1. Originally Posted by lowtherlore
Correct.

It would be perfect if you share the solution, as it has I think beautiful two-step logic in it.
Heh-heh, it's kind of embarrassing to tell you my solution, which involves million steps not necessarily logical.
First: I listed on a piece of paper all possible X + Y (i.e., every integer from 5 to 39)
Second: Because X and Y cannot be both primes, I deleted from the list all even numbers as well as numbers that equal to 2 + A (where A is a prime). So I ended up with a short list of 11, 17, 23, 27, 29, 35, and 37.
Third: I asked myself: "Gosh, how many possible combinations do I have to test through trial and error?" So I came up with a little formula: floor function (X + Y) ÷ 2 - 1. For instance, there are 4 or floor [11 ÷ 2 - 1] possible combinations for a X + Y total of 11, and there are 7 for a X + Y total of 17. And I frustratedly browsed through the short list and discovered that 17 stands out like a sore thumb because it is the only one that has an odd number of possible combinations.
Forth: I broke down 17 into (2, 15 =3*5), (3, 14 = 2*7), (4 =2*2, 13), (5, 12=3*2*2), (6 =2*3, 11), (7, 10=2*5), (8=2*2*2, 9=3*3). And I noticed that (4, 13) is unique because the rest involve either three primes or times more than once.
Fifth: I tested (4, 13) and yelled BINGO.

I know it's sheer luck. But I solved it in a timely fashion, didn't I?

2. Originally Posted by skatinginbc
Now here comes my puzzle challenge:
X denotes the total Olympic medals that Skater A and Skater B had won and Y the total World medals that the two skaters had won.
X * Y = 0
X + Y = 10
Who are the skaters?
Kurt Browning and Todd Eldredge? No, in that case X + Y = 11

I can also come up with several examples where X + Y = 9, but that doesn't help either

3. Kurt Browning and Gustave Hugel?

Todd Eldredge and Graham Sharp?

What is the next letter in this sequence: O T T F F S S E N?

4. Originally Posted by skatinginbc
How can I prove to my cat that 1 + 1 = 2?
This question was undertaken by Bertrand Russell and Alfred North Whitehead in Principia Mathematica, a remarkable (but as it turned out, doomed) exercise in reducing all of mathematics to formal logic. The theorem 1+1=2 occurs on page 378, the first 377 pages being required for the proof.

5. I doubt the cat would be willing to read all 378 pages. How about an executive summary?

6. Originally Posted by gkelly
Kurt Browning and Todd Eldredge? No, in that case X + Y = 11
Originally Posted by Mathman
Kurt Browning and Gustave Hugel? Todd Eldredge and Graham Sharp?
Kelly and Mathman are both right. One is a genius in reading my mind. I thought the answer would be Kurt (5) and Todd (5 ==> I mistakenly counted only Post-1993 world medals listed in the wikipedia). The other is a living encyclopedia that knows the skating history inside out.
Kurt (5 world medals) + Gustave (5 world medals) = 10
Todd (6 world medals) + Graham (4 world medals) = 10

Originally Posted by Mathman
What is the next letter in this sequence: O T T F F S S E N?
T as in TEN

Now, the next puzzle:

A is a male single skater who holds 3 Olympic medals. B is a female single skater who holds 2 Olympic medals. The total number of Olympic gold medals earned by the two skaters is 2, and the total number of silver is 3. Who are the two skaters?

7. Originally Posted by skatinginbc
Kelly and Mathman...
gkelly is the one who has encyclopedic knowledge of figure skating. I just have a quick finger on Wikipedia.

A is a male single skater who holds 3 Olympic medals. B is a female single skater who holds 2 Olympic medals. The total number of Olympic gold medals earned by the two skaters is 2, and the total number of silver is 3. Who are the two skaters?
But this one I know without looking anything up. Gene Plushenko and Carol Heiss. (Evgneni gave me his permission to call him Gene. )

(a) If you expand the fraction 1/41 as a decimal you get

.02439 02439 02439 02439 ...

The decimal expansion repeats in groups of 5 digits. 5 is the period of the repeating decimal.

Find another prime number P besides 41 for which 1/P also repeats with period 5.

(b) 1/239 = .0041841 0041841 0041841 ... (period 7).

Find another prime P for which 1/P has period 7.

8. Or Plushenko and T.Albright?
What are you playing here? I just discovered the thread.

9. Originally Posted by Mathman
(a) If you expand the fraction 1/41 as a decimal you get

.02439 02439 02439 02439 ...

The decimal expansion repeats in groups of 5 digits. 5 is the period of the repeating decimal.

Find another prime number P besides 41 for which 1/P also repeats with period 5.

(b) 1/239 = .0041841 0041841 0041841 ... (period 7).

Find another prime P for which 1/P has period 7.
you mean find by trial and error?

the other one is 271 (p=5) and the other 4649 (p=7) but I ve seen before the trick. You do 11111/41 and 1111111/239 but I dont know why.

10. Originally Posted by skatinginbc
...
I know it's sheer luck. But I solved it in a timely fashion, didn't I?
You sure did! With plenty of logic.

11. Originally Posted by seniorita
The other one is 271 (p=5) and the other 4649 (p=7) but I ve seen before the trick. You do 11111/41 and 1111111/239 but I dont know why.
If you expand 1/41 as a decimal by long division, the sequence of remainders that you get are the remainders when 10, 100, 1000, 10000, etc. are divided by 41. Stop when you get back to a remainder of 1. The remainder of 100000 when divided by 41 is 1. In the language of congruences, 105 is congruent to 1 mod(41) – that’s why the period of 1/41 is 5.

This means that 100000 – 1 is divisible by 41. But 100000 – 1 = 99999 = 9x11111, so we conclude that 41 is a factor of 11111. Any other prime factor of 11111 will have the same property.

In general, if P is a prime number greater than 5, then the period of 1/P will be the smallest positive integer k for which P is a factor of 11111…111 (k one’s).

-------------------------------------------------

Back to skating-related problems...

Beatrix Schuba was the worlds' greatest skater of school figures. She once traced a perfect circle on the ice. When the judges measured it they found that the radius was exactly one meter and the circumference was c = 6.283185307179560 meters.

What is the radius of the earth?

Hint: For an algebraic solution you will need to know (a) the decimal expansion of π out to 15 decimal places, (b) the formula from spherical geometry for the circumference of a circle of radius r drawn on a sphere of radius R, and (c) the first two terms of the MacLaurin series for the sine function.

12. i have to catch up with all of you here! I remember every genus and species name I learned in biology, but my algebra has vanished from my view.

13. I wish Toni would come and share her opinion!!!! Where are you Toni????

14. Thank you for posting Olympia and Dee. I thought I killed the thread with my last post.

15. Mathman
Beatrix Schuba was the worlds' greatest skater of school figures. She once traced a perfect circle on the ice. When the judges measured it they found that the radius was exactly one meter and the circumference was c = 6.283185307179560 meters.

What is the radius of the earth?

Hint: For an algebraic solution you will need to know (a) the decimal expansion of π out to 15 decimal places, (b) the formula from spherical geometry for the circumference of a circle of radius r drawn on a sphere of radius R, and (c) the first two terms of the MacLaurin series for the sine function.
I was just passing by and it called my attention. Since I wanted a distraction, I decided to give it a try.

Actually from the first sight I thought that it is impossible to do what you demand and may be that is true if I didn't make an error later.

Let c = 6.28318530717956 m, R = radius of the Earth, r = 1 m.

We do know from formula that c = 2*Pi*R*sin(r/R).

So R*sin(r/R) = c/2*Pi.

Let x = c/2*Pi.

Using McLauren series, we can determine that sin(r/R) is approximately r/R - r^3/(6*R^3) if we use only two first terms.

So R*(r/R - r^3/6*R^3) = x.

Which trivially means that R = square_root(r^3/6*(r - x)).

x = 0.9999999625374750275214946974942 (approximately)

Therefore R = 2109.23928... m.

In other words, the radius of this "Earth" is just 2.109 km which can't be right.

I assume that the reason for this is the fact that for a big sphere like this the ratio of circumference to a diameter for a small circle like this is almost Pi already and for a sphere with Earth's diameter the precision to detect deviations from 2*Pi*r of circumference length had to be insane.

That is, unless I made an error somewhere.

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