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
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.
Multiplication of the 1st & 2nd numbers, 5*3 = 15; 9*2 = 18…thusly, 7*2 = 14
Multiplication of the 1st & 3rd numbers, 5*2 = 10; 9*4 = 36…thusly, 7*5 = 35;
Multiplication of the 1st & the sum of the 2nd & 3rd numbers. The generated result is reduced by the value of the 2nd number, …thusly, 7*(2+5) = 49 - 2 = 47