Brain Teaser I can't quite wrap my head around

[TunezCottage]

I received a brain teaser in school which I solved but don't understand the logic behind it, I'm not sure I'm in the right forum for this one but I'll give it a go. Would really appreciate if someone could give me the solution so I can see whether or not I was right, and proof of said solution!

Question:

All the students at Tim's school have different amounts of hairstrands. None of them have exactly 2012 hairstrands. Tim is the one student in school with the most amount of hairstrands. The amount of students at the school is more than Tim's hairstrands. Which is the highest possible number of students at Tim's school.

Options:

a) 2010 b) 2011 c) 2012 d) 2013 e) 2014

 

[Ishuda]

I received a brain teaser in school which I solved but don't understand the logic behind it, I'm not sure I'm in the right forum for this one but I'll give it a go. Would really appreciate if someone could give me the solution so I can see whether or not I was right, and proof of said solution!

Question:

All the students at Tim's school have different amounts of hairstrands. None of them have exactly 2012 hairstrands. Tim is the one student in school with the most amount of hairstrands. The amount of students at the school is more than Tim's hairstrands. Which is the highest possible number of students at Tim's school.

Options:

a) 2010 b) 2011 c) 2012 d) 2013 e) 2014

It would be "interesting" to see a number and proof for the question as stated.

 

[Steven G]

Well, show us your solution....

That solution would just be an imagination of your figment.

 

[Ishuda]

That solution would just be an imagination of your figment.

There are n people and each have a different number of hair strands with Tim's being the most. At a minimum, how many hairs strands does Time have [it must be an integer and, if there are two people of more including Tim, remember that it must be a positive integer assuming someone could have zero hair strands]?.

 

[Ishuda]

I received a brain teaser in school which I solved but don't understand the logic behind it, I'm not sure I'm in the right forum for this one but I'll give it a go. Would really appreciate if someone could give me the solution so I can see whether or not I was right, and proof of said solution!

Question:

All the students at Tim's school have different amounts of hairstrands. None of them have exactly 2012 hairstrands. Tim is the one student in school with the most amount of hairstrands. The amount of students at the school is more than Tim's hairstrands. Which is the highest possible number of students at Tim's school.

Options:

a) 2010 b) 2011 c) 2012 d) 2013 e) 2014

Assume there are m students at the school. Letting H_T be the number of Tim's hairstrands [HS] then it is given that
(1) H_T \(\displaystyle \lt\) m.

Consider two cases:
(a) m \(\displaystyle \lt\) 2012
At most Tim can 2010 [see Jomo's comment below] HS [see (1) above]. Since Tim has the most HS and each student has a different number of HS, Tim must have at least m-1 HS (assuming some student could have zero HS), i.e.
m-1 \(\displaystyle \le\) H_T

(b) m \(\displaystyle \ge\) 2012
Since Tim has the most HS, doesn't have 2012 HS, and each student has a different number of HS not equal to 2012, Tim must have at least m HS (assuming some student could have zero HS), i.e.
m \(\displaystyle \le\) H_T

Therefore we can conclude that the maximum number of students in the school is ?.

EDIT: Fix that last \(\displaystyle \lt\), it should have been \(\displaystyle \le\). [Have to watch that copy and paste more carefully]. Later edit fix mistake [see red]

 

[Steven G]

Assume there are m students at the school. Letting H_T be the number of Tim's hairstrands [HS] then it is given that
(1) H_T \(\displaystyle \lt\) m. I agree with this

Consider two cases:
(a) m \(\displaystyle \lt\) 2012 OK
At most Tim can 2011 HS [see (1) above]. Since Tim has the most HS and each student has a different number of HS, Tim must have at least m-1 HS (assuming some student could have zero HS), i.e.
m-1 \(\displaystyle \le\) HT Since m< 2012 then the most m can be is 2011. You wrote at most Tim can have 2011 HS. Doesn't this contradict (1)????
I think that it should say that at most Tim can have is 2010 HS. I agree when you say
since Tim has the most HS and each student has a different number of HS, Tim must have at least m-1 HS. So then...
(b) m \(\displaystyle \ge\) 2012
Since Tim has the most HS, doesn't have 2012 HS, and each student has a different number of HS not equal to 2012, Tim must have at least m HS (assuming some student could have zero HS), i.e.
m \(\displaystyle \le\) H_T If m=2012, then Tim can have 2011 HS. But this contradicts what you said--Tim must have at least m HS

Therefore we can conclude that the maximum number of students in the school is ?.

EDIT: Fix that last \(\displaystyle \lt\), it should have been \(\displaystyle \le\). [Have to watch that copy and paste more carefully]

I am sure that you are correct but I still have a few questions. See red above. Sorry for not seeing this.

 

[Dale10101]

Is there enough information, something left out?

I received a brain teaser in school which I solved but don't understand the logic behind it, I'm not sure I'm in the right forum for this one but I'll give it a go. Would really appreciate if someone could give me the solution so I can see whether or not I was right, and proof of said solution!

Question:

All the students at Tim's school have different amounts of hairstrands. None of them have exactly 2012 hairstrands. Tim is the one student in school with the most amount of hairstrands. The amount of students at the school is more than Tim's hairstrands. Which is the highest possible number of students at Tim's school.

Options:

a) 2010 b) 2011 c) 2012 d) 2013 e) 2014

So I will not feel bad if this doesn't add up to me since nobody has actually given an answer.

I try rewriting the proposition.


I: There exists a set of distinct integers none of which equals 2012.
II: The set is finite, and has "m" elements.
III: One of the integers in the set, called Tim, is the greatest integer.
IV: The number of elements, m, in the set is greater then the value of the largest integer in the set, Tim.
V: What is the largest possible value for m?

Since there is no restriction on how many hairs are on Tim's head, it seems to follow that one cannot restrict m to any given number. One can only say that the size of the set is larger than the value if its greatest included integer.

(?)

 

[HallsofIvy]

But it is impossible for a set of distinct positive integers to have cardinality larger than its largest member!

 

[Dale10101]

OK

I need to go over the more recent posts to digest the latest thinking. On your point I suppose what I meant to say was that:

"I: There exists a subset of distinct integers none of which equals 2012."

However, as I say I must read and think more about this. Thanks for the insight, as written you make a fair point.

But it is impossible for a set of distinct positive integers to have cardinality larger than its largest member!

 

[Ishuda]

I am sure that you are correct but I still have a few questions. See red above. Sorry for not seeing this.

Jomo,

You are correct about the 2010 for case (a) m < 2012. It jumped out of the screen when I came back to read it today even before I read your comment [back to the corner for me for 2012 seconds]. If m is less than 2012 then m could be 2011. Since m is greater than Tim's HS, the most Tim's HS could be is 2010.

Adding in (1) for case (a) we have
m-1 [FONT=MathJax_Main]≤[/FONT] H_T < m
which implies H_T = m-1 and the largest m can be [the most students there can be in the school] is 2011.

Adding in (1) for case (b) we get
m [FONT=MathJax_Main]≤[/FONT] H_T < m
which is a contradiction. Therefore m can not be larger than or equal to 2012.

Putting (1), (a), and (b) together we have the most students there can be in the school] is 2011.