Logic riddles

cleanstorylogicclever

After recent events, Question Mark is annoyed with his brother, Skid Mark. Skid thought it would be funny to hide Question's wallet. He told Question that he would get it back if he finds it. So, first off, Skid laid five colored keys in a row. One of them is a key to a room where Skid is hiding Question's wallet. Using the clues, can you determine the order of the keys and which is the right key? Red: This key is somewhere to the left of the key to the door. Blue: This key is not at one of the ends. Green: This key is three spaces away from the key to the door (2 between). Yellow: This key is next to the key to the door. Orange: This key is in the middle.
The order (from left to right) is Green, Red, Orange, Blue, Yellow. The blue key is the key to the door.
73.60 %
86 votes
logicsimplecleanstory

A king has 100 identical servants, each with a different rank between 1 and 100. At the end of each day, each servant comes into the king's quarters, one-by-one, in a random order, and announces his rank to let the king know that he is done working for the day. For example, servant 14 comes in and says "Servant 14, reporting in." One day, the king's aide comes in and tells the king that one of the servants is missing, though he isn't sure which one. Before the other servants begin reporting in for the night, the king asks for a piece of paper to write on to help him figure out which servant is missing. Unfortunately, all that's available is a very small piece that can only hold one number at a time. The king is free to erase what he writes and write something new as many times as he likes, but he can only have one number written down at a time. The king's memory is bad and he won't be able to remember all the exact numbers as the servants report in, so he must use the paper to help him. How can he use the paper such that once the final servant has reported in, he'll know exactly which servant is missing?
When the first servant comes in, the king should write down his number. For each other servant that reports in, the king should add that servant's number to the current number written on the paper, and then write this new number on the paper. Once the final servant has reported in, the number on the paper should equal (1 + 2 + 3 + ... + 99 + 100) - MissingServantsNumber Since (1 + 2 + 3 + ... + 99 + 100) = 5050, we can rephrase this to say that the number on the paper should equal 5050 - MissingServantsNumber So to figure out the missing servant's number, the king simply needs to subtract the number written on his paper from 5050: MissingServantsNumber = 5050 - NumberWrittenOnThePaper
73.58 %
68 votes
trickylogic

A woman with no driver license goes the wrong way on a one-way street and turns left at a corner with a no left turn sign. A policeman sees her but does nothing... Why?
She is walking.
73.54 %
99 votes
logicmathcleanclever

Using only and all the numbers 3, 3, 7, 7, along with the arithmetic operations +,-,*, and /, can you come up with a calculation that gives the number 24? No decimal points allowed. [For example, to get the number 14, we could do 3 * (7 - (7 / 3))]
7 * ((3 / 7) + 3) = 24
73.48 %
54 votes
cleanlogicclever

You have two lengths of rope. Each rope has the property that if you light it on fire at one end, it will take exactly 60 minutes to burn to the other end. Note that the ropes will not burn at a consistent speed the entire time (for example, it's possible that the first 90% of a rope will burn in 1 minute, and the last 10% will take the additional 59 minutes to burn). Given these two ropes and a matchbook, can you find a way to measure out exactly 45 minutes?
The key observation here is that if you light a rope from both ends at the same time, it will burn in 1/2 the time it would have burned in if you had lit it on just one end. Using this insight, you would light both ends of one rope, and one end of the other rope, all at the same time. The rope you lit at both ends will finish burning in 30 minutes. Once this happens, light the second end of the second rope. It will burn for another 15 minutes (since it would have burned for 30 more minutes without lighting the second end), completing the 45 minutes.
73.48 %
90 votes
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.
73.48 %
90 votes
logictricky

If you had one match and entered a room in which there were a kerosene lamp, an oil burner, and a wood burning stove, which would you light first?
The match.
73.43 %
81 votes