Looking for examples of a kind of logic test

In My Humble Opinion
1

I know this has been discussed here, but the only thread I found provided only one example, and I would like to find some more. It’s a logic test, and what happens is that two people have between them all the information to solve a problem, if they got together, but for some reason they don’t/can’t get together. It ends this way:

Person A says, “I can’t get the answer with the information I have, but I know that Person B can’t get it either.”
Person B says, “I didn’t know the answer before, but when you said that, I realized that, in fact, I do know the answer.”
Person A says, “Okay, now I know the answer too.”

The example I found was a bunch of different dates and months, with the goal to find somebody’s birth date. I’ll bet there are other examples of this kind of logic out there, though, and I would like to see them. Particularly interested in answers without numbers, but using the same logical progression. Post them, link to them, whatever, TIA.

2

Here is a well-framed version of the classic of the genre.
https://www.xkcd.com/blue_eyes.html
https://xkcd.com/solution.html

3

Three people are painted with a dot on their forehead, which is either red or black; they are told this but not told what color of dot they have. They are told that at least one person will have a red dot.

They can see the dots on the heads of the other two.

{Insert a scenario to start the game. Either the person testing them brings them into the room so they’re together for the firs time, or turns on the light when previously they were in the dark. So the clock starts counting, in some sense}

The person I will refer to as the third person sees the other two each with a red dot.

A couple beats go by. The person testing them asks the first participant who admits they can’t figure out what color dot they’ve got. A couple more beats go by and the the person testing them asks the second participant the same question and gets the same answer.

At which point the third person says “I, too, have a red dot”.

If the third person had (hypothetically) had a black dot, (thinks the third person to themself), person two, would soon realizing that person one had not been staring at TWO people with black dots or they’d quickly realize they had to have a red dot, because someone had to have one. And person one had not done that (all this is the third person imagining the second person working this out). So therefore (says the third person to themself) since person two didn’t go through that process and realize their dot was red, that has to mean that their failure to do so means person three also has a red dot.

4

Three logicians walk into a bar. Bartender says, “You all want a beer?”
The first logician says, “I don’t know.”
The second logician says, “I don’t know.”
The third logician says, “Yes.”

5

1 < x < y < 50 are two distinct integers. A knows the sum x+y and B knows the product xy. A says to B: “I know you don’t know the two numbers.” B replies, “Now I know the two numbers.” A says, “now I do too.”

What were x and y?

6

Why do just the blue-eyed people leave? Couldn’t the brown-eyed people figure out the same thing?

7

All right, I will say right here that I don’t understand this one at all. It looks like it’s saying that 1 is less than x is less than y is less than 50 and if x is less than 1 I am lost right there. But I’m not really all that up on mathematical symbols. A little help?

8

There are two numbers between (but not including) 1 and 50. They are different numbers (I don’t think that it really matters which one is greater than the other one). Person A only knows what the numbers added together are. Person B only knows what the multiplied together are.

After that, I’m not sure how to figure it out.

9

I’ve started.

Let’s say the numbers are 3 and 4. A knows the sum is 7, right? But he doesn’t know what the numbers are: they could be:
2,5
3,4.

That means that he thinks the product could either be 10 (2x5) or 12 (3 x 4). He doesn’t know which.

If the product is 12, then B won’t know what the numbers are either (could be 3,4; or 2,6). However, if the product is 10, the B would know the numbers are 2,5, because there’s no other pair of numbers with a product of 10 that matches the rules.

So: if A is told “7,” he won’t make that declaration. With me?

But what if A is told “11”? The possible pairs are:
2,9
3,8
4,7
5,6

In each case, B’s product will have multiple factors.

So, here’s rule #1 for solving it:
A is given a number all of whose addend pairs include at least one composite number.

I’ll see where I can get from there.

10

Going a little bit further, I think that the numbers that work are the numbers that are two more than an odd composite number (I’ve gotten up to sums of 54, and for every one up to this point, I can find an addend pair in which both numbers are prime, e.g., 7 and 47, such that A couldn’t rule out B’s knowing the answer). So, A may have received:
11
17
23
27
29
35
37
47
51
53
57
65
67
71
77
79
83
87
93
95
97

I think that’s a complete list.

Now: if B hears A’s declaration, B must know that A has one of those sums. There must be a product that appears as a factor of addends for only one of those sums. For example, if A has 28, that could be 4 x 7 or 2 x 14. But if it were 2 x 14, then A would have 16, and wouldn’t have made that declaration. So it must be 4 and 7.

Which seems very plausible, but that’s the first pair I tried out, so maybe my reasoning is faulty.

11

“Is anyone here mute?”

12

https://curiosity.com/topics/can-you-solve-the-secret-word-logic-puzzle-curiosity/

13

Copying and pasting from above:

14

That’s one’s dirty because there’s more than one possible answer - meaning that you can never really say that any given answer is right. Might as well claim that “what number am I thinking of” is a logic puzzle.

15

You have a slight error in your calculations:

You are missing a few numbers, e.g., 41, 55, etc, also several even numbers like 96 which is the sum of two primes 13+83, but I (slightly arbitrarily, actually) said both numbers should be between 1 and 50.

The complete list is 11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 82, 83, 85, 86, 87, 89, 91, 92, 93, 94, 95, 96, 97.

4 and 7 does not work, since while A can indeed make the first statement that B cannot deduce the numbers, in each of the cases (2,9), (3,8), and (4,7) B can then figure out the numbers (e.g., in case (3,8) B knows the product 24 must be 2x12, 3x8, or 4x6, and only 3x8 is consistent with A’s declaration). Therefore A will not be able to subsequently make her final declaration.

Your reasoning is spot-on; you just need to
work through it carefully with pencil and paper (or program a computer to apply the logic).

16

You’ll have to spell that out for me.

If it’s not ‘dog’, then what letter do you think each one got?

17

I believe “cat” works as well.
Al has “c”, so he knows instantly, since c is a single-instance letter
Bernie has “t” - as soon as Al says he knows, Bernard also knows, because the only other possibility for him, “tag”, has no single-instance letters
Cheryl has “a” and knows that the only two words that could produce the pattern “Al yes then Bernie yes” are either cat or dog

18

[spoiler]Albert = c
Bernard = t
Cheryl = a

Bernard and Cheryl know that Albert must have had c,o,h,s,x, or i, based on his statement.

Bernard previously was waffling between cat and tag, but tag doesn’t have any of Albert’s possible letters. Cat it is.

Cheryl previously was waffling between cat, has, max, and tag, and Albert’s statement only eliminated tag. But Bernard’s statement eliminated has and max, because if it was ‘has’ then Bernard would have had h or s and wouldn’t have had to wait for Alfred, and if it was ‘max’ then Bernard would have had m and wouldn’t have been able to pick between max or dim based on Alfred’s statement. So cat it is.[/spoiler]

19

Yeah, I didn’t get all the way up to the top with even numbers. I don’t think 41 works: odd numbers are the sum of an odd number and an even number, and the only even number that’s prime is 2, and 41-2=39, and 39 is composite. But you’re right about 55.

Do 82 and 92 have two prime addends less than 50?

So if I’m thinking about it right, the two numbers:

  1. Must add up to a sum that has no two addends that are both prime.
  2. Must multiply to a product whose possible factors only have one way to meet rule 1.
    But I’m having trouble wrapping my head around that. I’ll keep thinking about it.

Thanks, I love this puzzle!

20

Hmm.

[spoiler]Strictly speaking, are we told that Bernard had to wait until Albert had spoken, or only that he did wait until Albert had spoken?

As far as I can tell, we aren’t told that he was asked to announce whether he knows the word; sure, that was the case in AHunter3’s problem, where people take turns admitting that they can’t answer the question that gets asked — but, here, what, he at some point happens to say something that’s true, even though for all we know nobody ever actually bothered to ask him about it?

In which case, he could have an ‘s’, and Albert could have an ‘h’, and they could each tell the truth; and Cheryl — who’d know of that possibility — would ponder her ‘a’ and think, “well, crap; there are multiple possibilities.”[/spoiler]