There are 100 doors. 100 strangers have been gathered in the adjacent room. The first one goes and opens every door. The second one goes and shuts down all the even numbered doors – second, fourth, sixth... and so on. The third one goes and reverses the current position of every third door (third, sixth, ninth… and so on.) i.e. if the door is open, he shuts it and if the door is shut, he switches opens it. All the 100 strangers progresses in the similar fashion.
After the last person has done what he wanted, which doors will be left open and which ones will be shut at the end?
Think deeply about the door number 56, people will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1, the 1st person will open the door; pass 2, 2nd one will close it; pass 4, open; pass 7, close; pass 8, open; pass 14, close; pass 28, open; pass 56, close.
Thus we can say that the door will just end up back in its original state for each pair of divisor. But what about the cases in which the pair of divisor has analogous number for example door number 16? 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is recurrent because 16 is a perfect square, so you will only visit door number 16, on pass 1, 2, 4, 8 and 16… leaving it open at the end. So only perfect square doors will remain open at the end.
A man desired to get into his work building, however he had forgotten his code.
However, he did recollect five pieces of information
-> Sum of 5th number and 3rd number is 14.
-> Difference of 4th and 2nd number is 1.
-> The 1st number is one less than twice the 2nd number.
->The 2nd number and the 3rd number equals 10.
->The sum of all digits is 30.
A wicked sorcerer felt enmity towards elf and thus he chooses four among the rest of the elf's and concealed them. The elves are concealed in the ground in a manner that apart from their head the rest of their body was underneath the ground. The elf's are unable to move their body and can only see in that direction that they are facing. All the elf's are concealed underground in such a way that they form a straight line and among those four elf's that are concealed underground one of the elf is detached form the other three elf's via wall. The entire elf's are in the same direction. The elf that is the furthest can only see the heads of its friends in front and a wall. The elf that is second last can only see one head of his friend and a wall. The second elf can only view the wall. The elf can see nothing.
The sorcerer understands the situation and tells the elf's that he has placed hats over their heads. Among the hats places two hat are blue and the other two are red. Among all the four elfs one of the elf has to guess that which colour hat is he wearing. If the elf answers correctly then he shall be set free or else he will have to dig beneath the ground till the very last.
f the last elf is taking some time to answer the question then it shall mean that the elf's before him are all wearing distinct coloured hats. However sufficient time shall be given to the last elf to give the answer.
If he views the similar coloured hats in front of him , he shall quickly tell the answer
In front of you, there are 9 coins. They all look absolutely identical, but one of the coins is fake. However, you know that the fake coin is lighter than the rest, and in front of you is a balance scale. What is the least number of weightings you can use to find the counterfeit coin?
The answer is 2. First, divide the coins into 3 equal piles. Place a pile on each side of the scale, leaving the remaining pile of 3 coins off the scale. If the scale does not tip, you know that the 6 coins on the scale are legitimate, and the counterfeit is in the pile in front of you. If the scale does tip, you know the counterfeit is in the pile on the side of the scale that raised up. Either way, put the 6 legitimate coins aside. Having only 3 coins left, put a coin on each side of the scale, leaving the third in front of you. The same process of elimination will find the counterfeit coin.
Imagine a picture in your mind. There is a rose plant. A few roses have grown up in the plant. There are a few bees hovering over them.
Now read the given statements carefully:
1) If every one of the bees lands on a rose, one of them won't get a rose.
2) If two bees share rise together, then there will be one rose left without any bee.
Can you find out the number of roses on the plant and the number of bees?
Suppose there is a Christmas tree and four angels are sitting on it amidst the other ornaments. Two of them have black halos and two of them have white halos. No body among them can see above their head. Angel A is sitting on the top branch and can see angels B and C sitting below him. B can see C who is sitting in a branch lower than his. Angel D is at the base of the tree and can't be seen due to the branches in between. Also, he can't see anybody as well.
If they are asked to guess the color of their own halo (they dont know that), who do you think will be able to deduce and speak up first with a right answer?
Now there can be two solutions to the given situation because there can be two situations:
Suppose if B and C have same colors, A will know that his color is the other one and he will be able to speak up the first.
Now if B and C do not have same colors, A will stay silent. This will tell B that his and C's colors are different. Thus he will speak up first.
So either A or B will speak up first depending on the situation.