Solution:

9

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:

Playing chess

Game needs two players so Brother-8 is playing chess with Brother-2

Solution:

The two numbers are 4 and 13 whose product is 52 and sum is 17.

Solution:

Since the knife was in her hand which implies that she was planning to kill her husband "Inspector Montalbano".

Solution:

At each draw, the number of strawberry candies are either decreasing by 2 or not decreasing at all. In the case of orange candies, at each draw, they are either increasing by 1 or decreasing by 1.

Thus on an assumed outset with at least one candy in the jar to begin with, if the number of strawberry candies are 0 or are even in numbers, they will finish off leaving an orange candy at the end. If otherwise, the remaining candy will be a strawberry one.

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:

It is best if the person sits on the back of the train. After the stop, the train will be accelerating as it goes into the tunnel. Thus the train will be much master when the back of the train enters the tunnel. In this way, he will have to spend much less time in the tunnel.

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:

You must put labels on the tablets as A and B before using. In that case, if you pour tablets together, you will get 3A, 2A 1B, 1A 2B or 3B. If they are from the same bottles you can take one from another bottle and save the remaining two for another day. If you get two from same and one from other, you can draw one from another bottle and you will have two pairs of which you can eat one and save the other.

Solution:

It will take the man 58 minutes to reach the top.

Since he is climbing 3 feet and falling 2 feet back every passing minute, he is actually climbing 1 feet every minute.
Therefore, he would have climbed 57 feet in 57 minutes. The remaining distance is 3 feet. He will take one more minute to climb it and reach the surface. Since he has reached the top surface, he won't slip back.