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Topic: ANSWERS TO MEDIUM PUZZLES -- Part 2 (Read 10041 times) |
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I.M._Smarter_Enyu
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ANSWERS TO MEDIUM PUZZLES -- Part 2
« on: Jul 26th, 2002, 9:38am » |
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RELATIVELY MEDIUM -- Part 2
>> POPULATION OF FUNKYTOWN In the city of Funkytown, the following facts are true:
(1) No two inhabitants have exactly the same number of hairs.
(2) No inhabitant has exactly 483,207 hairs.
(3) There are more inhabitants than there are hairs on the head of any one inhabitant.
What is the largest possible number of inhabitants of Funkytown? <<
We'll assume Funkytown has at least 1 person, otherwise this problem wouldn't be any fun.
As usual, the way to solve problems like this is to take it down to the trivial case and then start building up.
So Let's say rule (2) was "No inhabitant has exactly 1 hair." Rule (3) requires that there be at least 2 people. Neither can have 1 hair because of rule (2). Let's guess dude #1 has 2 hairs. (Must be Homer Simpson. Nice comb-over!) So rule (3) requires Funkytown has at least 3 people. So dude #2 must have at least 3 hairs... now rule (3) requires Funkytown have at least 4 people.... and so on, ad infinitum. Hmmm, so rule (2) could never say "No inhabitant has exactly 1 hair."
So let's try: Rule (2) is "No inhabitant has exactly 2 hairs." Does it work if Dude 1 has 1 hair? If true, then rule (3) says there must be at least 2 people. Dude 2 can't have 2 hairs because of rule (2). So Dude #3 must have at least 3 hairs... now rule (3) says there must be at least 4 people... again, on to infinity. So rule (2) could never say "No inhabitant has exactly 2 hairs."
See where this is going? The rules require an infinite number of people.
This is a case of Cantor's diagonalization argument.
>> CHOCOLATE MILK You have two thermoses. The first contains a liter of milk, the second contains a liter of pure chocolate syrup. You pour one cup of milk out from the first thermos to the second one. Then, after mixing that, you take one cup of the mixture from the second thermos, and pour it back into the first thermos. After completing these two operations, which thermos contains more milk? <<
I'm doing this by American, not British units, 1 cup = 8 oz = .24 liters
Thermos 1 Thermos 2 -- Breakdown of milk|chocolate by litres:
m|c m|c
1|0 0|1 ... now move 1 cup, or .24 litres:
.76|0 .24|1
Now, the 2nd thermos contains 1.24 litres of fluid, and a cup is .24 litres, which is .24/1.24 = 19% of the volume.
So the cup you pull is 19% * .24 litres milk (apx 0.05 litres), 19% * 1 litre chocolate (.19 litres):
.81|.19 .19|.81
So, as you might expect, the first thermos still contains more milk. I think the question is wrong.
>> WHERE'S THE FATHER? The mother is 21 years older than the child. In 6 years from now, the mother will be 5 times as old as the child. Question: Where's the father? <<
M = 21 + C
M + 6 = 5 * (C + 6) = 5C + 30
M = 5C + 24
5C + 24 = 21 + C
4C = -3
C = -3/4 = 9 months before the child is born...
So therefore, the father is ZIPPING UP HIS PANTS! Heh heh, that's a good one.
>> .999 ... Compare the numbers 0.99999... (infinitely many 9s) and 1. Which of the following statements is true? Why?
0.99999 ... < 1
0.99999 ... = 1
0.99999 ... > 1 <<
.99999 is EXACTLY equal to 1, this is ancient....
x = 0.99999...
10x = 9.99999...
10x - x = 9.9999... - 0.99999...
9x = 9
x = 1 Q.E.D
*** I.M. Smarter Enyu!!!! ***
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klbarrus
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Re: ANSWERS TO MEDIUM PUZZLES -- Part 2
« Reply #1 on: Jul 26th, 2002, 10:42am » |
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I don't see why the answer to Funkytown isn't 483207:
Consider a population arranged such that citizens #1 - #483206 happen to have 1 through 483206 hairs on their head. Citizen 483207 is bald. Thus:
1) every citizen has a different number of hairs
2) nobody has 483207 hairs
3) there are 483207 citizens, which is more than the hairs on the head of any one of them (since the max hairs is 483206).
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I.M._Smarter_Enyu
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Re: ANSWERS TO MEDIUM PUZZLES -- Part 2
« Reply #2 on: Jul 26th, 2002, 11:21am » |
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Yeah, I assumed nobody was bald! Puzzle doesn't say either way.
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H. Lee
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Re: ANSWERS TO MEDIUM PUZZLES -- Part 2
« Reply #3 on: Jul 26th, 2002, 1:07pm » |
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For chocolate milk, I answered "the same amount of milk" because milk mixed with chocolate syrup is chocolate milk, so now you have 2 thermoses full of chocolate milk, just in different concentrations.
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Frothingslosh
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Re: ANSWERS TO MEDIUM PUZZLES -- Part 2
« Reply #4 on: Jul 26th, 2002, 1:35pm » |
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Funnytown: One bald guy lives there. Doh!
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Nick
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Re: ANSWERS TO MEDIUM PUZZLES -- Part 2
« Reply #5 on: Jul 26th, 2002, 2:24pm » |
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For the chocolate milk question, the question is misphrased.
It should ask "Which flask's content is more pure?"
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phasemute
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Re: ANSWERS TO MEDIUM PUZZLES -- Part 2
« Reply #6 on: Aug 25th, 2002, 10:44am » |
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on Jul 26th, 2002, 9:38am, I.M._Smarter_Enyu wrote:RELATIVELY MEDIUM -- Part 2
What is the largest possible number of inhabitants of Funkytown? <<
We'll assume Funkytown has at least 1 person, otherwise this problem wouldn't be any fun.
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"the largest possible number of inhabitants" means that this assumption you make about there being at least one person is incredibly redundant.
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pottymouthed
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Re: ANSWERS TO MEDIUM PUZZLES -- Part 2
« Reply #7 on: Sep 3rd, 2002, 8:48am » |
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RE Funkytown.
At a quick glance this looks like a simple programming question.
a couple of loops --an incrementer and a counter and.
am I way off base...maybe I should look at it again.
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S. Owen
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Re: ANSWERS TO MEDIUM PUZZLES -- Part 2
« Reply #8 on: Sep 3rd, 2002, 3:28pm » |
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I don't believe it requires any programming - see the other thread dedicated to this problem for answers.
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