Best difficult 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.
MessengerWhatsapp
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93 votes
cleanpoemssimple

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

Even though the odds are always in favor of the gambling house, why does the establishment insist on a house limit on stakes?
Every casino in the world would go bankrupt without a house limit on stakes. Without it, gamblers would keep doubling their stakes until they won. No matter how bad a losing streak they were on, they would eventually win. For more information, search: Martingale
MessengerWhatsapp
72.84 %
66 votes
logictrickycleversimple

When Manish was three years old he carved a nail into his favorite tree to mark his height. Six years later at age nine, Manish returned to see how much higher the nail was. If the tree grew by five centimeters each year, how much higher would the nail be.
The nail would be at the same height since trees grow at their tops.
MessengerWhatsapp
72.70 %
83 votes
cleanlogicstory

100 men are in a room, each wearing either a white or black hat. Nobody knows the color of his own hat, although everyone can see everyone else's hat. The men are not allowed to communicate with each other at all (and thus nobody will ever be able to figure out the color of his own hat). The men need to line up against the wall such that all the men with black hats are next to each other, and all the men with white hats are next to each other. How can they do this without communicating? You can assume they came up with a shared strategy before coming into the room.
The men go to stand agains the wall one at a time. If a man goes to stand against the wall and all of the men already against the wall have the same color hat, then he just goes and stands at either end of the line. However, if a man goes to stand against the wall and there are men with both black and white hats already against the wall, he goes and stands between the two men with different colored hats. This will maintain the state that the line contains men with one colored hats on one side, and men with the other colored hats on the other side, and when the last man goes and stands against the wall, we'll still have the desired outcome.
MessengerWhatsapp
72.70 %
121 votes
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