Solution:

Let us first assume that the number of drops required are N.

If the egg breaks at maximum number of tries, we will have N - 1 drops till it does not break. Thus we must drop the first egg from the height N. Now if the first drop of the first egg does not break the egg, we can have N - 2 drops for the second egg if the first egg breaks in the second drop.

Let us put that into a valid example for better understanding. Suppose we need 16 drops. Now let us drop the egg from the sixteenth floor, if it breaks, we will try all the floors below sixteen. Suppose it does not breaks, then we have 15 drops left and we will drop it from 16+15+1 = 32nd floor. This is because if it breaks at 32nd floor, we can try all the way down to 17th floor in fourteen tries making the total tries to be 16.

Let us assume that 16 is the correct answer

1 + 15 16 if breaks at 16 checks from 1 to 15 in 15 drops
1 + 14 31 if breaks at 31 checks from 17 to 30 in 14 drops
1 + 13 45.....
1 + 12 58
1 + 11 70
1 + 10 81
1 + 9 91
1 + 8 100 we can easily do in the end as we have enough drops to complete the task.

Seeking the above case, we must note that we have achieved it in 14 or 14 drops but we still need to find out the optimal one. We know from above that the optimal one will require 0 linear trials in the last step.

Thus we can say that

(1+p) + (1+ (p-1)) + (1+ (p-2)) + .........+ (1+0) >= 100.

Let 1+p=q which is the answer we are looking for

q (q+1)/2 >=100

q=14

Thus the optimal number of drops required are 14.

Drops will be tried at floor number 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100. How? Just move up 14, 13, 12 .... Floor till it breaks or does not.

Solution:

You may solve this question with different methods but we are only giving away one.

What you have to do in that method is look for a number x to which all the numbers from 2 to 10 divide evenly. Multiplying the numbers together (2*3*4*5*6*7*8*9*10) will give you a big number but we have to find a smaller possibility. What we can do is find the prime factors, a subset of which we can use to form a number from 2 to 10.

2*2*2*3*3*5*7 will certainly give us what we need. The product is 2520 and it is the lowest possible number which can be evenly divided by 2-10.

Now if we add 1 to the number, the result will be 2521 which will satisfy all the conditions from 2 to 10. But will not hold true in case of 11.

If you calculate, you will see that you can multiply 2520 by any integer and add 1 to it to satisfy all the conditions except the divisibility by 11. What we need is an integer Y that can be multiplied by 2520 and if we add 1 to the product, it becomes divisible by 11.

2520/11 leaves a remainder 1. Now let us keep increasing the number of 2520. Suppose if we divide two 2520 by 11, the remainder then will be 1 + 1 = 2. If we keep going on like that. Ten 2520 when divided by 11 will give 10 as remainder and if we add 1 to them, it will be completely divisible. Also it will satisfy all the conditions from 2-10.

Thus our answer is 25201.

Solution:

We agree that the series is quite difficult to crack and much tricky as well.

1 can be written as one (consist of 3 letters)
2 can be written as two (consist of 3 letters)
3 can be written as five (consist of 5 letters)
4 can be written as four (consist of 4 letters)
5 can be written as five (consist of 5 letters)
6 can be written as six (consist of 3 letters)

Thus the next term in the series is 3.

Solution:

The human in front will count the number of green hats and if the count is an odd number, he will say green and if the number is even, he will say pink.

The next human behind him definitely saw either an odd or an even number of hats. If the number is still odd, then he has a pink hat. That human says Pink and survives. Thereafter, there will be an odd number of green hats forward and the third person can decide whether he is wearing a green hat or a pink hat.

If the second person sees an even number of green hats, then he must have a green hat as well and will be freed. In this fashion, the humans will just have to keep the track if the number of green hats in front of them are odd or even in number and then they can spontaneously conclude with the color of their hats.

Since the first human will be guessing, he will have a 50/50 chance of surviving but the other 9 people will definitely be saved.

Solution:

Since there are 100 logicians, the lights are turned on and off hundred times and after the 100th time, all the logicians leave the room.

If all of them sees 99 black foreheads, the lights get turned off. When the light is turned on again, everyone is able to see 99 black foreheads again. This happens 100 times and then every logician leaves the room.

Let us make it more simplified for you. Assume that only one person has a black forehead. The people who invited them said that at least one person has a black forehead and as they turn off the lights, that logician leaves.

Imagine the scenario with two logicians having black forehead. One of the logician sitting there must be thinking "I have a black forehead or I don't have a black forehead. If I am not the one with black forehead, the other logician with black forehead will deduce that he is having a black forehead and then he will leave."

"If I don't have a black forehead, the other logician will stay and thus I must be having a black forehead as well. Thus we must leave when the lights are turned off for the second time"

When you repeat the logic to 100 times, you will get the answer.

Solution:

This particular answer depends on what the third person says.

You know that the other two logicians have 20 and 30 painted on their foreheads. If the first logician is 10, the second will either say he has either 40 or 20 painted on his forehead. The second logician will say "I'm either 30 or 10 but I can't be 10 because every number is unique as explained earlier. If the third logician is able to deduce that he has 30 painted on his forehead, but the third logician is not able to deduce the number. Thus the first logician is not numbered 10 and he must be painted 50 on his forehead.

Solution:

They select one of the prisoner named Jason to do the trick and frame out a full proof plan. According to their plan, whenever a prisoner other than Jason is selected, they will follows some simple steps - If the bulb is off, they will switch it on but if the bulb is already lit, they won't do anything and sit idle.

If Jason is selected and he finds out that the bulb is lit, he will add one to his count and will switch off the bulb. If the bulb is already switched off, he will sit there idly. In such a manner when his count reaches to 99, he will tell the warden that every prisoner has now been to the room.

Let us craft out the solution in simpler terms. Whenever a prisoner goes inside the room, he simply switches on the bulb if it is off. Thus one prisoner will only light the bulb once. When Jason finds out that the bulb it on, he will know that this has been done by a new prisoner and he will add one to his count. He will keep doing this till his count reaches 99. 99 because, he already has been in the room which makes a total of 100. Thus it is a full proof plan that will definitely set them all free.

Solution:

Everyone will appoint any one person as a leader. Now everyone must follow some steps. Every prisoner will flip the first switch up at the first opportunity and also at the second opportunity. If the first switch is already up or they have flipped it up two times already, then they will flip the second switch. The leader however will have a different routine. He is the only one who will flip the switch down the first switch. Suppose if the switch is already down, he will flip the second switch. Also he will keep a count and when he has flipped the first switch down 44 times, he will declare that every prisoner has visited the room.

This is because once the leader has flipped down the switch 44 times, he will know that everyone has visited the room. If the switch was down initially, the other 22 prisoners will turn it up two times. If the first switch was already up, there will be one prisoner who will flip the switch up only once and the rest will do it twice.

Now this trick won't work when the leader has flipped the switch 23 times. Consider the fact that it is possible that 12 prisoners randomly and the leader have been taken to the switch room 24 times and the rest have not been taken till then. Remember that the first switch might have been up in the beginning itself. Thus the prisoners must flip the switch up twice. If they do it just once, the leader will never know whether to count till 22 or 23.

Solution:

Now this problem is a bit tricky. Before you start panicking about the solution let us give you a hint. You need to think in terms of binary numbers.

There are 1000 bottles which should be labeled from 1 to 1000 first. Now write those numbers in binary format.

Bottle Number 1 = 0000000001
Bottle number 2 = 0000000010
Bottle number 500 = 0111110100
Bottle number 1000 = 1111101000

Now the king will summon ten prisoners and label them from 1 to 10. The prisoner labeled as 1 will take a sip from those bottles that have a 1 in its least significant bit and the prisoner labeled as 10 will take a sip from every bottle that have a 1 in its most significant bit etcetera.

Let us frame an example. The wine from the bottle number 924 will be sipped by 10, 9, 8, 5, 4 and 3. Thus if that bottle will be poisoned, only those 6 prisoners will die.
After a month's time, all the king need to do is line up the prisoners in order and call each living prisoner as 0 and each dead prisoner as 1. N such a manner the number that the king will get will be of that bottle which is poisoned.

Now also note that if there would have been 1024 or more bottles, there would have been more prisoners required for the very same task.

Solution:

This is quite a tricky puzzle and quite hard to deduce at first impression.

What you have to do is write the number in English
For the first case, 1 and 2 produce 9
1 = One
2= Two
There are three characters in both one and two
3x3 = 9

Similarly for the second case, 3 and 4 produce 20
3 = Three
4 = Four
Three has 5 characters and four has 4 characters
5x4 = 20

Now for the question, seventeen has 9 characters and twelve has 6 characters, thus the missing number is the product of 9 and 6 which is 54.