Solution:

11550

Explanation:
There is just one way to find it out and that is the generic computing method.

The number of squares and rectangles formed = (Summation of column numbers)
= (1 + 2 + 3 +......+ 19 + 20) * (1 + 2 + 3 +......+ 9 + 10)
= 210 * 55
= 11550

Solution:

7 Races.

Solution:

mother

We know from (3) that the youngest person was not the victim, from (4) that the youngest person was not the helper and from (6) that the youngest person was not the killer. The youngest person can only have been the witness therefore. If we make up a chart there are now three possible combinations:

Oldest person (father) H H M

Next to oldest (mother) V M H

Next to youngest (son) M V V

Youngest (daughter) W W W

(H = Helper ; V = Victim ; M = Murderer ; W = Witness)

We can work out from (5) that the father was the oldest, from (2) that the youngest person must have been the daughter. Therefore the next to the youngest must have been the son and the next to the oldest, the mother.

Solution:

Karate Kid (car + at + tea + kid)

Solution:

Pages 19, 20, 23, 37, 38, & 42 will also be missing

Solution:

None

The sum of numbers from 0-9(0,1,2,3,4,5,6,7,8,9) is 45 and therefore can be divisible by 3 and 9.

Solution:

8 liters jar: 8; 5 liters jar: 0; 3 liters jar: 0
Fill 5 liters jar entirely.
8 liters jar: 3; 5 liters jar: 5; 3 liters jar: 0
Fill 3 liters jar with 5 liters jar.
8 liters jar: 3; 5 liters jar: 2; 3 liters jar: 3
Pour entirely from 3 liters jar to 8 liters jar.
8 liters jar: 6; 5 liters jar: 2; 3 liters jar: 0
Pour entirely from 5 liters jar to 3 liters jar.
8 liters jar: 6; 5 liters jar: 0; 3 liters jar: 2
Fill 5 liters jar with water from 8 liters jar.
8 liters jar: 1; 5 liters jar: 5; 3 liters jar: 2
Fill the 3 liters jar with water from the 5 liters jar.
8 liters jar: 1; 5 liters jar: 4; 3 liters jar: 3
Empty 3 liters jar in 8 liters jar.
8 liters jar: 4; 5 liters jar: 4; 3 liters jar: 0

Now you have 4 liters of water in 8 liters jar as well as 5 liters jar.

Solution:

Firstly: Policeman should take the thug across the river thames and escort the boat back with him. (Thug across, everyone else on this side)
Then, Policeman will take one cat along, leave it there, and on the way back, bring back the thug (since thug cannot be left with anyone without policeman's supervision).
Then Mrs. Mother will take the second cat on the other side. Drop it there and bring the boat back. For the 4th trip, Mrs. Mother And Mr. Father Will travel to the other side, and Mrs. Mother will stay there, and MR. Father will bring the boat back. (Now Mrs. Mother and 2 cats are on the other side.)
5th trip, the policeman takes the thug across. Both of them stay there and Mrs. Mother will come back. Then for the next trip, Mr. Father and Mrs. Mother will go on the boat, and Mrs. Mother will stay there, and Mr. Father will return back.
Then for the 7th trip, Mr. Father will travel with one dog. On the return, the policeman and the thug will return. Then in the next trip, the Policeman will take the second dog. (Now Mrs. Mother and her two cats; Mr. Father and his two dogs and the Policeman is here).
Now finally the policeman takes the boat on the other side and brings back the thug.
Tough one, right ?

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:

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.

Solution:

4

No Of circles in the number.