Assuming we cannot know the position we will have at the final circle, and that memorizing a number for each other prisoner with us can be tricky.

Just start by chosing one of the prisoners before going in the final circle. We will call him Mr. A

Method 1: CHANGING POSITIONS SIGNALS

Now all we need to so is to look at his number and then by counting clockwise starting from his position in the circle, count as many people as the number on A's hat and that guy raises his hand.

Example: if A has 13 on his hat, the 13th person on his left will raise his hand first. (Now if the number is 20, no hands will be raised after 5 seconds.)

After that, everybody will focus on the second guy to the left of Mr. A and the corresponding person to his number raise his hand this turn.

After 20 rounds, we have effectively told each prisoner the number on his hat.

Method 2: FIXED POSITIONS SIGNALS

In this method, I'm trying to avoid having the prisoners count their new position after each round.

So we can start by the guy on the left of Mr. A being #1 and continuing clockwise till #19. This number represented by the prisoners position will remain till the end of the exercise.

After showing Mr. A his number like in the previous method. Mr. A will then take #1 at the second round, #2 at the third... (In fact taking the place of the guy on which we are focusing in this round) and he will raise his hand if the prisoner's starting position matched the number on his hat by some coincidence.

After 20 rounds, we have effectively told each prisoner the number on his hat.

Oh and by the way,

1- Since the longest round would takes 5 seconds, they should be done before the 3 minutes are up. (and for that METHOD 2 would work a bit better in term on speed)

2- Since we have numbers from 1 to 20 and 20 prisoners (none of them having 20 as an index), there's no chance for all of them to raise their hands before the 20 rounds are up!

My reply comes a bit more than a year after the last reply.

Well, better late than never I suppose

Totally agree with DeeGee here.

If the first click STARTS the timer, Buzzy is the fastest clicker with 19 clicks in 20 seconds (19/20), followed by Emily with 9/10 then by Anthony with 4/5.

To better illustrate we need to consider the time between clicks and not the number of clicks in a period of time.

C = Click (no time)

T = Time period

For Anthony, he's clicking by following the series below

C T C T C T C T C in 5 seconds

4T = 5 seconds => T = 1.25

Anthony will then complete 40 clicks in 39xT => 39x1.25 = 48.75 seconds

Emily:

C T C T C T C T C T C T C T C T C T C in 10 seconds

9T = 10 seconds => T = 1.11 seconds

Emily will then complete 40 clicks in 39xT => 39x1.11 = 43.33 seconds

Buzzy:

C T C T C T C T C T C T C T C T C T C T C T C T C T C T C T C T C T C T C T C in 20 seconds

19T = 20 seconds => T = 1.05 seconds

Buzzy will then complete 40 clicks in 39xT => 39x1.05 = 41.05 seconds (THE FIRST TO GET TO 40 CLICKS)

I tried this as a test!!

I just threw the phone guide from the 3rd floor of my building and all 527 pages landed in a perfect pile!! Amazing, I wonder if people would actually pay money to see this incredible show!

I think I can get a job as a university professor now easy

What a long (and complicated) thread to read... So here's my contribution to this subject.

Going back to the original OP, the 2 players are masters of logic (and both love beer). They want to maximize the number of beers, they don't look to get more than each other. The word "sucker" will not affect their judgements either (emotional triggers do not affect masters of logic). In that case defecting would be stupid, illogical and an EMOTIONAL decision.

That being said, Masters of logic would always cooperate. Defecting even once is an emotional decision

Now I realize that, since this thread is taking so long, people are considering these masters of logic as regular people with feelings, who may seek EGO results and even vengeance.

Well, in that case, NOTHING BIG really changes either. The only chance of defecting would be at the last round (unless these masters of logic are plain stupid). If that round is defined before hand, they may both defect or both cooperate that round and the total number of beers they accumulated will change by a maximum of 1 beer.

Now if we don't take the "masters of logic" part literally, we will end up in what the humanity has been through since early history. And one of the 2 brothers will kill the second to maximize its winnings and wealth. We may even find a World War I or II scenario. In real life, this game will never end. People will end (being mortal), but the game won't, it will continue with our children and grand-children playing it.

That is why now we know that our decisions are much more emotional than rational, and some people (egoistic, short-term people), may at some point choose to defect.

I assume that this puzzle may have something to do with the situation the world is right now. And if you don't know what I'm talking about, just watch the news.

I don't intend to discuss politics, I hate this subject (everybody assumes they are on the "right" side of the conflict). But in my opinion, the intention behind the United Nations was to provide an arena for this game to be played rationally. (My opinion about the UN set aside).

Remember this: "We decide emotionally, then justify it rationally!"

In 2 words, No one wil be sober after that game ends

Is the second condition (divisible by 9) needed in this case? It goes without saying, doesn't it?

I thought they arrested the murdereD as by the OP.

A committee of five is to be chosen from a group of nine people.

Determine the total number of ways this can be accomplished, given that:

(i) Ted and Chuck must serve together or not at all, and:

(ii) Hannah and Cindy refuse to serve with each other.

I have no idea how to write the combinations notation in this box so explenation first:

7C5 means the number of combinations of 5 people in a group of 7

(i) Both in + both out = 7C3 + 7C5 = 35 + 21 = 56 ways

(ii) 2xonly one in + none of them in = 2 x 7C4 + 7C5 = 2x35 + 21 = 91 ways

(i) and (ii)

= None of the four characters in + the two men in none of the girls in + two men in one girl in x 2 + One girl in the two guys out x2

= 5C5 + 5C3 + 5C2 x 2 + 5C4 x 2 = 1 + 10 + 10 x 2 + 5 = 36 ways

I hope I didn't make any mistakes rushing into this...

Ooops!

$66.67 (rounded to the nearest penny from $[200/3])

Rule #1:

Never be the first to go for the average in your counter offer!!!!

In the previous solutions, it was assumed that the seller would start by asking for $75

In my opinion, the seller being more professional and experienced would have waited untill Sean agreed on paying $75, then he would ask for the new average.

superprismatic was right on both accounts I think.

This OP did produce a single solution!

This made me think of a twist:

Would we always have one single solution had K Sengupta chose any other triplet of 3 jobless candidates in that interiew?

N,Y,N,Y,N

If not, what is(are) the triplet(s) that would give more than one possible solution in that case?

Remember, the right sequence is fixed in the spoiler above

The jobless candidates have 2 or less right answers

I hope it's ok with K Sengupta that I posted this.

I hope you find it interesting to prove/disprove.

At least, it was for me!

I've been sitting here reading all these posts about ogre being stupid and I've had it!

I think you all owe my wife an apology!

In her own words: "Omf kuf tuug gsut gout!"

You guys should be ashamed of yourselves.

Proposed answer follows:

The key is to get to shore before the ogre meets you, since once there you can outrun him. I believe the solution is to begin by rowing in the exact opposite direction of the ogre. He can then be expected to begin running around the lake one way or the other. Once he commits, you then alter your course to continue rowing 180 degrees away from wherever he is at the moment, and you continue this until you reach shore. If he continues running in the same direction (clockwise or counterclockwise), your path will look somewhat like an arc from the center to the shore. If he reverses direction, your path will appear more like a zigzag.

I won't bother calculating exactly how close he is when you reach shore, but he'll have run sufficiently more than 1/2 of the way around the roughly 6.28 mile circumference of the lake.

I only have one issue with these solutions,

Since there is no information about my rowing speed in the OP (usually much less then my running speed), the proposed solutions may leave me running in circles, never getting to shore!! A very tight giant spiral will be my final trajectory.

In fact, there must be some minimum rowing speed where I can reach the shore with the ogre at least a feet away from me, so that I can outrun it. This would make a nice puzzle!

I think your count is too small.

You are correct that there are 24 rigid-motion

symmetries of the cube. But, in dividing

6⁶ by 24 you are tacitly assuming

that every coloring is counted 24 times.

This is not true for coloring the entire cube

with a single color, for example. Cubes which

are colored with 2 colors also have fewer than

24 symmetric brothers. So, to get a correct

count, you would have to break the problem up

into sets of colorings which have the same

number of symmetric copies. This seemed difficult

to me, which is why I wrote a program to brute

force the thing.

I think I agree...

And the Brain wins again... What a shocker!

I'm not sure why but...

6^6/4^2 = 2916

6^6 = total number of colored cubes with no conditions

4^2 = the number of different (identical) rotations for each cube.

Although I learned not to use my instincts in probabilities.

But this feels like the ultimate battle between Emotions & Machine (Computer code)

And the winner is ... again.

What paradox?

Unfortunately I haven't written a program but I think it would be quicker to do by hand anyway.

0000 1100 2200 3300 4400

0011 1111 2211 3311 4411

0022 1122 2222 3322 4422

0033 1133 2233 3333 4433

0044 1144 2244 3344 4444


0110 1210 2310 3410 4010

0121 1221 2321 3421 4021

0132 1232 2332 3432 4032

0143 1243 2343 3443 4043

0104 1204 2304 3404 4004


0220 1320 2420 3020 4120

0231 1331 2431 3031 4131

0242 1342 2442 3042 4142

0203 1303 2403 3003 4103

0214 1314 2414 3014 4114


0330 1430 2030 3130 4230

0341 1441 2041 3141 4241

0302 1402 2002 3102 4202

0313 1413 2013 3113 4213

0324 1424 2024 3124 4224


0440 1040 2140 3240 4340

0401 1001 2101 3201 4301

0412 1012 2112 3212 4312

0423 1023 2123 3223 4323

0434 1034 2134 3234 4334

Wow that took a while. Maybe I should write a program after all.

Also have you inputted the numbers correctly into your formula? It looks like the denominator should be 4.

I'm forgetting something. It seems there's something I'm not counting, or more precisely, removing more than once! See, my final formula in post#7 was: d^n/n. Very close to yours, but where did I go wrong?

Eureka! I simply multiplied by "d" instead of "n"... This would've got me the same results.

I am now sure that your formula stands... Great job...

Thanks for the effort...

For example the 27 4-digit base 3 IDs you can make are:

0000 1100 2200

0011 1111 2211

0022 1122 2222

0110 1210 2010

0121 1221 2021

0102 1202 2002

0220 1020 2120

0201 1001 2101

0212 1012 2112

Feel free to check any of this. If I have made a mistake I'd like to know.

Man do I wish you picked another example:

In this particular example, our 2 different formulas yield the same result!!!

Psychic N = d^(n-1) = 3^3 = 27

Rool's N = d^n/n = 3^4/3 = 3^3 = 27

What a coincidence!!

Still this doen't mean I am not convinced of your answer...

So if you've already written a program, can you please develop a 4-digit number base 5??

In fact I'm rather sckeptical about my formula since we cannot guarantee it's going to be a whole number in every case!!!!!!!!! And even more sckeptical about the spelling of the word skeptical!

I'm just trying to figure out the error in my , usually flawless, logic... and humility....

I think I underestimated the number there...

In fact, in my method, every ID# was removed 12 times!!!!

and so the final formula would be multiplied by 12:

(12^10/(12*10))*12 = 12^10/10

or 12^n/n n being the length of the ID#

in the 4-long ID# it's:

(12^4/(12*4))*12 = 12^4/4 = 5184

But still didn't get the 1344 Number there...

It seems I'm missing something...

Better sleep on it!

This was my system

Simply group consecutive pairs of digits (i.e the 1st & 2nd, 3rd & 4th etc) and these pairs will always have the same digit as each other. This would make it like a five digit number but it would actually have 10 digits and there would always be 2 differences from every other number. This gives 12⁵ combinations. I'm not sure if this is the maximum though.

At least one of us has misunderstood the OP

The way I saw it was that

ID#1: 1234567890

and

ID#2: 1234567891

and

ID#3: 2234567890

ID# 1,2 & 3 are accepted as one same ID (they only differ at 1 place!)

So,

You did find some identical IDs but not all of them, in fact you only studied one particular case instead of the millions others...

Example:

ID#1: 1122334455

and

ID#2: 2222334455

are actually different

but so is

ID#3: 1212334455 with 2 differences!

and your 12^5 does not cover the likes of ID#3

In my formula, the 12^10 possibilities are grouped into 10*12 groups of identical IDs => Number = 12^10/(10*12)

BTW I MISSED THE PARENTHISIS IN MY LAST POSTS...

Please remember that this is based on how I understood the OP...

JMust realized an error in my previous post and too late to change it.

Corrections in RED:

10^10 = Total possiblilities with 10 digits

Now for each one of the above IDs, the number of similar IDs (ones with only 1 digit changed) is 10*10 (each digit can be changed 10 different ways ALONE, either that digit OR another, so 10*10 and not 10^10)

Therefore:

TOTAL DIFFERENT IDs = 10^10/10*10 = 10^8

I just realized the duodecimal thing so:

12^10 = Total possiblilities with 10 digits of duodecimal thing

Now for each one of the above IDs, the number of similar IDs (ones with only 1 digit changed) is 12*10 (each digit can be changed 12 different ways ALONE, either that digit OR another, so 12*10 and not 12^10)

Therefore:

TOTAL DIFFERENT IDs = 12^10/12*10 = 12^9/10

Not a power of 12 I might add...

Hope I'm right...

As for the below

I exhausted on 4-long duodecimal numbers. I get 1344=2⁶*3*7, not a power of 12.

The numbers of IDs should be 12^4/12*4 = 432

Are you sure you're not counting similar IDs more than once? More than 3 times even?

10^10 = Total possiblilities with 10 digits

Now for each one of the above IDs, the number of similar IDs (ones with only 1 digit changed) is 9*10 (each digit can be changed 9 different ways ALONE, either that digit OR another, so 9*10 and not 9^10)

Therefore:

TOTAL DIFFERENT IDs = 10^10/9*10 = 10^9/9

I just realized the duodecimal thing so:

12^10 = Total possiblilities with 10 digits of duodecimal thing

Now for each one of the above IDs, the number of similar IDs (ones with only 1 digit changed) is 11*10 (each digit can be changed 11 different ways ALONE, either that digit OR another, so 11*10 and not 11^10)

Therefore:

TOTAL DIFFERENT IDs = 12^10/11*10

Not a power of 12 I might add...

Hope I'm right...

By moving one of the following digits, make the equation correct. 62 - 63 = 1

HAYET ALBEH!!!!!!

Either my sister or my cousin's cousine...

Or even...

Nope, that's about it

I love Calves' milk with my morning cereals...

If no more than 3 bills of each should be used,

1, 4, 16, 64 will cover up to 255

Adding 256 will cover up to 1023

Adding 1024 we can go as high as 4096

...

I think I've seen these numbers somewhere before... Hmmm...

[hummm this one's hard]'s siamese brother was actually the one behind the idea for the second post on this thread...