Solution:

1 (shown in the diagram below)

Explanation:
At every step number of joint goes down by 1.
i.e
8 have 6 joints
9 have 5 joints
5 have 4 joints
3 have 3 joints
7 have 2 joints
1 has 1 joint... the answer.

Solution:

The man asked the king that he wanted to die of old age.

Solution:

3,3,8

Lets break it down. The product of their ages is 72. So what are the possible choices?

2, 2, 18 sum(2, 2, 18) = 22
2, 4, 9 sum(2, 4, 9) = 15
2, 6, 6 sum(2, 6, 6) = 14
2, 3, 12 sum(2, 3, 12) = 17
3, 4, 6 sum(3, 4, 6) = 13
3, 3, 8 sum(3, 3, 8 ) = 14
1, 8, 9 sum(1,8,9) = 18
1, 3, 24 sum(1, 3, 24) = 28
1, 4, 18 sum(1, 4, 18) = 23
1, 2, 36 sum(1, 2, 36) = 39
1, 6, 12 sum(1, 6, 12) = 19

The sum of their ages is the same as your birth date. That could be anything from 1 to 31 but the fact that Torres was unable to find out the ages, it means there are two or more combinations with the same sum. From the choices above, only two of them are possible now.

2, 6, 6 sum(2, 6, 6) = 14
3, 3, 8 sum(3, 3, 8 ) = 14

Since the eldest kid is taking piano lessons, we can eliminate combination 1 since there are two eldest ones. The answer is 3, 3 and 8.

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:

The person on 73rd number will survive.

Explanation:
Till the number is the power of 2, the last person to survive will be the one who started it. But since the number here is not the power of 2, we will take the greatest power of 2 that is less than the number which is 64.
100 - 64 = 36
36 people are killed as 2, 4, 6, ..., 72. Thus the sword will now be given to the 73rd person. Now he is the first person to start in the remaining 64 people. Thus he will be the one to survive.

Solution:

Costa

Explanation:
c + 1 characters-> d
o + 2 characters-> q
s + 3 characters-> v
t + 4 characters-> x
a + 5 characters-> f

=> dqvxf = costa

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:

In the picture you can find the green matchsticks. We have moved them from the bottom and placed them in the middle. It was a bit tricky but now you have your desired result.

Solution:

This statement is true but only in a specific case i.e. if both of the siblings continue to live till the day the younger is as old as the elder one was during the time of their birth.

Solution:

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

Solution:

It can be solved as below.