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Let $P = x \cdot y$ and $S = x + y$.
If Proctor could not determine the two numbers offhand, then there must be at least two valid factorizations of $P$, which means that $P$ cannot be the product of two primes.
And if Summer knew that Proctor could not determine the two numbers, then every possible pair of $x$ and $y$ that add up to $S$ must have a ...
37
I think I have a faster solution than Rubio. (Or I did, when Rubio's solution took one day longer than mine; he's since incorporated my solution into his answer).
The answer:
The explanation:
Let's start on the time of release and follow the thought process.
35
The answer
Assumptions
We assume both have at least one bar on their window (or the window couldn't be said to be barred, and they're told their windows are the only barred ones).
We further assume that they must determine which total, 18 or 20, is correct.
If they are being asked to determine if it is either (18 or 20) or (not(18 or 20)), that's a whole ...
34
It will be
Both A and B can see, what C sees, and that's why they both know that
From there, the problem reverts to the earlier one:
26
There are only 6 possible configuration of hats.
wwbb
wbwb
bwwb
wbbw
bwbw
bbww
And
This would be a better question if you specify that every player is killed if he guesses wrong (my answer) or they must all answer at the same time ($1$ and $2$ always guesses opposite of $3$).
26
The numbers are
Alice: You can't know my sum.
Bob: Thanks to you, I know your sum.
Alice: Then I know your product.
EDIT: While I show the consistency of the two numbers with the provided statements, please see Mike Earnest's answer for a proof that this is the only possible pair of numbers consistent with the conversation.
22
The following table gives the numbers of one person and the corresponding possible numbers of the other person.
own other
0 5, 8, 15
1 4, 7, 14
2 3, 6, 13
3 2, 5, 12
4 1, 4, 11
5 0, 3, 10
6 2, 9
7 1, 8
8 0, 7
9 6
10 5
11 4
12 3
13 2
14 1
15 0
Round 1:
If one of them has a number greater than 8 he/she would ...
20
The crucial fact here (which I think makes the question kinda unfair since it's not exactly common knowledge) is that
And we had better
Now,
So
And in this case
20
Answer:
First, some observations
Alright. Now, let's take a look at different possibilities:
Generalizing this, we get the rule above:
Do note however that the actual answer is actually slightly more complicated, thanks to the effect of asking them in order:
18
I'll try another explanation(with same result):
Here are the steps:
Next step:
So: Who talks?
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If A doesn't know the colour of his hat, then at least one of B and C must be wearing a red hat, because if they were both wearing black, then A would know immediately that he was wearing a red one.
B knows this as well. But if C were wearing a black hat, then B would know that she was wearing a red one because of that. So since B doesn't know, C must be ...
17
This was trickier than it looked. I have a feeling that there should be a quicker way to find the answer than the list of cases I worked through.
Note: I interpret "two of which are factors of the third" to mean that two of the numbers divide the third, and that either of those two factors might be 1.
16
X's statement means that his number must be even (if it was odd, Y and Z would have numbers summing to an even number, so they could be the same).
In the first case, Y's statement means that Y has an even number, and neither X nor Z can have the same even number. Therefore Y must have a number that is at least 1008.
Z knows all three numbers, so X and Y ...
16
First observations
Since the puzzle is symmetric in white and black hats, I assume that the first man has a black hat. Then only three cases remain: (1) BWWB; (2) BWBW; (3) BBWW.
($\ast$) If a man sees two hats of color white, then he knows that his hat is either red or black. If a man sees two hats of color black, then he knows that his hat is either ...
16
The Answer:
My reasoning:
How I got the numbers:
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Alice: I know we have different numbers.
Bob: Aha, I got it. I found all the numbers.
Charlie: Me, too. I know our numbers, now.
Alice: Alas, I still don't know.
16
Reasoning:
15
In other words, we are looking for two positive integers $x$ and $y$ with $x\ge2$, $y\ge2$ and $4\le x+y\le40$ that fit with the conversation.
Let us denote the sum $s=x+y$ and the product $p=xy$.
(1) Danielle's first statement "I don't know what the dimensions of the farm are" gives the following information on the product:
$p$ is not the product of (...
15
Let $a$ and $b$ be the integers thought of by A and B, and let $N$ and $n$ be the integers given by the teacher, where $N>n$. Assume for contradiction that both A and B keep on saying "no" forever.
If $a\geq n$, then A knows $n$ cannot be the sum and will say "yes" the first time. Thus after A's first "no", B knows $a<n$.
If $b\leq N-n$, then $a+b<...
15
Let's start with some ASCII art as a reference:
+-----+ +-----+
| 4 | | 5 |
| mul | | add |
--+-----+-- --+-----+--
| ° ° | | ° ° |
| A | | B |
\___/ \...
15
You are
1) From my birth I want to rise
But I shall fall, its no surprise
I should display approval from afar
But pedantic snobs is what you are.
2) I could produce a rise in rank
But not if my cachet has sank
Uplift me please, by 288*
And my master will reach another state
What am I?
*time sensitive
14
The fact $3z\geq x+y+z>xyz$ implies that $3>xy$, so $x=1$ and $y=1$ or $x=1$ and $y=2$. In the former one, we get $2+z>z$. Otherwise we get $3+z>2z$, so $z<3$ and then in fact $z=2$.
So the possible tuples are:
However, Summo can only deduce what the tuple is if he knows that $P>S$. It follows that for his sum, there is exactly one ...
14
I'm no mathematician, but I think both Deusovi and Gareth can say:
I'm not sure about making it into a proof, but by the extreme example:
Basically (maybe/hopefully):
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Riley's answer proves that the numbers
are consistent with the conversation. I will prove that these are the only possible numbers.
After Alice's first statement,
After Bob's statement:
After Alice's second statement:
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Edit: Assuming $y$ is a positive integer, the below reasoning works. Without this assumption, it does not. See ricksmt's answer for the final word on this issue.
Sam: I don't know your number.
Now Peter knows that $x$ is a factor of $2002$: if $x$ didn't divide $2002$, Sam would know that $2002$ wasn't $xy$, so he would know $y=2002-x$.
Peter: I don't ...
12
The professor says:
I have written on Alice's paper an even ordinal and on Bob's paper an odd ordinal. Who has the lesser ordinal?
Then, after $\lambda$ iterations of the game, it is common knowledge that both players have an ordinal of at least $\lambda$. Obviously, if the player whose turn it is has $\lambda$, they immediately know that they have the ...
12
We have five statements to process:
"Two of us are truth tellers".
"None of us are truth tellers".
"Three of us are truth tellers".
"Only one of us is a truth teller".
"Three of us are truth tellers".
These five statements are all mutually contradictory except 3) and 5). So out of the first five speakers, either none, one, or two are telling the truth.
If ...
12
The following truth values should solve the puzzle:
Steps I used to find the solution:
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Along the lines of Glen O's answer, this answer attempts to explain the solvability of the problem, rather than provide the answer, which has already been given. Instead of using the meta-knowledge approach, which, as Glen stated, can get hard to follow, I use the range-base approach used in Rubio's answer, and specifically address some of the objections ...
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