There were 5 people, and let's just call them A, B, C, D and E for simplicity's sake. They were each given a distinct natural number 1, 2, 3, 4 or 5, and also a distinct roman numeral I, II, III, IV or V, but not necessarily in that order.

Note that each person received a different number from all other people person and also received a different Roman numeral from them.

Each person only knew what they were given and did not know what the others were given. Each turn one will say something about the numbers given to them. They were to raise their hand if they knew what everyone had (numbers and roman numerals) and if that person was correct , he/she will get candy, so everyone wanted to get it first. Everyone knew about the first paragraph beforehand, so they know that no one would get what they have. And also, they were only given one try.

This is how it went...

A: My roman numeral and natural number are consecutive (ascending or descending). I hope that isn't too discrete.

B: That is. Anyway, my 2 numbers are also consecutive... if you think of it like a cycle. For instance, 5 and 1 are consecutive. I hope everyone will now give easier clues. Basically, one of my number is 5 and one is a 1.

C: No way. I'm not giving easy hints. My natural number is even.

D: That's not very nice! But thanks to B for giving us that hint! That helped me know what B has. And there's my hint.

E: You all probably helped me best. I now know all the natural numbers' owners. That's probably a very good hint. Since I think I'm ahead, I will give you all another clue. Right after the second clue was given, I know that the sum of my natural number and roman number is the highest here, and no one draws with me. I did not know this before that. Wait, this is too large a hint!!

Just as they were about to continue, someone raised his hand and answered correctly. (He was 100% sure about it!)

Who was it?

The numbers / roman numerals each person had were:

A had 3 and IV.

B had 5 and I.

C had 2 and II.

D had 1 and III.

E had 4 and V.

Moderator
C only
A only
E only
B only
D only
No one
More or equal to 2 people

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This problem is best solved using elimination grids . Let's start by drawing out the grid of the numbers versus the alphabets:

$\begin{array} {| c | c | c | c | c| c | } \hline \text{Letter \ Number} & 1 & 2 & 3 & 4 & 5 \\ \hline A & & & & & \\ \hline B & & & & & \\ \hline C & & & & & \\ \hline D & & & & & \\ \hline E & & & & & \\ \hline \end{array}$

A's statement:

wasn't helpful at all.

However, B's statement:

is helpful, because we can rule out the possibility that B has either 2, 3 or 4. By updating the grid, we have

$\begin{array} {| c | c | c | c | c| c | } \hline \text{Letter \ Number} & 1 & 2 & 3 & 4 & 5 \\ \hline A & & & & & \\ \hline B & & \times & \times & \times & \\ \hline C & & & & & \\ \hline D & & & & & \\ \hline E & & & & & \\ \hline \end{array}$

Similarly, C's statement

was helpful, because only 2 and 4 are even numbers.

$\begin{array} {| c | c | c | c | c| c | } \hline \text{Letter \ Number} & 1 & 2 & 3 & 4 & 5 \\ \hline A & & & & & \\ \hline B & & \times & \times & \times & \\ \hline C & \times & & \times & & \times \\ \hline D & & & & & \\ \hline E & & & & & \\ \hline \end{array}$

Similarly, D's statement:

was helpful as well. This is only possible if D also have a 1 or 5 only.

$\begin{array} {| c | c | c | c | c| c | } \hline \text{Letter \ Number} & 1 & 2 & 3 & 4 & 5 \\ \hline A & & & & & \\ \hline B & & \times & \times & \times & \\ \hline C & \times & & \times & & \times \\ \hline D & & \times & \times & \times & \\ \hline E & & & & & \\ \hline \end{array}$

And because B and D must have the numbers 1 and 5 (not necessarily respectively), then the other users can't have the numbers 1 and 5.

$\begin{array} {| c | c | c | c | c| c | } \hline \text{Letter \ Number} & 1 & 2 & 3 & 4 & 5 \\ \hline A & \times & & & & \times \\ \hline B & & \times & \times & \times & \\ \hline C & \times & & \times & & \times \\ \hline D & & \times & \times & \times & \\ \hline E & \times & & & & \times \\ \hline \end{array}$

E's statement was helpful as well:

Because he/she can make such a bold claim, then both his/her numbers must be maximized. In other words, his natural number is 4 and his roman numeral is V.

$\begin{array} {| c | c | c | c | c| c | } \hline \text{Letter \ Number} & 1 & 2 & 3 & 4 & 5 \\ \hline A & \times & & & & \times \\ \hline B & & \times & \times & \times & \\ \hline C & \times & & \times & & \times \\ \hline D & & \times & \times & \times & \\ \hline E & \times & \times & \times & \checkmark & \times \\ \hline \end{array} \quad \Rightarrow \quad \begin{array} {| c | c | c | c | c| c | } \hline \text{Letter \ Number} & 1 & 2 & 3 & 4 & 5 \\ \hline A & \times &\times &\checkmark &\times & \times \\ \hline B & & \times & \times & \times & \\ \hline C & \times & \checkmark & \times & \times & \times \\ \hline D & & \times & \times & \times & \\ \hline E & \times & \times & \times & \checkmark & \times \\ \hline \end{array}$

Similarly, we construct the grid for alphabets versus roman numerals and filling up all the details previously obtained, we have

$\begin{array} {| c | c | c | c | c| c | } \hline \text{Letter \ Roman} & \rm {I} & \rm {II} & \rm {III} & \rm {IV} & \rm {V} \\ \hline A & \times & & \times & & \times \\ \hline B & \checkmark & \times & \times & \times & \times \\ \hline C & \times & & & & \times \\ \hline D & \times & & & & \times \\ \hline E &\times & \times & \times & \times & \checkmark \\ \hline \end{array}$

Now we just need to consider all the five cases that the author has provided:

Case one: A is the one raised his hand and answered correctly, and A had 3 and IV.Case two: B is the one raised his hand and answered correctly, and B had 5 and I.Case three: C is the one raised his hand and answered correctly, and C had 2 and II.Case four: D is the one raised his hand and answered correctly, and D had 1 and III.Case five: E is the one raised his hand and answered correctly, and E had 4 and V.It is impossible to be case two because there like the rest of the readers, we already know what B have and can't provide any additional details. This applies for case five as well.

If case one is true, then

$\begin{array} {| c | c | c | c | c| c | } \hline \text{Letter \ Roman} & \rm {I} & \rm {II} & \rm {III} & \rm {IV} & \rm {V} \\ \hline A & \times & \times & \times & \checkmark & \times \\ \hline B & \checkmark & \times & \times & \times & \times \\ \hline C & \times & & & \times & \times \\ \hline D & \times & & & \times & \times \\ \hline E &\times & \times & \times & \times & \checkmark \\ \hline \end{array}$

But we couldn't complete the grid, so it's impossible for case two to be true.

Similarly, if it's case four, we will obtain a similar inconclusive grid.

$\begin{array} {| c | c | c | c | c| c | } \hline \text{Letter \ Roman} & \rm {I} & \rm {II} & \rm {III} & \rm {IV} & \rm {V} \\ \hline A & \times & & \times & & \times \\ \hline B & \checkmark & \times & \times & \times & \times \\ \hline C & \times & & \times & & \times \\ \hline D & \times & & \times & \checkmark & \times \\ \hline E &\times & \times & \times & \times & \checkmark \\ \hline \end{array}$

This leaves us with case three, and the illustration in the grid below indeed checks out.

$\begin{array} {| c | c | c | c | c| c | } \hline \text{Letter \ Roman} & \rm {I} & \rm {II} & \rm {III} & \rm {IV} & \rm {V} \\ \hline A & \times & & \times & & \times \\ \hline B & \checkmark & \times & \times & \times & \times \\ \hline C & \times & \checkmark & \times & \times & \times \\ \hline D & \times & & & & \times \\ \hline E &\times & \times & \times & \times & \checkmark \\ \hline \end{array} \quad \Rightarrow \quad \begin{array} {| c | c | c | c | c| c | } \hline \text{Letter \ Roman} & \rm {I} & \rm {II} & \rm {III} & \rm {IV} & \rm {V} \\ \hline A & \times & \times & \times & \checkmark & \times \\ \hline B & \checkmark & \times & \times & \times & \times \\ \hline C & \times & \checkmark & \times & \times & \times \\ \hline D & \times & \times & \times & \checkmark & \times \\ \hline E &\times & \times & \times & \times & \checkmark \\ \hline \end{array}$

Hence our answer is C only.