Solution:

South

The series: south, east, north, south, west, east spirals clockwise from the top left-hand corner.

Solution:

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.

Solution:

Step one: First you will have to make 3 trips of 1,000 apples 333 miles. Then you will be left with 2,001 apples and 667 miles to go.

Step two: Next you will have to take 2 trips of 1,000 apples 500 miles. In this manner you will be left with 1,000 apples and 167 miles to go (you have to leave an apple behind).

Step three: At last, you will have to travel the last 167 miles with one load of 1,000 apples and you will be left with 833 apples when you reach the town

Solution:

For this answer is 3^0, 3^1, 3^2... That is 1,3,9,27,81,243 and 729.

Solution:

If you are thinking to hold one chocolate from each box in hand and then balancing weight in bare hands, you are thinking all wrong.

Let us begin by labelling boxes as 1, 2, 3 and so on till 10.

Now pick one chocolate from box 1, two chocolates from box 2, three from box 3 and so on. In total, you will have 55 chocolates now. (1 + 2 + 3 + ..... + 10)

The ideal weight of the chocolates should be 55 * 20 = 1100. However, somewhere in there are the defected chocolate/s.

You can judge that clearly by noting down the result of 1100 - total weight of chocolates. If the weight is less than 1 gram, the defected box is box 1, if the weight is less than 2 grams, the defected box is box 2 and so on.

Solution:

If you see the picture, they formed a pentagonal shape and achieved the required.

Solution:

On Your lap.

Solution:

You will ask, I have a number in my mind that i.e. either 2 or 3. Is the number that I have in my mind smaller or equal to the number that you are thinking of?

If Alan replies with no, he is thinking of the number 1.
If Alan replies with yes, he is thinking of the number 3.
If Alan replies with don't know, he is thinking of the number 2.

Solution:

20
Just count the nines in the numbers: 6, 16, 26, 36, 46, 56, 76, 86, 96, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69.
Note: 66 contains TWO nines

Solution:

Hockey Stick $49.50 & ball $0.50

Solution:

Put one black rock in one sack and put all the rest in the other one. In such manner you will get about 3/4 probability of surviving.