Kenneth Rosen Edition 7 Exercise 1.2 Question 39 (Page No. 24)

Question

Freedonia has fifty senators. Each senator is either honest or corrupt. Suppose you know that at least one of the Freedonian senators is honest and that, given any two Freedonian senators, at least one is corrupt. Based on these facts, can you determine how many Freedonian senators are honest and how many are corrupt? If so, what is the answer?

Comments

And rubbish answer here.

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And rubbish answer here

 :p

@Arjun Sir, For $3^{32} \mod 80,$ Are you using GPT-3 and generating the response multiple times (Asking “Compute $3^{32} \mod 80$” multiple times (at least 2-3 times) to GPT) ? 

I have asked 4 times to compute $3^{32} \mod 80$ and 3 times it gives output as 71 and one time, it gives 21 as output. There are some memes also there to show lack of performance of ChatGPT-3 for some mathematical questions. 

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@ankitgupta.1729 I’m using GPT-4

Answer 1

Suppose we assume that within those 50 person, A and B are honest and rest 48 are corrupt. Now in the question it is mentioned that whenever we take 2 person at least one of them is corrupted. Like it maybe possible that both are corrupted but there is no chance that both person are honest. So, when we will choose A and B it is not possible that both are honest because it is contradicting the question statement. It’s also mentioned that at least one person is honest.

So, it is clear that only 1 person is honest and rest 49 are corrupt.

Comments

Why can we not have honest-corrupt pairings? In that case also, at least one corrupt and at least one senator being honest condition is satisfied!!

@gatecse

@tamal03

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We can have exactly one H-C pair. If we have two such pairs say (H1, C1) and (H2, C2) that means (H1, H2) is also a possible pair and it violates the condition of “at least one corrupt in any pair”.

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Is it? What if we have 25pairs = 50 senators where every pair has one honest, one corrupt.

Which condition is violated here? (is what I am unable to understand)

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We have 50 senators and so we must consider ALL possible pairs out of them – there is no choice here. For each senator from $1..50$ he can be paired with any of the remaining senators. For the first senator he can be paired with any of the remaining $49$ senators. For the second senator he'll be having $48$ new senators to pair with and likewise the second last senator will have 1 new senator to pair with and the last senator will have 0 new senator to pair with. That’s how we get $49 + 48 + 47 + \ldots 1 + 0 = \frac{49.50}{2} = {}^{50}C_2$ possible pairs.

Answer 2

Based on the given information, we can determine that at least one senator is honest, but we cannot determine the exact number of honest and corrupt senators.

We know that at least one senator is honest, but we do not know how many honest senators there are. It could be that only one senator is honest, or it could be that multiple senators are honest.

We also know that given any two Freedonian senators, at least one is corrupt. This means that if we choose any two senators, we know that at least one of them is corrupt. However, we cannot determine the number of corrupt senators based on this information alone. It could be that all but one senator is corrupt or that a majority or minority of the senators are corrupt.

Therefore, while we know that there is at least one honest senator and that given any two senators, at least one is corrupt, we cannot determine the exact number of honest and corrupt senators.