Solution:

There are so many things that the child is miscalculating by using his mischievous mind. He is accounting all the hours he spend separately as well as in the holidays. The weekend are present in the summer holidays as well.

Then the school last maximum for 6 hours and in that case, even the 4 days that remains will comprise of at least 16 school days. In the similar fashion, the weekends off are not entirely to be excluded but only 6 hours each day to be excluded.

If you calculate gain, you will be left with so many more days/hours.

Solution:

Fred Odea won.

Just look at the announcement lines. The first sentence in 'Ok now everybody!' If you take out just the first letter of every word, it will form 'ONE'. In the same manner the second sentence 'The winner of…..' will give form 'TWO' and the third sentence 'The Hardest Riddle Ever Event.' Will form THREE.

Thus the final sentence must form FOUR. There is only one candidate who can suffice with the first two letter and he is Fred Odea. The final sentence of the mastermind must be, 'Fred Odea: Ultimate Riddler!'

Solution:

The clerk said that every day the password is CHANGED. Changed is the password.

Solution:

Remove the same 2 matchsticks from each figure and move figure towards left 45 degrees, It forms the letters V W X Y.

Solution:

Just 2

Day One: You make your first cut at the 1/7th mark and give that to the worker.
Day Two: You cut 2/7ths and pay that to the worker and receive the original 1/7th in change.
Day three: You give the worker the 1/7th you received as change on the previous day.
Day four: You give the worker 4/7ths and he returns his 1/7th cut and his 2/7th cut as change.
Day Five: You give the worker back the 1/7th cut of gold.
Day Six: You give the worker the 2/7th cut and receive the 1/7th cut back in change.
Day Seven: You pay the worker his final 1/7th.

Solution:

The last marble will be yellow.

Since marbles can only be taken out in pairs and you started off with an odd number of yellows there is always going to be one yellow left over that you'll keep putting back in the box until it's left on it's own.

Solution:

It can be solved as below.

Solution:

'five'

Its the count of alphabet

Solution:

A) Fill 5 ml gallon ( 5mlGallon - 5, 3mlGallon - 0)
B) Transfer to 3 ml gallon (5mlGallon - 2, 3mlGallon - 3)
C) Empty 3 ml gallon ( 5mlGallon - 2, 3mlGallon - 0)
D) Transfer 2 ml from 5 ml gallon to 3 ml gallon (5mlGallon - 0, 3mlGallon - 2)
E) Fill 5 ml gallon(5mlGallon - 5, 3mlGallon - 2)
F) Transfer 1 ml from 5 ml gallon to 3 ml gallons(5mlGallon - 4, 3mlGallon - 3)

Solution:

The numbers that have been written over the calendar are:
12, 4, 8 and 3

A few numbers have been circled under particular months. The curious thing is that the numbers that have been written are the same number of corresponding months under which the dates have been marked.

What we can judge by the overall calendar is that the dates that have been marked may refer to the letters at that position in the month's name. In addition, the written numbers can be the position of those letters. Let's simplify below:

12th month is December
The number marked under December = 5
The fifth letter of December = M

4th month is April
The number marked under April = 4
The fourth letter of April = I

8th month is August
The number marked under August is 6
The sixth letter of August = T

3rd month is March
The numbers marked under March are 4 and 5
The fourth and fifth letter of March = C and H

When you combine the letters, you get the name as MITCH.
Therefore, Mitch is the killer.

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.