5 people competition

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:

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

wasn't helpful at all.

However, B's statement:

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."

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

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

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:

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.

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:

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.

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.

Firstly, this question took "years" too finish, so yeah, hope you liked it.

I suggest that one may invite 5 people to do this as an experiment (and yes, the candy as well, but... moving on)

Part 1 . Why C?

1 . C knows that he has 2 and II.

2 . Also, C knows that E has a sum of 9. Why?

If it were 10, he would have known before the hints.

If it were below that, he would not have known he was the highest. This is left to the reader to figure out.

3.1 . Let E have 4 and V.

Then B has I and 5 since V is already taken.

Also, D will then have 1 since I, V and 5 are already taken.

Thus, A has a 3 since 1, 2, 4 and 5 are already taken. This implies A has a II or IV.

Therefore, A has IV, since C has the II.

So, we have: A , 3, IV. B , 5, I. C , 2, II. D , 1, III. E , 4, V.

3.2 . Let E have 5 and IV.

By logical deduction, you will get A , 4, III. B , 1, V. C , 2, II. D , 3, I. E , 5, IV.

4 . Why 3.1, not 3.2?

If it were 3.2, E will only get that B is 1 in addition to E is 5. He does not get anything else.

However, if it were 3.1, E can get who the natural values were given to.

Part 2 . Why not others?

Because the others cannot get the roman numerals. All have confusion over A and C (which has II and which has IV) while A has confusion over C and D (which has II and which has III).

Many things were omitted, and the reader is encouraged to figure these gaps out.