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.
You have two cubes with plane faces. You have to mark both the cubes with numbers in a manner that all the days of a month can be portrayed using those cubes. Also, note that you have to use both the cubes for displaying any date. Suppose if you have to display the 5th day of the month, you will have to display it as 05.
This one is a bit of tricky river crossing puzzle than you might have solved till now. We have a whole family out on a picnic on one side of the river. The family includes Mother and Father, two sons, two daughters, a maid and a dog. The bridge broke down and all they have is a boat that can take them towards the other side of the river. But there is a condition with the boat. It can hold just two persons at one time (count the dog as one person).
No it does not limit to that and there are other complications. The dog can’t be left without the maid or it will bite the family members. The father can’t be left with daughters without the mother and in the same manner, the mother can’t be left alone with the sons without the father. Also an adult is needed to drive the boat and it can’t drive by itself.
How will all of them reach the other side of the river?
Rohit is on his way to visit your Grandma, who lives at the end of the state.It's her birthday, and he want to give her the cakes that he has made.Between his place and her grandma house, he need to cross 7 toll bridges.
Before you can cross the toll bridge, you need to give them half of the cakes you are carrying, but as they are kind trolls, they each give you back a single cake.
How many cakes do rohit have to carry with him so he can reach his grandma home with exactly 2 cakes?
The husband had booked two tickets for departure and only one for the return. The travel agent told this information to the police and they immediately came to know that he had planned the murder right from the beginning.
Two companions Terry and Garry were talking about their families. Terry told some great stories about his courageous grandfather who fought for Britain in "World War I". Terry told that his grandfather is so brave that he was awarded a bravery honour medal with words "For our Courageous Soldiers In World War I" embedded into it.
Step1. Lets say y = x
Step2. Multiply through by x xy = x2
Step3. Subtract y2 from each side xy - y2 = x2 - y2
Step4. Factor each side y(x-y) = (x+y)(x-y)
Step5. Divide both sides by (x-y) y = x+y
Step6. Divide both sides by y y/y = x/y + y/y
Step7. And so... 1 = x/y + 1
Step8. Since x=y, x/y = 1 1 = 1 + 1
Step9. And so... 1 = 2
Step 5 is invalid, because we are dividing by (x-y), and since x=y, we are thus dividing by 0. This is an invalid mathematical operation (division by 0), and so by not following basic mathematical rules
A confectionary shop owner allows children to purchase a chocolate in exchange of five wrappers of the same chocolate. Children from the locality consumed 77 chocolates in a month. Now, they all collected them together and decide to buy back chocolates.
How many chocolates do you think they can buy using those 77 wrappers ?
The children can purchase 19 chocolates in return.
Out of 77 wrappers, 75 will be used to buy 15 chocolates and two will be left spare.
The 15 chocolates will create 15 empty wrappers that can be exchanged to get three chocolates.
Three chocolates will return three wrappers which will help them buy another chocolate.
Now the wrapper from this chocolate and the two spare that were left earlier will get them another chocolate.
15 + 3 + 1 = 19
One man was moving to and fro in worry when a bright woman noticed him. On asking he told her that while replacing his tire, he incidentally dropped all the four nuts into a deep drain. She told him what to do and he was then able to drive back successfully till his office.