I write your exact weight on this piece of paper then you have to give me $50, but if I cannot, I will pay you $50." The boy looked around and saw no scale so he agrees, thinking no matter what the carny writes he'll just say he weighs more or less. In the end the boy ended up paying the man $50. How did the man win the bet?
The man did exactly as he said he would and wrote "your exact weight" on the paper.
73.07 %
148 votes
simplelogicmathcleverclean

An infinite number of mathematicians are standing behind a bar. The first asks the barman for half a pint of beer, the second for a quarter pint, the third an eighth, and so on. How many pints of beer will the barman need to fulfill all mathematicians' wishes?
Just one.
73.06 %
110 votes
logiccleansimple

Shadow drove into the Speedy Service Station and pulled up to the pumps. "Fill it up, please," said Shadow. " This may sound strange," said the owner, "but I'd rather fill two cars from out of town than one car from this town." Shadow looked across the small town and replied, "I know just what you mean." Why would the owner feel this way?
The owner would rather fill two cars from anywhere than one car from town because he would make twice the amount of money.
73.02 %
53 votes
cleanpoemssimple

Die without me, Never thank me. Walk right through me, Never feel me. Always watching, Never speaking. Always lurking, Never seen.
Air.
72.93 %
217 votes
cleansimplelogic

I heard of an invading, vanquishing army sweeping across the land, liquid-quick; conquering everything, quelling resistance. With it came darkness, dimming the light. Humans hid in their houses, while outside spears pierced, shattering stones walls. Uncountable soldiers smashed into the ground, but each elicited life as he died. When the army had vanished, advancing northward, the land was gree and growing, refresh.
Rainstorm.
72.80 %
79 votes
cleansimple

We travelled the sea far and wide. At one time, two of my sailors were standing on opposite sides of the ship. One was looking west and the other one east. And at the same time, they could see each other clearly. How can that be possible?
The sailors had their backs against either ends of the ship.
72.80 %
48 votes