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Here's an original puzzle by me. I hope you enjoy!

A mother had infinitely many children, with the following names:

$\text{Albert}_1$, $\text{Albert}_2$, $\text{Albert}_3$, … , $\text{Beth}_1$, $\text{Beth}_2$, $\text{Beth}_3$, … , $\text{Clyde}$, $\text{Dolly}$, and $\text{Emily}$.

One day, this mother left to run errands, and she instructed her children not to eat any cake while she was away. When she came back, she interrogated them to find out whether they had obeyed her instruction. They made the following statements:

$\text{Albert}_1$: “One of your children ate cake.”

$\text{Albert}_2$: “Two of your children ate cake.”

$\text{Albert}_3$: “Three of your children ate cake.”

$\text{Albert}_n$: “$n$ of your children ate cake.”

...

$\text{Beth}_1$: “One of your children didn’t eat cake.”

$\text{Beth}_2$: “Two of your children didn’t eat cake.”

$\text{Beth}_3$: “Three of your children didn’t eat cake.”

$\text{Beth}_n$: “$n$ of your children didn't eat cake.”

$\text{Clyde}$: “None of your children ate cake.”

$\text{Dolly}$: “Infinitely many of your children ate cake.”

$\text{Emily}$: “All of your children ate cake.”

The $\text{Albert}_n$’s and the $\text{Beth}_n$’s were using ambiguous language, since it is unclear whether the statement “$n$ of your children ate cake” means that at least $n$ of them ate cake or that exactly $n$ of them ate cake. However, this mother is aware that her children, although mischievous, are consistent. She knows that all of her children used the same convention, whichever that was. Additionally, she knows, from experience, that everyone who ate cake told a lie and that everyone who didn’t eat cake told the truth.

So, which of the children ate cake, and which ones didn't?

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    $\begingroup$ Hi, Welcome to Puzzling! Could you please share with us where you got this problem from, or if it is yours, could you please tell us? This is to stop plagiarism onour site. Please edit your question to put in the appropriate attribution. Thanks! $\endgroup$
    – Stevo
    Jul 7, 2023 at 0:23
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    $\begingroup$ Great puzzle - how on earth did you come up with it?! $\endgroup$
    – Vicky
    Jul 7, 2023 at 11:40
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    $\begingroup$ great puzzle, indeed! Seems quite fresh $\endgroup$
    – justhalf
    Jul 7, 2023 at 13:51
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    $\begingroup$ @Vicky Thank you! After seeing one of Joel Hamkins's transfinite epistemic puzzles (jdh.hamkins.org/cheryls-rational-gifts), I was inspired to think about how to make "infinitary versions" of other kinds of logic puzzles. Then, I saw Problem 2 from the 2004 Singapore Junior Math Olympiad (a puzzle like this one but involving only finitely many kids), and I tried to make an infintary version. I tried a couple different combinations of statements, and then found one that one worked. I added the "ambiguity" twist at the end to make it a bit harder. $\endgroup$ Jul 7, 2023 at 16:13
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    $\begingroup$ I will try to make more infinitary versions of classical logic puzzles in the future, and I encourage others to try too! $\endgroup$ Jul 7, 2023 at 16:14

2 Answers 2

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Answer:

All of the children except for $\text{Beth}_2$ and $\text{Dolly}$ ate cake.

Explanation:

Consider two possible cases.

Case 1 - the children used the "at least" convention:

Let's consider two subcases.

Case 1a - only a finite number of people ate cake

Let $k$ be the number of people who ate cake. Then every $\text{Albert}_i$ with $i > k$ made a false statement and so ate cake. But there are infinitely many such $\text{Albert}$s. Contradiction.

Case 1b - an infinite number of people ate cake

Every $\text{Albert}$'s statement is true, so none of them ate cake. Since there are infinitely many $\text{Albert}$s, an infinite number of people didn't eat cake. Thus, every $\text{Beth}$'s statement is true too, so none of them ate cake either. But now there's only $\text{Clyde}$, $\text{Dolly}$, and $\text{Emily}$ who could have eaten any cake, and we said there were an infinite number of people who did. Contradiction.

Wrapping up case 1

Both subcases led to contradictions, so we must not be in case 1.

Case 2 - the children used the "exactly" convention:

Since all of the $\text{Beth}$s' statements are mutually contradictory, either all or all but one of the $\text{Beth}$s ate cake (an infinite number of people either way).

Since an infinite number of $\text{Beth}$s ate cake, all of the $\text{Albert}$s' statements are false, so they all ate cake.

Since all of the $\text{Albert}$s ate cake, $\text{Clyde}$'s statement is false, so he ate cake too, but $\text{Dolly}$'s statement is true, so she didn't eat cake.

Since $\text{Dolly}$ didn't eat cake, $\text{Emily}$'s statement is false, so she ate cake.

Since everyone besides $\text{Dolly}$ and at most one of the $\text{Beth}$s ate cake, either 1 or 2 people didn't eat cake. If only 1 person didn't eat cake ($\text{Dolly}$), then there's a contradiction because $\text{Beth}_1$ must have eaten cake but her statement must be true. Thus, 2 people didn't eat cake: $\text{Dolly}$ and $\text{Beth}_2$. Everyone else did.

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This has already been answered, but I think that my approach is somewhat different from Joseph Sible-Reinstate Monica's.

Solution

The children all use the "exactly $n$ convention", and all of the children ate cake except for Beth$_2$ and Dolly.

Preliminaries

Emily ate cake. If $E$ is telling the truth, then she is one of the "everybody" who had cake. But cake-eaters are liars, so it is impossible for $E$ to have eaten cake and to be telling the truth. Thus she must be lying, from which we learn that (1) $E$ is a cake-eating liar, and (2) there is at least one person who did not eat cake.

Clyde ate cake. We already know that $E$ is a cake-eating liar, therefore $C$ is a cake-eating liar.

"Exactly $n$" convention

Assume that the children are using the "exactly $n$" convention. That is, if a child says "$n$ children did/did not each cake", this means exactly $n$ children either did, or did not, eat cake.

Every Albert ate cake. By induction, $A_n$ is a cake-eating liar for every $n \in \mathbb{N}$. First, note that $A_1$ claims that exactly one child ate cake, but we already know that both $C$ and $E$ are cake-eaters. Therefore $A_1$ is a cake-eating liar. For (complete) induction, suppose that $A_1$ through $A_k$ are all cake-eating liars. $A_{k+1}$ claims that there are exactly $k+1$ cake-eaters, but it is possible to identify $k+2$ cake-eaters (all of the $A_j$ prior to $A_{k+1}$, plus $C$ and $E$). Therefore $A_{k+1}$ is a cake-eating liar.

Dolly did not eat cake. $D$ claims that infinitely many children ate cake, and all of the $A_n$ ate cake, so $D$ is telling the truth.

Every Beth ate cake, except for Beth$_2$. Each $B_n$ claims that exactly $n$ children ate cake. At most, only one of them can be telling the truth, as their statements are all mutually exclusive. Suppose that none of them are telling the truth. Then exactly one child, $D$, did not eat cake. But then $B_1$ must be telling the truth. Contradiction. Therefore exactly one of the $B_n$ is telling the truth, which means that there are two cake-abstainers ($D$, and $B_n$ for some $n$—every other child is a cake-eating liar). Hence $B_2$ is telling the truth, and the rest are cake-eating liars.

"At least $n$" convention

Assume that the children are using the "at least $n$" convention. That is, if a child says "$n$ children did/did not eat cake", this means, "at least $n$ children did/did not eat cake".

The children are not using the "at least $n$" convention. This is shown by considering the two cases related to Dolly: either she is telling the truth, or she is lying. Both cases lead to contradictions.

Case 1:

Dolly cannot be telling the truth. For contradiction, suppose that $D$ is telling the truth, and infinitely many children ate cake. Then for any $n$, the statement "at least $n$ children ate cake" is a true statement, which implies that $A_n$ is telling the truth. Therefore all of the Alberts are telling the truth, and ate no cake.

But then there are infinitely many children who did not eat cake, which means that for any $n$, the statement "at least $n$ children did not eat cake" is a true statement, which implies that $B_n$ is telling the truth. Therefore all of the Beths are telling the truth, and ate no cake.

But then exactly two children ate cake: $E$ and $C$, which is a contradiction to the assumption that $D$ is telling the truth.

Case 2:

Dolly cannot be lying. For contradiction, suppose that $D$ is lying. Then only finitely many children ate cake—let us say that exactly $k$ children ate cake. But then every $A_n$ is a liar for $n > k$. This implies that there are an infinite number of cake-eating liars among the Alberts. This contradicts the assumption that $D$ is lying.

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  • $\begingroup$ rot13("Gura rknpgyl bar puvyq, $Q$, ngr pnxr." Qb lbh zrna "qvqa'g rng" vafgrnq bs "ngr" gurer?) $\endgroup$ Jul 9, 2023 at 2:37
  • $\begingroup$ @JosephSible-ReinstateMonica Yes. Thank you. Corrected. $\endgroup$ Jul 9, 2023 at 4:26

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