Miscellaneous trivia, riddles, puzzles and other games

Problem 8

solution

The wording is a bit awkward, since the pronoun “she” is used ambiguously, but that particular word is crucial since I think the answer relies on the specific letters of the words corresponding to chemical elements. The key hint is calling it a science riddle.

“she” = S (sulfur, 16) + He (helium, 2) = 18
“cat” = C (carbon, 6) + At (Astatine, 85) = 91

“mouse” = Mo (molybdenum, 42) + U (uranium, 92) + Se (selenium, 34) = 168

Problem 9

solution

This one isn’t really a riddle, but rather just a simple grade school math problem.

The wording is a bit clumsy, but I am assuming that you are essentially just asking for the difference between the area of a circle and that of a square inscribed inside of it, divided by four (since we are just looking at a quarter of the picture, where it is just a right triangle inside of a quarter of a circle).

The area of the quarter circle is simply pi, and the area of the triangle is 2, so the area of the section between the hypotenuse and the arc is pi minus two.

The arc length is pi (a quarter of the circumference).

Problem 10

Solution

This type of problem just requires working out all of the possibilities and eliminating the cases that are excluded.

Out of all of the ways to assign numbers, there are only eight scenarios that Albert could have seen that would have left uncertainty to him about what number was on his forehead:

  1. B = 1, C = 2 (A = 1 or 3), B must be uncertain
  2. B = 2, C = 1 (A = 1 or 3)
  3. B = 1, C = 3 (A = 2 or 4)
  4. B = 3, C = 1 (A = 2 or 4), B must be uncertain
  5. B = 1, C = 4 (A = 3 or 5)
  6. B = 4, C = 1 (A = 3 or 5)
  7. B = 2, C = 3 (A = 1 or 5)
  8. B = 3, C = 2 (A = 1 or 5)

Essentially, there are four observable pairs {(1,2), (1,3), (1,4), (2,3)} that would leave uncertainty. Only in scenarios 1 and 4 would Albert know with certainty that Bernard does not initially know (note that Albert’s statement immediately reveals to Bernard what his number was). In the other scenarios, it is possible that Bernard observes a pair that makes his number certain.

Expanding these scenarios, we have four total cases left:

  1. B = 1, A = 1, C = 2, (excluded, since C would have known she had 2)
  2. B = 1, A = 3, C = 2 (Cheryl was initially uncertain, could have been 4)
  3. B = 3, A = 2, C = 1 (Cheryl was initially uncertain, could have been 5)
  4. B = 3, A = 4, C = 1, (excluded, since C would have known she had 1)

However, cases 1 and 4 are excluded since Cheryl would have known before Albert spoke. In the two remaining cases, Cheryl does not have the largest number, so she wrote “no” on the paper.

The information in the bonus question reduces the possibility to only case 3:
A = 2, B = 3, C = 1
since in case 2, it is not possible to move just one person to the front to get the numbers in ascending order.

Another puzzle

The “blue-eyed islanders” puzzle is really neat

xkcd presentation: https://xkcd.com/blue_eyes.html
solution: https://xkcd.com/solution.html

Terence Tao wrote a detailed blog post: https://terrytao.wordpress.com/2011/04/07/the-blue-eyed-islanders-puzzle-repost/
Further discussion: https://terrytao.wordpress.com/2011/05/19/epistemic-logic-temporal-epistemic-logic-and-the-blue-eyed-islander-puzzle-lower-bound/

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I present to you another epistemic puzzle, my personal favourite:

Problem #12: A Rational Problem


Albert and Bernard are being tested by their teacher once more. He gives them both a little note with a number on it, and he tells them both numbers are different from each other and of the following form for n and k natural numbers larger than 0:

1/2n + 1/(2k+1)×2n


Then the following conversation happens:

Teacher: "Who of you has the smallest number?"
Albert: "I don't know."
Bernard: "I also don't know."
Albert: "I still have no clue."
Bernard: "No, nothing for me either."
Teacher: "You two can keep going like this for a long time, but you will never find the answer like this."
Albert: "Ah, that's very interesting information, yet still I don't know if I'm the smallest."
Bernard: "Neither do I."
Teacher: "Again, continuing like this will not lead any of you two to know their number."
Albert: "That is truly remarkable, but I do not know if I have the smallest number."
Bernard: "Nor do I."
Albert: "Ah! But now I suddenly know who is smaller!"
Bernard: "Fantastic, then I know both of our numbers!"

Which numbers are written on Albert’s and Bernard’s notes?

3 Likes

I’ve hidden the name of a go player in the following picture (top 500, with his / her name spelled as it is on goratings.org).

Who is it?

(please don’t spoil the method to this one in the comments)

4 Likes

Problem #18: Matches on a Goban


I have a box with 21 matches and a single coin. I discovered that the coin fits (almost) perfectly inside one of the squares of the board, and that the matches have a length that they take up three squares if placed vertically or horizontally.

Find a way to put all the matches and the coin on the board such that each square is touched by exactly one match or has the coin on top of it.

For example, one could start as in the following picture:

[N.B.: It is not a trick question: the matches don’t have to lie partly outside of the board, nor do any of the matches have to cover more than three squares, nor do they have to lie in a position that isn’t horizontal or vertical]


Bonus

Is there a way to fill the board with the matches when the coin is placed as in the picture above?

I’m stumped by this one:

Holmes once had a visit from an old acquaintance whom I had never met before. I found them chatting freely in a way he seldom did with me as I came in to offer them some tea.

“And how are your children?” Holmes was asking. “You have three, if I remember rightly, although I must admit I don’t quite recall their ages.”

“You always enjoyed deduction, didn’t you, old chap?” his acquaintance replied. “What if I told you that the product of their ages was 40?”

“That’s not quite enough information for me to deduce their ages,” said Holmes.

“Alright, well I shall add that the sum of their ages is the number of years we’ve known each other.”

Holmes considered this. “I’ll still need a tad more.”

“Finally, the youngest was our first summer baby, born in July.”

“Ah, I see.”

Holmes suddenly turned to me. “Why don’t you tell this gentleman the ages of his children, then, Watson?”

I balked. “But I don’t know how long you’ve known each other!”

“That doesn’t matter, Watson! You now have enough information to deduce it.”

And indeed I did. How old are the three children?

3 Likes

Do you want this published in the group news? We haven’t updated in a long time

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Sure, I don’t mind that you do. I’m not the author of this riddle, of course, and I don’t know the answer.

I love it! It reminds me of Conway’s bus wizards.

Last night I sat behind two wizards, Azemelius and Bartholomew, on a bus. I heard this conversation:

Azemelius: I have a positive integer number of children, whose ages are positive integers. The product of their ages is my own age, and the sum of their ages is the number on this bus.

Bartholomew (looking at the number of the bus): Perhaps if you told me your age and how many children you had, I could work out their ages?

Azemelius: No, you could not.

Bartholomew: Aha! At last I know how old you are! (Bartholomew had been trying to find Azemelius’s age for a long time.)

What is the number of the bus?

(shamelessly copied from this site)

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I still can’t get over that he passed away. RIP

Here’s one:

Alice and Bob, two perfect logicians, are captured by an evil logician, who believes that the world should be rid of all those who are not perfect logicians.

He puts them in seperate prison cells, with no communication to one another. However, he offers a way to escape.

He gives Alice the sum of two distinct positive integers greater than 2 and less than 100, and Bob the product of them.

He explains:

“I gave Alice the sum, and Bob the product, of two distinct positive integers strictly between 2 and 100. Every night, at 8:00, I will come to each of you, first Alice, then Bob, and ask if you know what the two numbers are. You may pass, or guess. If one of you guesses correctly, you will both be freed. If one of you guesses incorrectly, you will both be stuck here forever! Good luck!”

The first night, Alice passes, and then Bob passes. Bob wasn’t sure if Alice would be able to figure it out, but the next night, Alice does get it right and both are freed.

Bob says, “So the sum was odd, right?” Alice says yes.

Assuming Alice and Bob would escape at the first possible opportunity, and do not take any chances, what are the numbers? What would they be if the sum was even (2 solutions)?

(edited for clarification)

(another edit: I forgot to state that Bob wasn’t sure whether Alice would know. That’s quite an important point, otherwise there would be 2 more solutions for both the odd riddle and the even riddle.)

4 Likes

For clarification, is it

  • 2 < n,m < 100

or

  • 2 ≤ n,m ≤ 100 ?

It is 2 < n,m < 100.

In fact, it is even 2 < n < m < 100, since the numbers are distinct.

1 Like

Ok guys, so, there’s been quite a lot of activity during the past few days (thanks @yebellz for resurrecting our old forgotten thread). If you guys don’t mind, your contributions will be appearing as Problem of the week as we used to in our main group.

The protocol there was let each one simmer for seven days before posting the next one, so they’ll be appearing in the next few weeks. You can also check out the old problems in the news section, there are quite a few interesting ones.


The Egg Hunt

In celebration for this newfound interest, I’ve put a :old_key::moneybag:hidden treasure:moneybag::old_key: somewhere on OGS. It could be anywhere

it could be here (it is not though).

Although you could find it by guessing where it is—and there is more than one way to get there—I’ll give you a couple of things to help you get started:

  • Here are two codes: 2875*8, 313312; have at thee!
  • If you’ve ever had an indestructible old brick, it might help you decode them.
  • Remember, GIYF (search this if you don’t know, you’ll see why).

That’s it. Good luck!

2 Likes

Ok, here’s another one (Content warning: trypophobia).

The queen bee is building a honeycomb to accommodate her larvae. She builds following the usual rules:

  • Each cell is hexagon shaped;
  • Each cell has to be completely surrounded by other cells;
  • And she builds 3 cells around each intersection.


Image: “Honeycomb” by justus.thane is licensed under CC BY-NC-SA 2.0

However, sometimes she can make mistakes: she may put a pentagonal cell instead of a hexagonal one. This is the only kind of mistake she allows herself to make though; barring that, any botched nests would be thrown away.

Eventually, she realizes she’s finished—and in her first try!


  • How many mistakes did the queen make?

  • What is the largest number of larvae that is impossible to accommodate exactly one-to-one on each cell?


Edits for clarification:

Very important omission: Hexagons and Pentagons need not be regular
Second question: some nest sizes are impossible to build. What is the biggest one?

2 Likes

So Vsotvep (or do you want this link?) asked me the other day something along the lines of “What if the evil logician is sadistic? How long can he keep Alice and Bob in, if they can still escape?”

So I figured that out.

Same scenario as before. The evil logician captures Alice and Bob and puts them in separate prison cells. He tells Alice the sum of two numbers 1 < m < n < 100 (makes no difference) and Bob the product. Each night, he comes to Alice, then Bob, asking if they know the numbers. They may pass or guess. If one guesses correctly, they will both instantly be freed. If one guesses incorrectly, they will both be stuck forever.

For the first 6 nights, both pass, and nothing happens. On the 7th night, Bob guesses correctly and both are free.

What are the numbers?

2 Likes

I shared this problem with a friend and after giving the answer he said the question is flawed. To quote:

Hmm, “first summer baby” rules out twins, but how do you rule out kids born 9-12 months apart?
E.g. suppose kid B was born in September 2010 and kid C was born in July 2011. In August 2013 they both will be 2.
But I guess if information was sufficient for holmes, this dialogue did not happen during July-October when such situation can exist?

1 Like

Yes, this was also my conclusion. We know that information was sufficient for Holmes, thus we know the conversation must have been held outside of July - October.

I don’t see it as a flawed puzzle, just one where the truth is a little more obscure than apparent at first glance.

2 Likes