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!