Solution:

180

Solution:

step 3 is wrong

Solution:

White should move the rook from
A1 to D1 and hence complete the castling move.

Solution:

The figure tells that B is located to the North-East of E.

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:

one fish, one cat and one dog.

Explanation:
The question might appear a bit difficult in the starting but if you analyze the statements, you will realize that it is just a tricky one.

All except two were fishes and all except two were cats. With these statements we can assume that two of the animals were not fishes and two were not cats. Now one of those animals that are not fishes can be a cat and one of those two animals that are not cats can be a fish. Just carry out the same analysis for the statement that all except two animals were not dogs and you will come across the result i.e.:
In that competition, there was one fish, one cat and one dog.

Solution:

It can be down as shown below.

Solution:

Solution is below

Solution:

Read the question carefully again. New Year do falls after Christmas but that happens if two different years. The question is put up against the year 2050 and thus there will be 51 weeks and 2 days in between them as New Year will appear on 1 January 2050 and Christmas will happen on 25 December 2050.

Solution:

111111111

Explanation:
It's a maths magical square root series as :
Square root of number 121 is 11
Square root of number 12321 is 111
Square root of number 1234321 is 1111
Square root of number 123454321 is 11111
Square root of number 12345654321 is 111111
Square root of number 1234567654321 is 1111111
Square root of number 123456787654321 is 11111111
Square root of number 12345678987654321 is 111111111 (answer)

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.