Hard Riddles

54
7^2 + 2^2 + 1^1
6^2 + 3^2 + 3^2
2^2 + 5^2 + 5^2

Suppose that there is just one black hat. In such a scenario, the person wearing it will see the other two hats as white and then as the king had announced, he will be easily able to judge that he is wearing a black hat himself. But the other two men will see a black hat and a white hat and will not be able to judge. Thus this will be an unfair situation, so we can rule it out.

Now suppose that there are two black hats. Then, both the person will see a white hat and a black hat. And they have already realized that there must be two black hats, thus they can easily deduce that they are also wearing a black hat. But the one wearing white hat will see two black hats and thus he will not be able to identify on a sure note. Thus the competition will not be fair.

Thus the only situation where the competition can be fair is when all the hats are black. The first one who recognize the fact will stand up and say black.

He was Mass Murderer.

Density multiplied by volume equals mass.
Therefore, Luis Garavito killing mass would mean Luis Garavito was a MASS murderer

First the logicians will have to take a leap of logic. The master has told them that it won’t be impossible for any logician to solve the puzzle. Thus it is assure that any of the color can’t exist only once. If it did, the one wearing it will have no clue about that color which would be unfair for him.

Now every logician will look around in the circle and count the number of times they see a particular color. If a color is seen only once, then the logician will know that the color on his band must be of the same color (as per the leap of logic). And then the logician will leave on the first bell.

In the similar fashion, any logician who see any color just once will be able to identify his own color and they will leave when the bell rings or they will be disqualified and asked to leave. Equivalently, any color for which there are two bands, will be eliminated after the first bell has rung. Thus there must be three bands of any remaining color at least.

Assume that a logician don’t see any color once, but sees a color twice. If they were the only bands of this color then the two logicians must have left at the first bell. But they did not. Thus it means that his band color is the same and he will leave on the second bell.

DAVID

Talking in roman numerals,
D = 500
At the center is 5 = V
The first number is I
The first letter of the alphabet is A

Thus they combine to form DAVID.

The date is 12 December 2013. For any other date, the checkers would have been able to identify the papers. Why? Because the British people write dates in Day/Month/Year Format and the American people write date in Month/Day/Year format. On 12 December, the date and month will be the same and thus there is no way to distinguish when the names are same as well.

43

If you analyze, then you will find that all the 6 numbers divisible by 3 can be ordered. Why? Because you can break them own as the sum of 6 and 9.
Now after 26, all the numbers that are divisible by 3 if subtracted by 40 can be obtained.
After 46, all the numbers will fit into one of the above categories and thus all such numbers can be obtained. But 43 will be last number that will not fall into one of the above categories.
44 = 20 + 6 * 4, 45 = 6 * 6 + 9

Product is 52 and sum is 17. The numbers are 4 and 13.

The code word is TIE.

Suppose if you are told a character of MOD, then you can’t identify if the number of vowels are one or two. Suppose that the character you are told of is M, then you can associate two words with it i.e. AIM (that has two vowels) and MOD (that has 1 vowel). You can say the same regarding other characters O and D as well.

Thus all those words that comprises of M, O or D in them can be ruled out. This points us to TIE. If you check with the characters of TIE, you will agree that it stands true for each of the characters of the word. Thus this is the code word.

Let us assume that there were x flowers present in the basket of the pilgrim and he offered y flowers at each temple. Now as per the details given in the question, the number of flowers when he came out of the third temple.

But the number of flowers when he comes out of the third temple must be (8x – 7y)

Thus (8x – 7y) = 0
8x = 7y

Now the minimum values that you can put by hit and trial are 7 and 8 for x and y respectively.

Thus, the pilgrim had 7 flowers in the beginning and he offered 8 flowers at each temple.