# How many persons do you think are liar?

There are 10 person. First person says: At least one of the person is liar. Second person says: At least two of the person is liar. Third person says: At least three of the person is liar. Fourth person says: At least four of the person is liar. .. .. Tenth person says: At least a ten of the person is liar.

How many persons do you think are liar?

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"at least" ruffles my huffles. – Sabyasachi Mar 19 '14 at 17:19
spikedmath.com/445.html – rlartiga Mar 19 '14 at 17:20
According to Dr. House, all of them. – Batman Mar 19 '14 at 17:24
I think there is confusion in the answers if one person affirm there exist at least one liar, don't means he is lyng at that moment. Or am I wrong? – rlartiga Mar 19 '14 at 17:33

First: it can't be the case that they are all lying: if they are all lying, then it is true that "at least $n$ people are lying" for each $1\leq n\leq 10$, which in turn implies that they are all telling the truth: a contradiction.

If it is the case that exactly $n\geq1$ of them are telling the truth, then it must necessarily be the first $n$: if the person saying that "at least $j$ are lying" is telling the truth, then the people saying "at least $j-1$", "at least $j-2$", and so on must also be telling the truth.

So, if there are exactly $n$ of them telling the truth, then the $(n+1)$st person must be lying. So, it cannot be the case that "at least $n+1$ people are lying"; since we know that "at least $n$ people are lying", it must then be the case that exactly $n$ people are lying.

So, we have $n$ people lying, and $n$ people telling the truth; so, in this case, it must be the case that $n=5$. The first five people are telling the truth, and the last ten people are lying.

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Why you assume if one of the person says "there is at least one liar" means that the person is lying at that moment? – rlartiga Mar 19 '14 at 17:30
@rlartiga Where did I assume that? – Nick Peterson Mar 19 '14 at 20:04
The statement says the person $n$ says "there is at least n liars". But it don't means "in the previous affirmation at least $n$ persons are not telling the truth". What I'm trying to say is, the condition to be a liar don't ensure that they can't say the truth. Maybe is just a gramatical issue. – rlartiga Mar 19 '14 at 21:19
@rlartiga For purposes of a puzzle like this, "being a liar" just means "the statement they are making is not true". – Nick Peterson Mar 19 '14 at 21:29

You can solve this problem by 'ping-ponging':

Assume that the 10th person is telling the truth. Then all ten people are liars, which implies that the 10th person is also a liar, a contradiction; this means that the 10th must be a liar. Now, we know there is at least one liar, so the first person is telling the truth. This means there are at most 9 liars.

Next, consider the 9th person. If they were telling the truth, then there would be at least 9 liars - but since we know that the first person is a truth-teller, this implies that the 9th person would be a liar, in the same way that we argued for the 10th person above. So there are at least two liars, and now we know that the 2nd person is also a truth-teller.

Can you see how to continue from here?

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There are only 12 possible answers, so lets get through it.

Say $p_i$ is the Person that states "There are at least $i$ persons that lie." with $i \in 1, \dots, n = 10$

• Exactly 0 liars: Not possible, because of $p_{10}$.
• Exactly 1 liar: So $p_j$ where $j \in 2, \dots, 10$ is a liar $\Rightarrow$ There are $10 - 1 = 9 \neq 1$ liars $\Rightarrow$ ⚡
• Exactly 2 liars: So $p_j$ where $j \in 3, \dots, 10$ is a liar $\Rightarrow$ There are $10 - 2 = 8 \neq 2$ liars $\Rightarrow$ ⚡
• Exactly 3 liars: So $p_j$ where $j \in 4, \dots, 10$ is a liar $\Rightarrow$ There are $10 - 3 = 7 \neq 3$ liars $\Rightarrow$ ⚡
• Exactly 4 liars: So $p_j$ where $j \in 5, \dots, 10$ is a liar $\Rightarrow$ There are $10 - 4 = 6 \neq 4$ liars $\Rightarrow$ ⚡
• Exactly 5 liars: So $p_j$ where $j \in 6, \dots, 10$ is a liar $\Rightarrow$ There are $10 - 5 = 5 = 5$ liars $\Rightarrow$ ⚡
• Exactly 6 liars: So $p_j$ where $j \in 7, \dots, 10$ is a liar $\Rightarrow$ There are $10 - 6 = 4 \neq 6$ liars $\Rightarrow$ ⚡
• Exactly 7 liars: So $p_j$ where $j \in 8, \dots, 10$ is a liar $\Rightarrow$ There are $10 - 7 = 3 \neq 7$ liars $\Rightarrow$ ⚡
• Exactly 8 liars: So $p_j$ where $j \in 9, \dots, 10$ is a liar $\Rightarrow$ There are $10 - 8 = 2 \neq 8$ liars $\Rightarrow$ ⚡
• Exactly 9 liars: So $p_j$ where $j \in 10, \dots, 10$ is a liar $\Rightarrow$ There are $10 - 9 = 1 \neq 9$ liars $\Rightarrow$ ⚡
• Exactly 10 liars: So $p_j$ where $j \in 11, \dots, 10$ is a liar $\Rightarrow$ There are $10 - 10 = 0 \neq 10$ liars $\Rightarrow$ ⚡

So there is exactly one possible answer: There are $5$ liars.

It's quite obvious that this can be generalized for any $n \in \mathbb{N}_0$ people:

• If $n = 2t$, there are $t$ liars.
• If $n = 2t+1$, there is no valid solution.
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