Challenge Riddles and Puzzles: Extend the following to a valid equation

[fresh_42]

1. Extend the following to a valid equation, using only mathematical symbols!

Example: $1\; 2\; 3 \;=\; 1 \longrightarrow - (1 \cdot 2) + 3 = 1$. Solutions are of course not unique.

$9\;9\;9\;=\;6$
$8\;8\;8\;=\;6$
$7\;7\;7\;=\;6$
$6\;6\;6\;=\;6$
$5\;5\;5\;=\;6$
$4\;4\;4\;=\;6$
$3\;3\;3\;=\;6$
$2\;2\;2\;=\;6$
$1\;1\;1\;=\;6$
$0\;0\;0\;=\;6$

 

[DavidSnider]

Spoiler: Spoiler

-(9 / sqrt(9)) + 9
8 - sqrt(sqrt(8+8))
7-(7/7)
(6-6) + 6
5+(5/5)
sqrt(4)+sqrt(4) + sqrt(4)
3 * 3 - 3
(2 * 2) + 2
(1 + 1 + 1)!
((0!) + (0!) + (0!))!

 

[lpetrich]

For the three 2's, it ought to be (2 * 2) + 2. The others are correct.

 

[fresh_42]

2. On a round playing field with radius $R$, two players place coins of the same radius $r < R$ without moving coins. The first who doesn't find a place to position his coin has lost. For which ratios $R/r$ is the game fair?

 

[fbs7]

On a round playing field with radius $R$, two players place coins of the same radius $r < R$ without moving coins. The first who doesn't find a place to position his coin has lost. For which ratios $R/r$ is the game fair?

Wow, what an exquisite puzzle! It beats my little brain by a long, long distance - I look forward to reading the solution!

 

[fresh_42]

Wow, what an exquisite puzzle! It beats my little brain by a long, long distance - I look forward to reading the solution!

You are able to find the solution!

 

[Orodruin]

On a round playing field with radius $R$, two players place coins of the same radius $r < R$ without moving coins. The first who doesn't find a place to position his coin has lost. For which ratios $R/r$ is the game fair?

Spoiler

Trick question. It is never fair. The first player can always win by starting to play dead center and then playing to maintain a $C_2$ symmetry (i.e., playing diametrically opposed to the other player). By doing so the first player can always play.

 

[SSequence]

Nice ... Also, I think, strictly speaking, one also has to show that the game always terminates. It is somewhat trivial though (pigeonhole principle etc.).

But I think a good variation of the question might be that we aren't allowed to play the center position as first move. Or is that easy too?

 

[fbs7]

Trick question.

Brilliantly thought! Holy Choo-Choo - many times the simplest arguments are the most eloquent ones! People here are really exceptional - it's like waiting for a lightning, it comes sudden and when it comes it amazes you!

Even the nature of the question befuddled me - how can a non-probabilistic game that can't draw be fair? If a game can't draw (as the one above), then there must be at least one path starting of the 1st players 1st turn that ends in his win whatever the 2nd does, or there must be multiple paths for the 2nd player to always win whatever the 1st player does. So I couldn't reconcile the ideas of "deterministic", "non-drawing" and "fair".

Anyone has an example of a deterministic game that can't draw and is proven fair?

 

[fresh_42]

Anyone has an example of a deterministic game that can't draw and is proven fair?

No, but you assumed finiteness in your argument. If it is infinite at prior, and I think it has to be, i.e. of arbitrary finite length to be exact, then the argument for the existence of a strategy isn't obvious anymore. And another hidden assumption was, that we only consider two player games. You might say of course, I say: needs to be stated.

 

[fresh_42]

3. One line is wrong.

\begin{align*}
f(87956)&=4\\
f(82658)&=5\\
f(11111)&=0\\
f(46169)&=4\\
f(84217)&=3\\
f(57352)&=1\\
f(18848)&=7\\
f(27956)&=\;?
\end{align*}

 

[fbs7]

No, but you assumed finiteness in your argument. If it is infinite at prior, and I think it has to be, i.e. of arbitrary finite length to be exact, then the argument for the existence of a strategy isn't obvious anymore. And another hidden assumption was, that we only consider two player games. You might say of course, I say: needs to be stated.

Ah! I see! Thank you! Fbs7 = learned something today!

 

[fresh_42]

Nobody with an idea for number 3 in post 11? I know it's one of the kind preschoolers are better than mathematicians

 

[Charles Link]

Nobody with an idea for number 3 in post 11? I know it's one of the kind preschoolers are better than mathematicians

@fresh_42 I just looked at it for over 1/2 hour. Apparently the obvious must elude me.

 

[fresh_42]

@fresh_42 I just looked at it for over 1/2 hour. Apparently the obvious must elude me.

I found the clue as I read "preschoolers".

 

[Charles Link]

I found the clue as I read "preschoolers".

The preschoolers must be super-smart. I still don't see anything that stands out... I trust the function isn't random...

 

[fresh_42]

No, not random. However, formal languages might be of greater help than calculus. The puzzle was posed without the function $f$. I have added it in order to avoid endless discussions about sloppiness.

And don't make any nonsense if you will have figured it out. I seriously warn of

@WWGD should be quick at it.

 

[Charles Link]

Here is a joke I once heard, but I don't think it will help in solving this: "Why is 6 afraid of 7?" Answer: "Because 7 comes after 6, and 7 ate 9." LOL

 

[DavidSnider]

Is it as simple as "One line is wrong", meaning the line with ones is wrong?

 

[fresh_42]

Is it as simple as "One line is wrong", meaning the line with ones is wrong?

The question has two parts: which result has the last line and which line is the wrong one. It is simple, but there is a logical explanation.

 

[mfb]

Got it after the preschool hint. The wrong line:

Spoiler

The line f(57352)=1 is wrong. f(57352)=0

 

[Charles Link]

I tried a google: https://en.wikipedia.org/wiki/Preschool Am I dense? I still don't have clue...

 

[fresh_42]

I tried a google: https://en.wikipedia.org/wiki/Preschool Am I dense? I still don't have clue...

Maybe it is easier mathematically by the following claim (but I haven't checked the details, I only think it's true):

$f$ is a homomorphism from the Kleene star over $\{\,0,1,2,3,4,5,6,7,8,9\,\}$ into the half group $(\mathbb{N}_0,+)$.

 

[Charles Link]

I give up. I wonder if the answer may lie outside my realm of experience.

 

[fresh_42]

Since @mfb has solved number 3, here comes the next one:

4. In one of three urns there are two white balls, in another a white and a black ball, and in the third two black balls. The urns are labeled: one sign says WW, one WB and the third BB. But someone has switched the signs so that none of them specify the contents of the individual urns anymore.

One may take a ball from one of the urns, one after the other (without looking into the urn) until it is clear which urns contain which of the three ball pairs. How many balls do you have to take out at least to reach this goal?

 

[mfb]

I give up. I wonder if the answer may lie outside my realm of experience.

Forget numbers. Count circles.
f(0)=1, f(1)=0, ..., f(8)=2.

 

[fresh_42]

Forget numbers. Count [topological] circles.
f(0)=1, f(1)=0, ..., f(8)=2.

 

[Charles Link]

Forget numbers. Count circles.
f(0)=1, f(1)=0, ..., f(8)=2.

It's no wonder it eluded me=I don't feel so bad. It's quite clever, but somewhat un-mathematical.

 

[fresh_42]

It's no wonder it eluded me=I don't feel so bad. It's quite clever, but somewhat un-mathematical.

No it is not, just unconventional. It counts the genuses of concatenated topological geometric objects. On the left we have words of a formal language without grammar, and every letter has a weight. I wonder if we could make a baric algebra out of it.

 

[fbs7]

Got it after the preschool hint.

Holly Choo-Choo! You are a genius!

 

[Orodruin]

Since @mfb has solved number 3, here comes the next one:

4. In one of three urns there are two white balls, in another a white and a black ball, and in the third two black balls. The urns are labeled: one sign says WW, one WB and the third BB. But someone has switched the signs so that none of them specify the contents of the individual urns anymore.

One may take a ball from one of the urns, one after the other (without looking into the urn) until it is clear which urns contain which of the three ball pairs. How many balls do you have to take out at least to reach this goal?

Spoiler

One ball.

Let small letters denote true content and capital letters labelled content.

The urn WB is not wb, so it must be bb or ww. Taking a ball out of it will therefore tell you which it is. If we get white out of WB, then it is ww. We are now left with the urns WW and BB. We know that those labels are wrong and that one of those urns is bb != BB. Therefore bb = WW and wb = BB. The corresponding argument goes if we get black out of WB, but exchanging the roles of the blacks and whites.

This can also be seen as follows: There are two possibilities for labels:
{WW = wb, WB = bb, BB = ww} or {WW = bb, WB = ww, BB = wb}
Clearly they cannot be discriminated without taking a ball out. If we take a ball out of WB, it will distinguish between the two possible cases.

 

[fresh_42]

5. Let's have some fun. We will perform a 4 step manager test. I will post 4 questions step by step, i.e. question 2 if the first will be answered correctly etc. It is an old joke, but here I will add a "lessons learnt" comment after each question.

a. How do you get a giraffe into a fridge?

 

[mfb]

Open the fridge, put the giraffe in, close the fridge.

 

[fresh_42]

Open the fridge, put the giraffe in, close the fridge.

This question investigated whether you tend to find much too complicated solutions of very easy problems.

5.b. How do you get an elephant into the fridge? (@mfb please pause on this one.)

 

[jbriggs444]

This question investigated whether you tend to find much too complicated solutions of very easy problems.

5.b. How do you get an elephant into the fridge? (@mfb please pause on this one.)

Open the fridge, take the giraffe out, put the elephant in, close the fridge.

[Zoning into the apparently expected level of abstraction]

 

[fresh_42]

Open the fridge, take the giraffe out, put the elephant in, close the fridge.

This question investigated whether you are aware of the consequences of your doings.

5.c. The Lion King is holding up his yearly conference of all animals. However, one is missing. Which one?
(this time @jbriggs444 please pause)

 

[Orodruin]

5.c. The Lion King is holding up his yearly conference of all animals. However, one is missing. Which one?

The elephant. He is in the fridge.

 

[fresh_42]

The elephant. He is in the fridge.

This question checked your memory.

5.d. You have to cross a river where dozens of crocodiles live. How do you resolve this situation?
(@Orodruin to pause).

 

[jbriggs444]

This question checked your memory.

5.d. You have to cross a river where dozens of crocodiles live. How do you resolve this situation?
(@Orodruin to pause).

Oooh oooh, can I go again?

 

[fresh_42]

Oooh oooh, can I go again?

Yes, time to finish this one.

 

[jbriggs444]

Yes, time to finish this one.

Use the bridge.

 

[fresh_42]

Use the bridge.

Nope. No bridge anywhere.

 

[Orodruin]

Nope.

I know! I know!

 

[jbriggs444]

Oh silly me. The crocs are at the annual get together.

 

[fresh_42]

Oh silly me. The crocs are at the annual get together.

Yes, you swim.

This question was about whether and how fast you learn from mistakes. (Although we had none.)

According to a study by Andersen Consulting, around 90% of all tested executives worldwide have answered all questions incorrectly.

On the other hand, several preschool children had correct answers.

Andersen claims that this clearly disproves the thesis that executives have the mental faculties of a four-year-old.

 

[fresh_42]

6. How many years of life did somebody complete on the first day of the year 30 AD, if he was born on the last day of the year 20 BC?

 

[mfb]

More of a history question - there was no year 0. The last day of 1 BC he was 19 years old, the next day was the first day of 1 AD. The last day of 1 AD he was 20 years old, the last day of 29 AD he was 49 years old -> 49.You find a historic letter dated 5 BC talking about an event that happened 20 BC. What is wrong?

 

[fresh_42]

More of a history question - there was no year 0. The last day of 1 BC he was 19 years old, the next day was the first day of 1 AD. The last day of 1 AD he was 20 years old, the last day of 29 AD he was 49 years old -> 49.

Correct analysis, wrong count.

You find a historic letter dated 5 BC talking about an event that happened 20 BC. What is wrong?

The five is wrong. The description of the twenty is not specific enough to be wrong, since it could also mean fifteen years prior to the letter, i.e. relates the dating to the impossible gauge.

 

[mfb]

Correct analysis, wrong count.

Oops, 48 of course. 20+28 is 48, not 49.

The five is wrong.

Right. You can make the same puzzle with a letter e.g. 20 AD, but then you need to figure out when the AD counting was introduced (long after that), so the puzzle is easier with BC.

 

[lpetrich]

You find a historic letter dated 5 BC talking about an event that happened 20 BC. What is wrong?

That dating system did not exist at the time, so that document could not have referred to such dates. So such dates would be a present-day interpretation of whatever dates that it used.