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A notebook contains only hundred statements as under:

1 . This notebook contains 1 false statement. 2 . This notebook contains 2 false statements . . .
. 99 . This notebook contains 99 false statements.
100. This notebook contains 100 false statements.

Which of the statements is correct? 1. 100th
2. 1st
3. 99th
4. 2nd

Robert Israel
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Harry
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  • 99 is true. All others false, i.e. the other 99 are false. – jdods Jul 27 '16 at 18:11
  • Where is your attempt at a solution? You should post your thought process for an attempt at a solution in addition to a clear question that states what about the problem you are having trouble understanding (as opposed to just a c&p of a homework problem without commentary). – Lame-Ov2.0 Jul 27 '16 at 19:10
  • What does this have to do with probability? – John Coleman Jul 27 '16 at 20:24

3 Answers3

2

If the $n^{th}$ statement is correct, that means that there must be $n$ false statements. However, note that two of the statements cannot both be true at the same time, as otherwise for $m \neq n$ this would imply that $m = n$, a contradiction. So, we must have at most one true statement. Also, note that if we had no true statements, statement 100 would be true, a contradiction. So there is exactly one true statement, so that $n + 1 = 100$ and $n = 99$, meaning that the $99^{th}$ statement is true.

Legendre
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    You should also mention that at least one statement must be true (if all statements are false, statement $100$ is true). This immediately shows that we have EXACTLY one true statement, hence $n=99$. Nevertheless, +1. – Peter Jul 27 '16 at 18:22
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    Thank you, I have edited my post to include your clarification. – Legendre Jul 27 '16 at 18:24
  • @yushwuth thanks for lucid response. – Harry Jul 27 '16 at 18:29
  • @peter thanks . – Harry Jul 27 '16 at 18:30
  • I wish to know that exactly which area of math the above type of question is usually asked as generally I have found it not covered in quantitative book that I currently have. I am from India and preparing for entrance exams . – Harry Jul 27 '16 at 18:34
  • It seems to be in the category of a logic puzzle, that is the best classification that I can provide at the moment – Legendre Jul 27 '16 at 18:36
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Yushwuth's answer is correct in the sense that it gives the only consistent assignment of truth values to the statements. However, self-referential statements such as these are tricky. My notebook contains the following $100$ statements.

1) $1$ of these statements is false and you owe me \$100.

2) $2$ of these statements are false and you owe me \$100.

...

99) $99$ of these statements are false and you owe me \$100.

100) $100$ of these statements are false and you don't owe me \$100.

If you don't owe me \$100, statements (1) to (99) are all false, but then (100) leads to a contradiction whether it is true or false. So that can't be right.

If you do owe me \$100, statement (100) is false, and at most one of the others is true. It could be that (99) is true or that none is true; I don't care which, but I would like that \$100 please...

Robert Israel
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At first we will show that more than one statement cannot be correct simultaneously. This is easy to see since any two persons among the 100 contradict each other in their statements. Now there arises two possibilities: (a) All statements are false (b) exactly 1 statement is true We can reject (a) as the 100th person says so (so 100th person must be correct). Hence only one statement is correct. Let the nth statement be correct. nth statement says exactly n statements are incorrect and we know 1 statement is correct. So n + 1 = 100. Hence n = 99. Thus the 99th is true.

abhinav
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