algebra problem

[skatinginbc]

Kurt Browning and Todd Eldredge? No, in that case X + Y = 11

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 :thumbsup:

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?

 

[Mathman]

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.

 

[seniorita]

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

 

[seniorita]

(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.

 

[lowtherlore]

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

You sure did! With plenty of logic. :thumbsup:

 

[Mathman]

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, 10[SUP]5[/SUP] 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.

 

[Olympia]

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.

 

[Dee4707]

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

 

[Mathman]

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

 

[Daniel5555]

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.

 

[skatinginbc]

I thought I killed the thread with my last post.

You beat me with that for sure. When I tried to use my calculator and it wouldn't take in 15 decimals, I automatically gave up, not to speak of the MacLaurin series, which I've never heard of before.

 

[PolymerBob]

Mathman

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)).

I got R = sqrt( r^3 / ( 6r - 3C/pi ) ) , only I got 76.7 kilometers. ( It's a small world. )
It could be solved with a calculator with enough digits.

 

[Mathman]

Well, Daniel and PolymerBob both got it algebraically. Yet when they put the numbers in their calculators the numerical answer was either 2.9 km or 76.7 km. This seems like a big difference, considering that the algebraic answers are identical.

Let me try it without a calculator. The circumference of the circle was measured with an accuracy of 10[SUP]-15[/SUP] meters, so let’s estimate pi to the same accuracy in our calculations:

Pi = 3.141592653589793

2Pi = 6,283185307179586

c = 6.283185307179560

So the difference between the Euclidean circumference 2 pi r and the spherical circumference 2 pi R sin (r/R), for r = 1, is

2 Pi – c = 2.6 x 10[SUP]-14[/SUP]

(Note that we have at best only two significant figures in this problem, so the answer can turn out to be all over the place, but at least should have the right order of magnitude however we compute it.)

Now solve for R:

c = 2 pi R sin(r/R) = 2 pi R (r/R—r[SUP]3[/SUP]/6 R[SUP]3[/SUP]) = 2 pi r – (pi/3)(1/R[SUP]2[/SUP]), so

1/R[SUP]2[/SUP] = (3/pi)(2pi – c) = (3/pi)x(2.6x10[SUP]-14[/SUP]) = 2.5 X 10[SUP]-14[/SUP]

the last approximation holding because pi is just a little bigger than 3.

Thus R[SUP]2[/SUP] = (1/2.5) x 10[SUP]14[/SUP] = .4 x 10[SUP]14[/SUP]

So R = .63 x 10[SUP]7[/SUP] = 6,300,000 meters. (The square root of 40 is somewhere between 6 and 7 – let’s guess 6.3 )

So the radius of the earth is 6,300 kilometers, which isn’t too bad an estimate. (Correct order of magnitude and at least one significant digit is right. If you do it algebraically and put the numbers in at the end, without using scientific notation, then you must divide by a number that is very close to zero.)

By the way, the error in using the polynomial approximation for the sine is vanishingly small in this example, because the earth is very much bigger than the circle on the ice.

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

Here is another one in the same spirit.

According to the Bible (Second Chronicles 4:2),

Also King Solomon made a molten sea of ten cubits from brim to brim, round in compass, and five cubits the height thereof; and a line of thirty cubits did compass it round about.

In other words, the circumference of Solomon’s bowl was 30 cubits and the diameter was 10 cubits. (A cubit is the length of your forearm from elbow to fingertips. L. cubitum = elbow).

What was the circumference of the earth back in Solomon’s time?

Solution: We have c = C sin (pi d/C). Put in c = 30, d = 10, and solve for C by inspection.

 

[PolymerBob]

When I tried to use my calculator and it wouldn't take in 15 decimals, I automatically gave up, not to speak of the MacLaurin series, which I've never heard of before.

Actually, I found a better way. The calculator function that comes with Microsoft Windows goes out to 32 digits when in scientific mode. So using the nuclear weapon of calculation .......

3C / PI = 5.9999999999999747163987764178222

6r ( 6 x 1 ) minus this is 2.5283661223582177753345612998071 x 10E-14

The reciprocal of this is 39551727801646 ( rounded off )

The square root is 6,288,585 meters, or 6,289 kilometers. ( The correct value is 6,371 kilometers. )

 

[Mathman]

6r ( 6 x 1 ) minus this is 2.5283661223582177753345612998071 x 10E-14

This is the key. If you use a small calculator that stores only a few digits of pi, this number will be off by many orders of magnitude, resulting in an answer that is way too small.

 

[Daniel5555]

Mathman
PolymerBob
I don't know what happened in my case, because I actually used scientific mode of Windows Calculator. I calculated it once again, and it gave me the correct answer (6,288,984.something... m). I guess I just pressed wrong button or there were some other silly errors somewhere since it was late at night.

Well, the solutions we had: R = square_root(r^3/6*(r - x)) and R = sqrt( r^3 / ( 6r - 3C/pi ) ) are equivalent to what Mathman presented, so I think it's generally correct, 5 out of 10, I guess

 

[Olympia]

I can't contribute to the discussion at hand. But I just read the following joke online. I think it will play well to this crowd:

A physicist and a mathematician watch a man walk into a phone booth. Shortly afterward, two men walk out of the booth.

The physicist says, "There must be something wrong with our measurements."

The mathematician says, "If one more person walks into the booth, it will be empty."

 

[skatinginbc]

The mathematician says, "If one more person walks into the booth, it will be empty."

That sounds like something Sheldon (The Big Bang Theory) would say.

 

[Mathman]

A physicist and a mathematician watch a man walk into a phone booth. Shortly afterward, two men walk out of the booth.

The physicist says, "There must be something wrong with our measurements."

The mathematician says, "If one more person walks into the booth, it will be empty."

That sounds like something Sheldon (The Big Bang Theory) would say.

My hero!

Sheldon's version's of the circle problem goes like this. (And this really is how extra-galactic astronomers attempt to measure the curvature of the universe. So far all measurements are consistent with the Euclidean model -- curvature = 0.)

Suppose we could draw a circle of radius 2 meters in empty space, far from any gravitating matter. Suppose that we could measure the circumference so accurately that we could compute pi = C/D out to 52 decimal places. Suppose that the 52nd decimal place of pi computed by this method turned out to be 7 instead of 8.

What is the curvature of the universe?

Hint: c = [2 (textbook pi) / k ] sin (rk). Solve for k as above. (Answer: .07 per billion light years. )

Sheldon told me that.

 

[Olympia]

I told the joke to my math supervisor and he was laughing uproariously. There's something so satisfying about it as an illustration of negative numbers.