Solution:

So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8.

There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second one possibility is to weigh 9,10,11 against 1,2,3

(1) They balance, in which case you know 12 is the different coin, and you just weigh it against any other to determine whether it is heavy or light.
(2) 9,10,11 is heavy. In this case, you know that the different coin is 9, 10, or 11, and that that coin is heavy. Simply weigh 9 against 10; if they balance, 11 is the heavy coin. If not, the heavier one is the heavy coin.
(3) 9,10,11 is light. Proceed as in the step above, but the coin you're looking for is the light one.

That was the easy part.

What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings.

Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. The direction of the tilts can tell us whwther the offending coin is heavier or lighter.

Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin.

For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.

Solution:

13 numbers

Explanation:
Since we are looking to find odd reminder which implies that the remainder will be either 1 or 3
Therefore number must follow below two conditions:
A) We can list number 5n + 1 or 5n + 3. (where n = 0,1,2,3.....)
B) The number must end in 1 or 3.

List of such numbers less than 100 are:
1, 3, 11, 13, 21, 23, 31, 33, 41, 43, 51, 53, 61, 63, 71, 73, 81, 83, 91, 93

Let's remove the nonprime ones which are 1, 21, 33, 51, 63, 81 and 91

So, I will be left with 13 numbers which fit the bill.
3, 11, 13, 23, 31, 41, 43, 53, 61, 71, 73, 83 and 93

Solution:

Anybody will prefer the shop in the eastern market for sure. There are two reasons for it:
Firstly, since there are only two barbers in the town, the one running the shop in eastern market must have gotten his hair done by the one running the shop in the northern market and vice versa. Thus, if the barber in the eastern market has a shabby haircut, it must be due to the incompatibility.
Secondly, the neat and clean shop can be due to the scarcity of customers as well. While the untidy place refers to the fact that so many customers get their hair done in that shop.

Thus any one will opt for the barber shop in the eastern market.

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:

Tuesday

We know that Christanio Ronaldo tells the truth on only a single day of the week. If the statement on day 1 is untrue, this means that he tells the truth on Monday or Tuesday. If the statement on day 3 is untrue, this means that he tells the truth on Wednesday or Friday. Since Christanio Ronaldo tells the truth on only one day, these statements cannot both be untrue. So, exactly one of these statements must be true, and the statement on day 2 must be untrue.
Assume that the statement on day 1 is true. Then the statement on day 3 must be untrue, from which follows that Christanio Ronaldo tells the truth on Wednesday or Friday. So, day 1 is a Wednesday or a Friday. Therefore, day 2 is a Thursday or a Saturday. However, this would imply that the statement on day 2 is true, which is impossible. From this we can conclude that the statement on day 1 must be untrue.

This means that Christanio Ronaldo told the truth on day 3 and that this day is a Monday or a Tuesday. So day 2 is a Sunday or a Monday. Because the statement on day 2 must be untrue, we can conclude that day 2 is a Monday.

So day 3 is a Tuesday. Therefore, the day on which Christanio Ronaldo tells the truth is Tuesday.

Solution:

Yes it is possible in Roman Numerals. In Roman Numerals,
9 = IX
10 = X
11 = XI

Thus by removing one from 9, you are getting 10 and by removing one from 11, you are getting 10 again.

Solution:

Take 10 number of cards in a new deck and change their face direction. For example- You create a new deck of 10 cards and out of 10, 3 faces up in the new deck. So remaining 7 faces up are in the old deck. But hey! while creating the new deck you reversed the face direction of new cards. So actually the 3 cards which were facing up are actually face down in the new deck while 7 faces up.

Solution:

17 mins

The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 mins. Is that it? No. That would make this question too simple even as a warm up question.

Let’s brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let’s put all this together.

1 and 2 go cross
2 comes back
7 and 10 go across
1 comes back
1 and 2 go across (done)

Total time = 2 + 2 + 10 + 1 + 2 = 17 mins

Solution:

See the colored sticks in the picture. We have moved those sticks in the position given in the second image corresponding with the color.

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 wise man told them to switch the horses and then race. In this manner, both of them will be willing to race as fast as they can to prove that the other horse (his own horse) is slow.