Logic riddles

cleanlogic

Three playing cards in a row. Can you name them with these clues? There is a two to the right of a king. A diamond will be found to the left of a spade. An ace is to the left of a heart. A heart is to the left of a spade. Now, identify all three cards.
Ace of Diamonds, King of Hearts, Two of Spades.
80.14 %
48 votes
logicsimplecleanclever

You have two jugs, one that holds exactly 3 gallons, and one that holds exactly 5 gallons. Using just these two jugs and a fire hose, how can you measure out exactly 4 gallons of water?
Fill the 5-gallon jug to the top, and then pour it into the 3-gallon jug until the 3-gallon jug is full. You now have 2 gallons remaining in the 5-gallon jug. Pour out the 3-gallon jug, and then pour the 2 gallons from the 5-gallon jug into the 3-gallon jug. Finally, fill the 5-gallon jug to the top and pour it into the 3-gallon jug until it's full. Since there was only space left for 1 more gallon in the 3-gallon jug, you now have exactly 4 gallons in the 5-gallon jug.
80.14 %
48 votes
logicstorytricky

Two Japanese people who have never seen each other meet at the New York Japanese Embassy. They decide to have drinks together at a nearby bar. One of them is the father of the other one's son. How is this possible?
The Japanese are husband and wife and both blind since birth.
80.14 %
48 votes
logiccleansimpleclever

A man needs to send important documents to his friend across the country. He buys a suitcase to put the documents in, but he has a problem: the mail system in his country is very corrupt, and he knows that if he doesn't lock the suitcase, it will be opened by the post office and his documents will be stolen before they reach his friend. There are lock stores across the country that sell locks with keys. The only problem is that if he locks the suitcase, he has no way to send the key to his friend so that the friend will be able to open the lock: if he doesn't send the key, then the friend can't open the lock, and if he puts the key in the suitcase, then the friend won't be able to get to the key. The suitcase is designed so that any number of locks can be put on it, but the man figures that putting more than one lock on the suitcase will only compound the problem. After a few days, however, he figures out how to safely send the documents. He calls his friend who he's sending the documents to and explains the plan. What is the man's plan?
The plan is this: 1. The man will put a lock on the suitcase, keep the key, and send the suitcase to his friend. 2. The friend will then put his own lock on the suitcase as well, keep the key to that lock, and send the suitcase back to the man. 3. The man will use his key to remove his lock from the suitcase, and send it back to the friend. 4. The friend will remove his own lock from the suitcase and get to the documents. Search: Man-in-the-middle attack
80.14 %
48 votes
logiccleanclevermath

At a dinner party, many of the guests exchange greetings by shaking hands with each other while they wait for the host to finish cooking. After all this handshaking, the host, who didn't take part in or see any of the handshaking, gets everybody's attention and says: "I know for a fact that at least two people at this party shook the same number of other people's hands." How could the host know this? Note that nobody shakes his or her own hand.
Assume there are N people at the party. Note that the least number of people that someone could shake hands with is 0, and the most someone could shake hands with is N-1 (which would mean that they shook hands with every other person). Now, if everyone at the party really were to have shaken hands with a different number of people, then that means somone must have shaken hands with 0 people, someone must have shaken hands with 1 person, and so on, all the way up to someone who must have shaken hands with N-1 people. This is the only possible scenario, since there are N people at the party and N different numbers of possible people to shake hands with (all the numbers between 0 and N-1 inclusive). But this situation isn't possible, because there can't be both a person who shook hands with 0 people (call him Person 0) and a person who shook hands with N-1 people (call him Person N-1). This is because Person 0 shook hands with nobody (and thus didn't shake hands with Person N-1), but Person N-1 shook hands with everybody (and thus did shake hands with Person 0). This is clearly a contradiction, and thus two of the people at the party must have shaken hands with the same number of people. Pretend there were only 2 guests at the party. Then try 3, and 4, and so on. This should help you think about the problem. Search: Pigeonhole principle
80.14 %
48 votes
interviewlogicmath

Four people need to cross a rickety bridge at night. Unfortunately, they have only one torch and the bridge is too dangerous to cross without one. The bridge is only strong enough to support two people at a time. Not all people take the same time to cross the bridge. Times for each person: 1 min, 2 mins, 7 mins and 10 mins. What is the shortest time needed for all four of them to cross the bridge?
It is 17 mins. 1 and 2 go first, then 1 comes back. Then 7 and 10 go and 2 comes back. Then 1 and 2 go again, it makes a total of 17 minutes.
80.14 %
48 votes