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# Baldyville

78 replies to this topic

### #51 Sue

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Posted 26 March 2008 - 09:22 PM

Flawed logic. The answer is infinity-1.

No one has EXACTLY 517 hairs. But you can have someone with 0, 1, 2,...516, 518.... eleventy billion... infinite

So given that your question allows for infinite hairs, there can be infinity-1 people.

Nooo! Look:

For a minute, let's pretend the "No person has exactly 518 hairs" condition DOESN'T EXIST.

If there were 519 people, no-one could have more than 518 hairs.
Each of the 519 people must have between 0 and 518 hairs.
No two people can have the same amount...
And there are only 519 possible values between (and including) 0 and 518
So this works....but every value has been used.

BUT...

We bring back the condition.

The value "518" cannot be used...

So if there were 519 people, let's count all the possible integers from 0 to 518 not including 518.

Oh. Now there are only 518 possible values.

518 values for 519 people...

And "518" is gone for good...

So there will be 519 values for 520 people -- doesn't work.
And 520 values for 521 people -- doesn't work.
And 1000000 values for 1000001 people -- doesn't work.

Therefore, you cannot go past 518.
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### #52 largeneal

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Posted 26 March 2008 - 11:14 PM

So an inhabitant can not have 518 hairs. It is no where specified that all inhabitants have to have any specific integer values. If there are 1000 inhabitants, then how many different values exist for hair count? 1001, because 0 is an option. Therefore, all 1000 inhabitants can have a different number of hairs AND include the rule of not have 518 hairs. For this, I say the answer is infinity since you have the extra number "0" to allow for hair count.
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### #53 largeneal

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Posted 26 March 2008 - 11:18 PM

Crap - I need to read the puzzle putting up a post sometimes. I retract my argument, because by rule #3, you lose that top endpoint value (so 1000 people, you not only lose the 518 value, but you lose the 1000 value, so you only have 999 possible values for 1000 people).
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### #54 Lost in space

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Posted 26 March 2008 - 11:18 PM

Perhaps they could have rabbits tatooed on their heads, people will think they are hares (hairs).
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### #55 Talent

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Posted 16 April 2008 - 03:56 PM

Ok, someone explain to me why there can't be infinity inhabitants? As 'there are more inhabitants than any inhabitant's hair in the town' = infinity, 'No inhabitant has exactly 518 hairs' = that's wonderful news(sarcasm), 'No two inhabitants have the same number of hairs on their head' = a useless fact as well.

It doesn't say there is a limit on 518 hairs, just that no one has exactly 518. Great?

Can anyone convince me otherwise?

So an inhabitant can not have 518 hairs. It is no where specified that all inhabitants have to have any specific integer values. If there are 1000 inhabitants, then how many different values exist for hair count? 1001, because 0 is an option. Therefore, all 1000 inhabitants can have a different number of hairs AND include the rule of not have 518 hairs. For this, I say the answer is infinity since you have the extra number "0" to allow for hair count.

Nice to see someone agrees.

I love how people start to insult others for not seeing their point of view, which is wrong so far as I can see.

Edited by Talent, 16 April 2008 - 03:57 PM.

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### #56 DebbieKoolBabez

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Posted 24 April 2008 - 07:54 PM

518
and one must be a bauldy otherwise no-one can live in the town
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### #57 momentrider

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Posted 01 May 2008 - 03:32 PM

The ANSWER is 518. you cant have 519 or more. We all can agree 0-517 are valid amounts of hair. There can't be 518 so your next amount would be 519 but that number and any number larger will not work Take number 519 for example. There is 0-517 and 519 which makes 519 people total and this is the same number of hairs on the head of 519. However rule three states there are more inhabitants than any inhabitants hair in town. Not the same or less therefore numbers 519 and up do not work.
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### #58 TySim

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Posted 02 May 2008 - 01:35 PM

T = Total Inhabitants
H = Maximum Hair Count (T-1)

Th = total unique integers > -1 and <=H
By definition Th >= T (if you have less combinations than inhabitants, you are in violation - you can however have more based solely on this one rule for example 1 inhabitant has 3 hairs does not violate the uniqueness rule (although it does violate the 3rd rule)

So right now we have rule 1 covered (Th disallows duplicates) and Rule 3 ( Because H = T-1)
Rule 2 means that Th can not include 518 in the calculations.
To show mathematically this means:
If Th <= 518 it is calculated as H + 1 (0.1.2.3....H)
If th >518 it is calculated as H (0.1.2.3...516.517.519.520....H)

When Th = H we have violated one of the conditions of the problem.

Th >=T
H = T-1
Since Th = H (in our hypothetical) we can sub and get
T-1 >= T which we know to be false.

At th less than or equal to 518 we have
Th >=T
H = T-1
Th = H+1
H+1 >= T
(T-1)+1 >= T
T >= T which is true.....

Therefore 518 total combinations is 517 max hairs and 518 max inhabitants

I could have done it more accurately with H <= T-1 and do progession, but the problem called for Max, and therefore dictates H = T-1 (max possible inhabitants)

Edited by TySim, 02 May 2008 - 01:43 PM.

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### #59 Theo

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Posted 02 May 2008 - 03:55 PM

Baldyville - Back to the Logic Puzzles
These are the conditions in Baldyville:
1. No two inhabitants have the same number of hairs on their head.
2. No inhabitant has exactly 518 hairs.
3. There are more inhabitants than any inhabitant's hair in the town.
What is the highest possible number of inhabitants?

Spoiler for Solution

Solution is simple. Let me try to explain this as well:

Always keep in mind that you are trrying to get the maximum nuber of inhabitants.

First condition is esy to understand. No 2 can be equal. Second condition is important, cant have exactly 518, keep in mind for the third condition. Third condition says to get the maximum number of inhabitants, knowing that the number of inhabitants is higher than the maximum number of hairs, on any one person in town.

True it does not say the numbers dont have to be consecutie, for example, inhabitant 1, 0 hairs. Inhabitants 2, 1 hair. and so on. People do it like this just to keep orginized in their brain.

Remember in order to meet the condition 3's requierments, there has to be a bald person.

You can say, inhabitant 1 has 20 hairs. But remember condition 3, you will now have at least 21 inhabitants in baldyville. So you would continue, inhabitant 2 has 15 hairs, inhabitant 3 has 9 hairs and so on, until you meet the requerment of condition 3.

Now, seeing that we can not use the number 518, saying there could be an infinit number if inhabitants would be incorrect. Say there is 1 inhabitant with 635 hairs on his head, you would now need at least 636 inhabitants to meet the requierments of condition 3. You are forced to use every number in the number line in order to get your answer. But seeing that we can not use 518 for the number of hairs on anyones head, any number greater 518 will give you a matching number, because of the fact we can not use 518 and we have to skip over it.
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### #60 Deviate Hampster

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Posted 02 May 2008 - 09:44 PM

The aside in the solution makes the given solution false. If one inhabitant is bald, then any number of non-repeating hair counts would equal the population of the city. If no one is bald, and there is no one with 518 hairs, then there could be an infinite number of people in the town, given that the person with the most hair has one less hair the the number of people in town

I disagree, consider having 517 people lined up with the number of hair on their heads in numerical order starting with one and going to 517. If you were to add one more person who is bald then you have 518 people with the largest amount of hair on one person 517. Thus you have one person bald, 518 people and the largest hair count is 517. For your second statement, if no one is bald and you have 517 people, and you add one more with 519 hairs (which is your only option) then you have 518 people and the largest hair count is 519. Your solution therefore, unfortunately, does not hold.
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