Solution:

There are 7 people.

Solution:

The answer is very simple. All she had to do is take the fifteen cards from the top and reverse them. This would make another pile out of that and there will be two piles - one of 15 cards and one of 37 cards. Also both of them will have the same number of inverted cards.

Just think about it and if the mathematical explanation will help you understand better, here it is.
Assume that there were p inverted cards initially in the top 15 cards. Then the remaining 37 cards will hold 15-p inverted cards.
Now when she reverses the 15 cards on the top, the number of inverted cards will become 15-p and thus the number of inverted cards in both of the piles will become same.

Solution:

23rd day

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21 + 22 + 23 = 276

Thus our anniversary falls on the 23rd day of the month.

You can apply other formulas to shorten the process but that is the simplest way to do it.

Solution:

1

Explanation:
4 + 9 = 1
The trick to solve is to assume the first number as the time in AM and the second number is the number of hours you have to add to it. You'll find the time in PM when you do that.

For example
7 AM + 7 hours = 14:00 or 2 PM

4 AM + 9 hours = 13:00 or 1 PM.

Solution:

This puzzle has no answer if you are considering this figure in a 2d domain. However if you follow it on a three-dimensional plane, you can do it like shown in the figure given.

Solution:

The driver is actually the wife and mother.

Solution:

73

Explanation:
It's a very famous puzzle known as "Josephus Problem Puzzle". It can be solved in various ways but the trick to solve this problem is illustrated below.

Binay number of 100 is: 1100100
complement pf 100 is 0011011, whose decimal equivalent is 27.
The desired answer will be 100 - 27 = 73

Solution:

The three errors are as follows:
a) fynd is spelled wrong. The correct spelling is 'find'.
b) that is not appropriate to use here. It should be 'this'.
c) There are only two errors in the question and that is the third error because the questions asks us to find three.

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:

16

Explanation:
9 [1x1] squares
5 [2x2] squares
1 [4x4] square
1 [4x4] square

9 + 5 + 1 + 1 = 16 squares.

Solution:

7, 4, 6, 5, 8