If a shopkeeper can only place the weights on one side of the common balance. For example, if he has weights 1 and 3 then he can measure 1, 3 and 4 only. Now the question is how many minimum weights and names of the weights you will need to measure all weights from 1 to 1000? This is a fairly simple problem and very easy to prove also.
There are people and strange monkeys on this island, and you can not tell who is who (Edit: until you understand what they said - see below). They speak either only the truth or only lies.
Who are the following two guys?
A: B is a lying monkey. I am human.
B: A is telling the truth.
There are two arch enemies Messi and Ronaldo who hate each other to an extreme. One day both were going together and a Jeanie appeared in front of them. Jeanie grants 3 wishes to Ronaldo and one to Messi.
Messi replied smartly 'Give me twice whatever Ronaldo demands'.
Ronaldo asked his 1st wish 'Give me 10000 billion dollars. Soon Messi gets 2000 billion dollars.
Ronaldo asked for his 2nd wish 'Give me one mansion in every country in the world. Soon Messi gets two mansions in every country of the world?
It's pretty hard to give up.
If you remove a part of it, you will be left with a bit.
Even if you remove another part, the bit still remains.
Remove one more and it still remains.
See the given image carefully. What you have to do is move the blue checkers in the position of the black checkers and vice versa. You are only allowed to move the checker to an adjacent empty space. Do it in the least possible moves.
A mules travels the same distance daily.
I noticed that two of his legs travels 10km and the remaining two travels 12km.
Obviously two mules legs cannot be a 2km ahead of the other 2.
The mules is perfectly normal. So how come this be true ?
