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# Baldyville

### #11

Posted 03 July 2007 - 04:51 PM

Take these 2 conditions -

1) There are more inhabitants than any inhabitant's hair in the town. Thus, no person can have hair which number more than the total amount of people.

2) No two people have the same amount of hair.

And the 3rd condition -

3) No inhabitant has exactly 518 hairs

Now, if we consider that there is only 1 person in baldyville, he has 2 be bald, according to condition 1. If there are 2 people, 1 of them has to be bald and the other should have 1 hair. We can continue this sequence, till say 20, where there will be 20 people. Thus, the max amt of hair that one can have will be 19 (cond. 1) and, since no 2 people can have the same amount of hair (cond 2), there need to be 20 distinct amounts of hair. This can only be satisfied by the values 0,1,2,...,19.

Now, let us consider 518 people. If we extrapolate the previous example, there will be 517 diff amounts of hair, ranging from 0 to 518. But, if a 519th person to be present, there has to be atleast 1 person with 518 hairs. This goes against the 3rd condition, and hence, the max number of people is 518.

### #12

Posted 12 July 2007 - 04:56 PM

### #13

Posted 26 July 2007 - 08:55 PM

That's not true. It says there are more people than on ANY inhabitants head. Therefore, the logic is incomplete, and there are infinite possibilites for the answers.

Those who are saying 518 or 519 are operating under a false assumption that the number of people is limited, or that the hairs are in a numbered sequence.

What you are providing are possible answers, not an absolute answer proven by logic.

Can someone dispute this?

### #14

Posted 30 July 2007 - 04:43 PM

In other words, for the solution to be true, one more stipulation is required "The number of hairs on the inhabitants hairs is consecutive" or the like.

Thoughts?

### #15

Posted 30 July 2007 - 05:01 PM

Rules:

1) No two inhabitants have the same number of hairs on their head. (someone may be bald - 0 hairs)

2) No inhabitant has exactly 3 hairs on their head

3) The maximum number of hairs on any one person's head may not be greater than or equal to the number of inhabitants. (simply re-wording the original rule)

- So say there is only one inhabitant, then that person must be bald due to rule 3.

- Now say there are 2 inhabitants. Then one must have 0 hairs, and the other must have exactly 1 hair (they can't be the same by rule 1). If one of them had 2 hairs or more, rule 3 would be violated.

- Now say there are 3 inhabitants. Then one must have 0 hairs, another must have 1 hair, and the third must have 2 hairs. If one of them had 3 hairs or more, rule 3 would be violated.

- Now say there are 4 inhabitants. Then one must have 0 hairs, another must have 1 hair, and another must have 2 hairs. That leaves one more inhabitant who cannot have 0, 1, or 2 hairs on their head (rule 1). Rule 2 states they cannot have 3 hairs on their head, and rule 3 states they cannot have more than 3 hairs on their head. Therefore this situation cannot exist according to the rules since, by assumption, negative numbers of hair cannot exist.

- Now say there are 5 inhabitants. Then one must have 0 hairs, another must have 1 hair, another must have 2 hairs, another must have 4 hairs. But the 5th inhabitant cannot have 3 hairs (by rule 2), and they also cannot have 5 or more hairs. Therefore this situation cannot exist either.

Continue on to infinity and you'll realize that the max number of inhabitants for this problem is 3. So whatever number is specified in rule 2 (whether it's 3 or 518 or X), that is the max number of inhabitants that will satisfy all rules.

### #16

Posted 31 July 2007 - 07:52 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 think you're missing an essential element of this puzzle. The number of inhabitants must be more than the greatest number of hairs on a single inhabitants head. Since no two inhabitants can have the same number of hairs and the number 518 is forbidden, the series stops at 519.

in VBA for Excel, not the greatest language but ubiquitous, it looks like this:

Sub countHairs() Dim h Dim p h= 0 p = 1 While h<> 518 h= h+ 1 p = p + 1 ActiveCell.Value = p Wend End Sub

Note that 'p', the number of people, starts out one greater than 'h', the number of hairs, and that 'h' cannot equal 518.

### #17

Posted 01 August 2007 - 01:30 AM

You guys still aren't thinking of it correctly. Let me modify the problem to see if it is a little easier for you.

What you did was change the problem. By rule 3

3) The maximum number of hairs on any one person's head may not be greater than or equal to the number of inhabitants.

was

3. There are more inhabitants than any inhabitant's hair in the town.

Your explanation changed the parameters.

There is nothing in the rules to say there has to be an order and to skip a number voids a rule.

Someone might not have had 314 hairs.... the number of inhabitants based on these parameters is infinite.

Thoughts?

### #18

Posted 01 August 2007 - 01:45 AM

I cannot see the difference. On the one hand, the first version above, we're saying 'not (h >= p)' while the second version says that 'p > h'.What you did was change the problem. By rule 3

3) The maximum number of hairs on any one person's head may not be greater than or equal to the number of inhabitants.

was

3. There are more inhabitants than any inhabitant's hair in the town.

Your explanation changed the parameters.

This must be the most common fallacy in this kind of problem. I missed it myself on this one. It's not stated but it is implicit. Try it. Let's say there are three people. p > h so the greatest number of hairs in the population of 3 is 2. That being the case, someone has 2, someone has 1, and someone has 0. Try it with any number; you can't skip any.There is nothing in the rules to say there has to be an order and to skip a number voids a rule.

### #19

Posted 14 August 2007 - 03:23 AM

### #20

Posted 28 August 2007 - 03:09 AM

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