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## Math problem

After throwing out the highest and lowest there a re seven judges scores to determine the GOE.

If 1 judge out of seven gives +1 and the rest give 0, the average GOE is .142857.
If 2 judges give +1, the average is .285714.
If 3 judges give +1, the average is .428571
If 4 judges give +1, the average is .571428.
If 5 judges give +1, the average is .714285.
If 6 judges give +1, the average is .857142.

Note that all of these average GOE’s have the same six decimal digits, and in the same order, just starting at a different place.

Challenge. This works for a panel of n = 7 judges. Find another panel size n for which this also works.

N = 17 works. Can you find another?

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Mathman, those are a wonderful set of numbers! However, unless I have misunderstood, I don't find 17 works the same way at all. 1/17, 2/17, 3/17........ don't yield such a set of rotating digits for me, not for 6 decimal digits anyway. Don't give me problems with a string of digits greater than my little calculator can accomodate!

You will eventually post the formula after stumping everybody, right?

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Originally Posted by SkateFiguring
Mathman, those are a wonderful set of numbers! However, unless I have misunderstood, I don't find 17 works the same way at all. 1/17, 2/17, 3/17........ don't yield such a set of rotating digits for me, not for 6 decimal digits anyway. Don't give me problems with a string of digits greater than my little calculator can accomodate!
17 works for 16 digits. 1/17 gives the repeating decimal

.0588235294117647 05882352941176470 588235294117647...

All the other fractions 2/17, 3/17,...16/17 give exactly the sixteen cyclic permutations of these same 16 digits. (So, yes, you need a bigger calculator. )

In general, the number of repeating digits should be n-1.

You will eventually post the formula after stumping everybody, right?
I wish. It is an unsolved problem in number theory to prove that this trick works for about 37.395% of all prime numbers n.

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So it's not another golden ratio.

And how many decimal digits of Pi can you recite?

eta I don't care what n you give me, I know the last answer is 1 with however long string of decimal digits (0) you want! You can't trump me with that!

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Get rid of the GoEs. Use the scale other judged sports use. 0-10, zero meaning the jump didn't happen; 1-9 judges the faulty skate; 10 means perfection. Then multiply the consensus of the judges' input by the element's base value.

Example: A 3salchow may show from the judges: 6; 9; 5; 5; 5; 6; 6; 6; 7. Then drop the 9 and one of the 5s. The consensus of the judges would be 46 divided by 7 = 6 and 2/23 multiplied by base value of 4.5(?) gets a skater 29,97 for the 3salchow. Too high? but so what, there's no downgrading; no automatic deductions; no measuring of landing faults; etc. Just the consensus of the Judges on what that element deserves. (Rules on Rounding scores may have to be worked out.)

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A chance for Mathman to win \$1 million:

http://www.newscientist.com/article/...s-problem.html

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13 works.

1 / 13 = 0.076923076923077
2 / 13 = 0.153846153846154
3 / 13 = 0.230769230769231
4 / 13 = 0.307692307692308
5 / 13 = 0.384615384615385
6 / 13 = 0.461538461538462
7 / 13 = 0.538461538461538
8 / 13 = 0.615384615384615
9 / 13 = 0.692307692307692
10 / 13 = 0.769230769230769
11 / 13 = 0.846153846153846
12 / 13 = 0.923076923076923

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Originally Posted by SkateFiguring
A chance for Mathman to win \$1 million:

http://www.newscientist.com/article/...s-problem.html
I had a master's student a couple of yers ago who wrote her thesis on this problem. (She didn't win the million dollars, but she investigated a particular attempt at a solution and analyzed why it didn't work.)

There are actually six of these "Millenium Problems" that the Clay Institute is offering a million dollars for. There were seven, but one of them -- the Poncare conjecture -- was solved in 2003. The other six remained unsolved.

The guy who solved the Poincare Conjecture, Russian matematician Grigor Perleman, refused to accept the million dollar prize on the grounds that others who laid the ground work for his proof were equally deserving to be honored.

First we compared P versus NP with 18 mathematical problems, from Fermat's last theorem to the Poincaré conjecture, that were not solved until more than a decade after their "births".
Fermatis Last Theorem was "born" in 1637 and solved in 1995-96, while the Poincare Conjecture was formulated a little before 1904 and soleved in 2002-03.

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Calling George Bernard Dantzig.

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Originally Posted by PolymerBob
13 works.

1 / 13 = 0.076923076923077
2 / 13 = 0.153846153846154
3 / 13 = 0.230769230769231
4 / 13 = 0.307692307692308
5 / 13 = 0.384615384615385
6 / 13 = 0.461538461538462
7 / 13 = 0.538461538461538
8 / 13 = 0.615384615384615
9 / 13 = 0.692307692307692
10 / 13 = 0.769230769230769
11 / 13 = 0.846153846153846
12 / 13 = 0.923076923076923
Not exactly. For P = 13 the fractions fall into two classes, each of which repeat in cycles of the same six digits, rather than one class with all possible permutations of P-1 digits, as is the case for 7, 17, 19 and 23, among many others.

1/13 = .076923 (repeated)
3/13 = .230769
4/13 = .307692
9/13 = .692307
10/13 = .769230
12/13 = .923076

2/13 = .153846
5/13 = .384615
6/13 = .461538
7/13 = .538461
8/13 = .615384
11/13 = .846153

What is is about the fractions 1/13, 3/13, 4/13, 9/13, 10/13, and 12/13 that make them "belong together?

Here is another strange thing. Can you find two prime numbers P for which the decimal expansion of the fraction 1/P repeats with period five?

Yes. P = 41 and P = 271.

1/41 = .02439 02439 02439 02439 ...

1/271 = .00369 00369 00369 00369 ...

Now multiply these two denominators together:

41x271 = 11111 (five ones).

Let's try another. Here are the only two primes P for which 1/P repeats with period seven.

1/239 = .0041841 0041841 0041841 ...

1/4649 = .0002151 .0002151 0002151 ...

Multiply the denominators:

239x4649 = 1111111 (seven ones).

(OK, I'll stop now. )

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Originally Posted by SkateFiguring
Calling George Bernard Dantzig.

The most remarkable thing about Dantzig's simplex method is that it is just about the only mathematical algorithm I know that works better in practice than it does in theory.

Usually, mathematical ideas look great in textbooks but as soon as you carry them into the real world, they fall apart. Sort of like the CoP in figure skating. (I wonder if the problem of n judges trying to assign ordinals to m skaters properly is "P". )

But for some reason -- nobody really knows exactly why -- in linear programming the practical problems that one would really like to solve in the real world turn out to be exactly the small subset of problems that the simplex method works most efficiently for. It is quite hard, actually, to invent an artificial problem that frustrates Dantzig's method.

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George also didn't recognize unsolvable problems.

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The only thing worse than dreaming about your Math Final is.......... waking up from the dream during the Final.

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Originally Posted by SkateFiguring

The only thing worse than dreaming about your Math Final is.......... waking up from the dream during the Final.
so true!

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