Solution:

W02

F21: Friday the 21st.
S23: Sunday the 23rd.
T25: Tuesday the 25th.
T27: Thursday the 27th.
S29: Saturday the 29th.
M31: Monday the 31st.

Avoiding every other day, the next term must be 'Wednesday the 2nd' i.e. 'W02'

Solution:

He can ask any one of them. The question he will ask is - "Which door will the other guy say is the correct one?"

If he asks this question to the truth teller, he will get the incorrect door as he will honestly tell him what the liar will say. If he asks the liar, he will get the incorrect door again as he will lie about what the truth teller will tell him (the truth teller will obviously tell him the right door). Thus, by asking any of them, Sheldon Cooper will get the incorrect door only and he can choose the other one to find the treasure.

Solution:

Sherlock will ask any one of them "Are you the truth teller?" There can be two responses to this question - Yes or No.

If the person says yes, then he is the truth teller or a lying toggle.
He will ask the second question to the same person "Who is the truth teller?"
If he is the truth teller, he will admit it. If he is a lying toggler, he will speak truth on the second question if he lied on the first, telling you who the truth teller is.

If the person says no, Sherlock will be confirmed of the fact that he is not the truth teller and thus he is a toggle who has told truth and will say a lie on the second question.
The next question he will ask to the same person "Who is not the truth teller?"
Since the person told truth in the first question, he will speak a lie this time and in order to tell a lie, he will tell Sherlock exactly who the truth teller is.

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:

252 ways.

Think how symmetric the problem is. Pick up any random cell. Suppose that you took 'n' steps in reaching the particular cell above it and 'm' steps to reach the particular cell placed left to it. In such circumstances, the number of ways to reach to the random cell is equal to the sum of two steps 'n+m'. Thus, keep that logic in mind and fill out the six by six grid. In that manner, there are 252 ways to get to the destination that is the bottom right step.

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:

10 Bulbs

For this tricky puzzle, you must check how many light bulbs in the row are having an odd number of factors. The first one surely has odd number of factors, the second has even, four has odd. Thus the bulb four and one will remain lit. The bulbs that are going to remain lit are perfect squares as they have an odd number of factors - 1, 4, 9, 16.

Since there are 100 passes, you can go up to 10 times 10 i.e. the square of 10. There are 10 perfect squares available to you - one, two, three, four, five, six, seven, eight, nine and ten. They corresponds to the bulb number 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100. All of them will remain lit and thus total ten bulbs will remain lit after 100 passes.

Solution:

90

Split the cube into three by three grids. You will get three layers. Now get to one square from the cell adjacent instead of counting the path. You can now count the paths from the left (x), the right (y) and the level above (z). So the path for any square will be x + y + z.

Solution:

Keep one coin from the first set and place it on the scale along with the two from the second set etc. If the weight is off by one hundredth of an ounce, then you will know that it is the first set that is faulty and if the weight is off by two hundred of an ounce, then the second set is faulty and so on.