The Mathematics Thread!

[yittz]

OK here is how I solved Dyson number 4:


First we would need to check how many digits it would have. This I did by the hack method.

Let's say it's 2 digits. What would we get? When we take the last digit and place it right in front, we are supposed to get 4 times the original digit.


Breaking it down into the 'tens' and 'ones' placing, here's what we get:

10a + b = 4(10b + a)
10a - 4a = 40b - b
6a = 39b


But 6 is not divisible by 39 without leaving a remainder, so the 2-digit case is out. Now let's try the 3-digit case.


100a + 10c + b = 4(100c + 10b + a)
96a = 390c + 39b


Still doesn't work. Let's try 4 digits.

996a = 3900d + 390c + 39b

Nope. How about 5 digits?

9996a = 39000e + 3900d + 390c + 39b

Still no. How about 6 digits?

99996a = 390000f + 39000e + 3900d + 390c + 39b


Yep! 99996 is divisible by 39! We have a Dyson number for 4 at 6 digits. OK, our work is not done yet. Let's continue.


99996 / 39 = 2564

So, 2564a = 10000f + 1000e + 100d + 10c + b


Now all we need to find are values of a, b, c, d, e & f that will fit the equation above. Sounds daunting? Not really.

You see, the b, c, d ... etc are already in the 'ones', 'tens', 'hundreds' places respectively!

If we know a, which actually corresponds to its respective Dyson number - in this case 4, we can solve the Dyson number with just one computation in the calculator.


Thus, 2564 x 4 = 10256.

Putting back the 4 in the original place, we now have 102564.


We have just solved Dyson Number 4!

Thankyou! Could not put that in words. I did the same with 2. 9999...8 divide by 19. Only divisible when its 18 digits.

Apart from 5, all the other Dyson numbers start with 10..

When I used the 4, (1)64, (6)564, (26)2564, (105)02564. ie 4+4x40^1+4x40^2 etc. etc. until the it's 0 in the front with a 1 remainder.

Han solo, you lied!! Having a calculator definitely helps.

I use 2 as the last number for 2, 3 for 3 etc. because the division of it = 1, which is what I want for my first digit. Not sure why it's 7 for 5. Other than 5 x 14 = 7.

EDIT: oh you had more posts.

[remember_Cedric]

Anyways, here's the official solution.

 click to show spoilers

For Dyson Number 2:

2 x 2 = 4
4 x 2 = 8
8 x 2 = 16 (6 carry 1)
.
.
.


So we have:

1 ... 36842

(To get the 3, we have 1x2 for that place and 1 from the carry from 16, so 2 + 1 = 3)

Finally we should get : 105,263,157,894,736,842


Yup. The whole process starts with 2 and keep multiplying until you get 1 right at the end without any carry.


Still, 2 things to note:


1. This method also doesn't say why we should start with 2. Same as my method, you either guess it right or try every single digit. (Seems quite arbitary to me actually...)

2. This method doesn't say why it should end in 1. So yes, I agree that you can just teach this method to the 4th-grader to compute all the Dyson numbers but he/she won't be able to figure out how it works or why it even works.

Gawd! Icyfox, you're the god of Mathematics!

I love physics! Can one really have the capability of bending a spoon?

[IcyFox]

I love physics! Can one really have the capability of bending a spoon?

Yup, anyone can bend a spoon. Just apply enough pressure.

[Nefertari]

Yup, anyone can bend a spoon. Just apply enough pressure.

Good answer! Icyfox. You make me laugh!

I thought you'd want to go into all that elasticity vs strength talk. Heh~

Or perhaps just make sure that spoon is made of rubber =)

[Nefertari]

On Han Solo's Dyson number.

No way I can do it in 2 secs lol. It has something to do with 2, then 42, then 842, 16842, 336842, 6736842 etc. But this takes soo long. Takes me 2 sec just to find the next number in the series. I think you need a 1 as the first digit, but that doesn't include the non-single digit numbers. Ba hard to explain.

So for dyson 3, it will be 3, 93, 363, 8463 until we meet 1 again. But I dont know why I started with 3 as last digit, if you use 6, does that mean you are looking for a 2?

Don't like this maths question, takes too long.

Hahaha, i'm sure you can once you have coffee early in the morning, and your mind is working at a ultra-fast mode.

Dyson's mind probably calculated most of the numbers by the first beat, then on the second, he already got his answer. One must remember that two beats in the past does not exactly mean 2 sec. One beat could be 5 sec apart from another.

Remember how Cao Cao's son made up the famous poem in 7 steps? Well, 7 steps is like 20 seconds today. Making up poems is also a everyday pastime back then, so it's not surprising for him to accomplish that.

Same for Dyson, he lives with Maths everyday. So no need to be so surprised. But he is indeed brilliant. Too bad he never gotten a Nobel.

[Han Solo]

Congrats on the solving of the Dyson Number.

The following is a copy+paste from another website.
http://mathforum.org/isaac/problems/zeno1.html
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The great Greek philosopher Zeno of Elea (born sometime between 495 and 480 B.C.) proposed four paradoxes in an effort to challenge the accepted notions of space and time that he encountered in various philosophical circles. His paradoxes confounded mathematicians for centuries, and it wasn't until Cantor's development (in the 1860's and 1870's) of the theory of infinite sets that the paradoxes could be fully resolved.
Zeno's paradoxes focus on the relation of the discrete to the continuous, an issue that is at the very heart of mathematics. Here we will present the first of his famous four paradoxes.
Zeno's first paradox attacks the notion held by many philosophers of his day that space was infinitely divisible, and that motion was therefore continuous.

Paradox 1: The Motionless Runner

A runner wants to run a certain distance - let us say 100 meters - in a finite time. But to reach the 100-meter mark, the runner must first reach the 50-meter mark, and to reach that, the runner must first run 25 meters. But to do that, he or she must first run 12.5 meters.

Since space is infinitely divisible, we can repeat these 'requirements' forever. Thus the runner has to reach an infinite number of 'midpoints' in a finite time. This is impossible, so the runner can never reach his goal. In general, anyone who wants to move from one point to another must meet these requirements, and so motion is impossible, and what we perceive as motion is merely an illusion.

Where does the argument break down? Why?

[remember_Cedric]

Hey folks! Check this:-

http://www.math.com/

Hope you'll find it helpful.

BTW, Matlab, anyone?

P/s: Gawd, Han, which level of Math is that?

[IcyFox]

The great Greek philosopher Zeno of Elea (born sometime between 495 and 480 B.C.) proposed four paradoxes in an effort to challenge the accepted notions of space and time that he encountered in various philosophical circles. His paradoxes confounded mathematicians for centuries, and it wasn't until Cantor's development (in the 1860's and 1870's) of the theory of infinite sets that the paradoxes could be fully resolved.
Zeno's paradoxes focus on the relation of the discrete to the continuous, an issue that is at the very heart of mathematics. Here we will present the first of his famous four paradoxes.
Zeno's first paradox attacks the notion held by many philosophers of his day that space was infinitely divisible, and that motion was therefore continuous.

Paradox 1: The Motionless Runner

A runner wants to run a certain distance - let us say 100 meters - in a finite time. But to reach the 100-meter mark, the runner must first reach the 50-meter mark, and to reach that, the runner must first run 25 meters. But to do that, he or she must first run 12.5 meters.

Since space is infinitely divisible, we can repeat these 'requirements' forever. Thus the runner has to reach an infinite number of 'midpoints' in a finite time. This is impossible, so the runner can never reach his goal. In general, anyone who wants to move from one point to another must meet these requirements, and so motion is impossible, and what we perceive as motion is merely an illusion.

Where does the argument break down? Why?

Isn't this logically the same as the 0.999... = 1 problem?

[KeongJai]

Since space is infinitely divisible, we can repeat these 'requirements' forever. Thus the runner has to reach an infinite number of 'midpoints' in a finite time.

That assumption is incorrect. Time can be infinitely broken down too.

[IcyFox]

Since space is infinitely divisible, we can repeat these 'requirements' forever. Thus the runner has to reach an infinite number of 'midpoints' in a finite time.

That assumption is incorrect. Time can be infinitely broken down too.

And so not only is the runner unable to move, time is unable to pass as well.

Claiming that the assumption is incorrect by bringing up a totally different and irrelevant assumption solves the problem perfectly well.

[KeongJai]

well if you look at each point of time as a reference to the distance he moves, then he's moving when you sum up the infinite points.

[KeongJai]

so saying he's not moving is given since time isn't moving. nothing moves in a single instance in time - therefore there no problem or paradox.

[hkopinions]

keongjai solved it.
This is a common problem through out by college professors in introductory math courses. usually someone solves it right away cause it is such a classic problem.

[IcyFox]

Yup. Math is all about popular opinion.

[yittz]

Well, if you've got a problem with his answer just kindly explain why.

the runner has to reach an infinite number of 'midpoints' in a finite time. This is impossible, so the runner can never reach his goal.

It's possible to reach an infinite number of midpoints in a finite time, as the respective times taken to reach those midpoints add up to that finite time it takes to reach 100m. So similar to the infinite series 1/2+1/4+1/8...=1

[Han Solo]

Congrats Keong Jai.

Maybe i should start giving out cookies for correct answers.

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Easy one this time.


Who Owns the Crocodile
There are 5 girls in a long row in maths.

Each girl has a favourite chocolate bar, colour, pet, hobby, and would like to go on a certain holiday.

All the girls like different things.

Your task is to solve the following clues - "who owns the crocodile"

Jo likes the Wispa Bite
The person with the hamster likes swimming
Hannah eats Dairy Milk
Jessica is on the left of Georgina
Lucy is the first on the left
The first person on the right likes swimming
The person who eats Milky Bars owns a horse
The person in the middle eats Dairy Milk
Jessica likes green
The person on the left of the middle wants to go to Tobago
The person who wants to go to the Maldives likes lilac
The person who likes Wispa Bites sits next to the person who wants to go to Florida
The person who likes pink wants to go to Florida
the person who sits first on the left likes lilac
The girl that likes blue owns a puppy
The person who likes skiing sits next to the person who has a hamster
The girl on the right of the girl who likes tennis likes horse riding
The girl next to the girl who likes Milky Bars likes Boost
The girl who likes purple wants to got to Canada
The girl who likes Crunchies owns a rabbit
The girl who likes skiing sits next to the girl who plays ten-pin bowling
Jessica wants to go to Australia


This puzzle was created by pupils from Wadebridge School in Cornwall.

[yittz]

If you own a crocodile, I guess you won't worry about alligators if you meet one on holiday.

[Han Solo]

If you own a crocodile, I guess you won't worry about alligators if you meet one on holiday.

I had seen those 6 feet croc in northern territories, definitely not something to forget in a hurry, they are fast, mean, and full of teeth.

Han Solo
P/s: you are meant to solve the puzzle

[yittz]

i got Hannah, didn't double check. I just thought it was interesting that someone who owns a croc wants to visit florida, and the the horse rider don't own a horse.

[Han Solo]

i got Hannah, didn't double check. I just thought it was interesting that someone who owns a croc wants to visit florida, and the the horse rider don't own a horse.

Ding Ding, Correct, i think.

Han Solo