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Old 03-18-2010, 08:22 PM   #1
Juniper
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Math Homework Help for 8th Grader

Got a problem to solve. I worked on this with my daughter last night, and I got an answer, but apparently I'm wrong!

There are six busybodies in town who like to share information. Whenever one of them calls another, they both know everything that the other one knew by the end of the conversation. One day, each of the six women picks up a juicy piece of gossip. What is the minimum number of phone calls required before all six of them know all six of these tidbits?

I got 10, but I guess it's not right!

Help?
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Old 03-18-2010, 09:42 PM   #2
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10's the best I got, sorry.
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Old 03-18-2010, 09:45 PM   #3
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It's the worst problem ever. 5, no 3 no....oh I don't know.
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Old 03-18-2010, 09:51 PM   #4
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I got nine.

ABC DEF

B calls A and C (2)
E calls D and F (2)

B calls E (1)

B calls A and C (2)
E calls D and F (2)

Total: 9
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Old 03-18-2010, 10:15 PM   #5
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I think I got 8.

Each caller is a letter and each number is a bit of gossip.

A B C D E F
1 2 3 4 5 6

A calls F (A has 1,6 F has 1,6)
B calls E (B has 2,5 E has 2,5)
C calls D (C has 3,4 D has 3,4)
A calls E (A has 1,2,5,6 E has 1,2,5,6)
A calls D (A has 1,2,3,4,5,6 D has 1,2,3,4,5,6)
A calls B (A has 1,2,3,4,5,6 B has 1,2,3,4,5,6)
C calls E (C has 1,2,3,4,5,6 E has 1,2,3,4,5,6)
E calls F (E has 1,2,3,4,5,6 F has 1,2,3,4,5,6)
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Old 03-18-2010, 10:33 PM   #6
Juniper
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I had amended my answer to 9 also, but I think Squirrel's right!

Thanks!

Every quarter daughter gets these "critical thinking" worksheets for math - things like the family crossing the river in a boat that can only hold two, the frogs and lily pads, various puzzles . . . and for us non-math types, they are HARD!!!
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Old 03-18-2010, 11:31 PM   #7
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If a hen and a half can lay an egg and a half in a day and a half, how long does it take a cricket with a wooden leg to kick all the seeds from a dill pickle?
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Old 03-19-2010, 01:33 AM   #8
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an infinite amount of time
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Old 03-19-2010, 09:26 AM   #9
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You get partial Credit. It takes half as long as it takes for an elephant with a wooden leg to bore a hole in a bar of soap.
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Old 03-19-2010, 09:40 AM   #10
monster
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Quote:
Originally Posted by Juniper View Post
Got a problem to solve. I worked on this with my daughter last night, and I got an answer, but apparently I'm wrong!

There are six busybodies in town who like to share information. Whenever one of them calls another, they both know everything that the other one knew by the end of the conversation. One day, each of the six women picks up a juicy piece of gossip. What is the minimum number of phone calls required before all six of them know all six of these tidbits?

I got 10, but I guess it's not right!

Help?
Why don't they conference call? Then it's 5 calls, right?
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Old 03-19-2010, 10:48 AM   #11
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No two card shuffles will ever be the same.
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Old 03-19-2010, 11:01 AM   #12
monster
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but strangely, two flint posts can be, despite the fact that the number of possible word/idea combinations in that head must outstrip the number of possible card shuffles.
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Old 03-19-2010, 01:01 PM   #13
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Quote:
Originally Posted by Flint View Post
No two card shuffles will ever be the same.
No, but eight will.

See it:
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Old 03-19-2010, 01:22 PM   #14
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Mathematical closed form solution: in this example, k=2, n=6.
f(n,k)=2[(nk)(k−1)] =2[(4)(1)] = 8

Squirrel for the win!
ETA, Juni, this is neither a trivial or a simple problem. Kudos to your kid's math teacher!
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per·son \ˈpər-sən\ (noun) - an ephemeral collection of small, irrational decisions
The fun thing about evolution (and science in general) is that it happens whether you believe in it or not.
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Old 03-19-2010, 01:34 PM   #15
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Interesting that ONR is supporting this research. Network topology is still a hot topic!
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per·son \ˈpər-sən\ (noun) - an ephemeral collection of small, irrational decisions
The fun thing about evolution (and science in general) is that it happens whether you believe in it or not.
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