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?

10's the best I got, sorry.

It's the worst problem ever. 5, no 3 no....oh I don't know.

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

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)

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!!!

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?

an infinite amount of time

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.

Why don't they conference call? Then it's 5 calls, right?

No two card shuffles will ever be the same.

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. :rollanim:

No, but eight will.

See it:

Mathematical closed form solution: in this example, k=2, n=6.
f(n,k)=2[(n−k)(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!

Interesting that ONR is supporting this research. Network topology is still a hot topic!