Solution:

One In Million

Solution:

There is no reason whatever why the customer's original deposit of Rs.100 should equal the total of the balances left after each withdrawal.
The total of withdrawals in the left-hand colum may equal Rs.100, but is is purely coincidence that the total of the right-hand column is close to Rs.100.
Let us show another example, but starting with Rs.200 in the bank:
Withdrawals Balance left
Rs.50 Rs.150
Rs.25 Rs.125
Rs.10 Rs.115
Rs.8 Rs.107
Rs.5 Rs.102
Rs.2 Rs.100
Rs.100 Rs.699
Moral of this story? Don't Total Balances

Solution:

The matchsticks that are marked with red are to be removed and the final figure will have just two squares. You can remove either pair from the possible four ways.

Solution:

10,15,9,20,12

Explanation:
The score obtained on the first throw is 10.
The score obtained on the second throw is 15.
The score obtained on the third throw is 9.
The score obtained on the fourth throw is 20.
The score obtained on the fifth throw is 12.

Solution:

   9567
+ 1085
======
 10652

Solution:

Zouma is 3 and has two footballs.
Ake is 1 and has one football.
Kalas is 4 and has three footballs.
Aina is 2 and has four footballs.

Solution:

For this answer is 3^0, 3^1, 3^2... That is 1,3,9,27,81,243 and 729.

Solution:

To do it, the marbles must be arranged as the vertices of an octahedron. If you are unsure, check the image below.

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:

28 =>1+2+4+7+14

cannot be 1 (as number cannot be included)

Solution:

Move and then pour all juice from second glass to fifth glass.