The Mathematics Thread!

[remember_Cedric]

"Maths can perform wonders!"

For the love of Maths, I present you this thread!

We don't solve your school assignments, but we want to exercise the logical side of your mind by practising some Math problems! Isn't that what part of Math teaches? Practice makes perfect!

Let's start with this appetitizer

Solve this!

Problem 1:

Tom has a saving of $924 (savings A). Tom owes Jerry loan $X. $Y of Tom's savings (savings A) is 20% of loan $X owing to Jerry. The balance of $924, which will be $Z, Tom needs to spend for his necessity.

1) Define Loan $X

2) Define Value $Y

3) Define Value $Z

Have fun!

[dictionary]

Yesterday my math teacher was lecturing us students about "practice makes perfect." He said it's not practice that makes perfect but rather "perfect practice makes perfect" 'cause if you practicing the wrong way then you're screwed. Hehe.

Yikes! You called that appetizer?

[IcyFox]

The original problem has no exact solution.

[1 0 0 | x ]
[0 1 0 | 0.2x ]
[0 1 1 | 924 ]

The solution will be in a parametric form.

[yittz]

Must be something wrong with the wording, or my intepretation of it. Or else it's what icy said.

12 weights. 1 of them is wrong (don't know if lighter or heavier than the rest).
You have a balance scale to compare weights one each side.
Using only 3 weighings, can you devise a way that will ensure you will find the reject and whether it's lighter or heavier.

[IcyFox]

Select 8 weights at random, 4 on each side. See which side is heavier.

Case 1 : If 1 side is heavier than the other.
Split the heavier pile into 2. Now you have 2 on each side.
See which side is heavier. Split into 2 again, 1 on each side.
You're done.

Case 2 : If both are equally heavy.
Take the 4 weights that you didn't touch yet and split into 2 sides.
See which side is heavier. Split into 2 again, 1 on each side.
You're done.


Flaw : This method assumes that weight is heavier than the rest. If it turns out to be lighter, then you will need to weigh 1 more time.

I tried a few other combinations and found no way around this problem. Trying 6 on each side makes it worse. Trying 4 at a time gives the same thing. Trying odd-numbered combo's is out of the question.

Essentially you need a min. of 3 tries, and if you're not too lucky, 4 times.

[remember_Cedric]

Yesterday my math teacher was lecturing us students about "practice makes perfect." He said it's not practice that makes perfect but rather "perfect practice makes perfect" 'cause if you practicing the wrong way then you're screwed. Hehe.

Yikes! You called that appetizer?

LOL Relax.... too complicated? We can skip it, as it'd require quite alot to solve this one.

The original problem has no exact solution.

[1 0 0 | x ]
[0 1 0 | 0.2x ]
[0 1 1 | 924 ]

The solution will be in a parametric form.

Well, the original problem would require "reverse engineering" done at a certain level.

Must be something wrong with the wording, or my intepretation of it. Or else it's what icy said.

I'll be back to deal with that when I have time... and mode.

Thank you people, for contributing your love~...

[IcyFox]

No amount of reverse engineering will give you the exact solution dude...


$x - the loan
$y - 20% of x
$z - balance

So :
(a) 5y = x
(b) y + z = 924

But x is NOT related to his SAVINGS!

That means $y can be anywhere from $0 to $924...

Let's say $y = $20. Then he owes the other bugger $100. His balance is then $904.

But now let's say he spends $4 instead, on say toothpaste. Then $y = $920 and the loan $x would be $4600.


SO WHICH SOLUTION IS IT??


There simply isn't enough information to tell, reverse engineering or upside-down engineering.

The rules of linear algebra state clearly what kind of systems have no solutions, 1 exact solution and a family of solutions. In this case the solution is a line (which can be expressed in parametric form).

[remember_Cedric]

No amount of reverse engineering will give you the exact solution dude...

// interesting solutions...//

SO WHICH SOLUTION IS IT??

There simply isn't enough information to tell, reverse engineering or upside-down engineering.

The rules of linear algebra state clearly what kind of systems have no solutions, 1 exact solution and a family of solutions. In this case the solution is a line (which can be expressed in parametric form).

My initiation problem works based on, solving one to get to another. Rightfully, in that order I've asked.

Should take quite a while. I have a concept of how it can be solved. Not now though , til my side of load is lighten, I'll be in full focus to get this done! Hasta la vista!

P/s: BTW, Icyfoxy, you seem to have a passion in Maths. Do you know about that theory of Maths, Matter, Mind? Under psychics.

Do you frequent your profile visitor messages, BTW?

[Han Solo]

The following was published in NY Times.

It’s adapted from a scene at Jason, a gathering of elite scientists near San Diego in the summer, described in the recent Times Magazine profile of Mr. Dyson by Nicholas Dawidoff:

At Jason, taking problems to Dyson is something of a parlor trick. A group of scientists will be sitting around the cafeteria, and one will idly wonder if there is an integer where, if you take its last digit and move it to the front, turning, say, 112 to 211, it’s possible to exactly double the value. Dyson will immediately say, “Oh, that’s not difficult,” allow two short beats to pass and then add, “but of course the smallest such number is 18 digits long.” When this happened one day at lunch, William Press remembers, “the table fell silent; nobody had the slightest idea how Freeman could have known such a fact or, even more terrifying, could have derived it in his head in about two seconds.” The meal then ended with men who tend to be described with words like “brilliant,” “Nobel” and “MacArthur” quietly retreating to their offices to work out what Dyson just knew.
Don’t worry — you don’t need any kind of award (or computer) to do this calculation. “Dyson wasn’t being coy when he said the problem isn’t difficult ­ once you know how to solve it,” says Pradeep Mutalik, a scientist at the Center for Medical Informatics at Yale. “In fact, the procedure to find the answer requires no more than 4th-grade arithmetic skills. I know, because I actually showed my fourth-grader daughter, Maya, how to do it, and she had no problem whatsoever in computing the answer.”

So how would a fourth-grader figure it out? And, for extra credit, try answering any or all of these questions from Dr. Mutalik about “Dyson numbers”:

Let’s call the smallest possible number that doubles its value when its last digit is moved to the front, the “Dyson number” for 2. The Dyson number for 3 would be one that tripled its value; the Dyson number for 4 would be one that quadrupled its value, and so on.

Now the questions are:

1. Do there exist Dyson numbers for all the digits from 2 to 9? If so, how many digits does each have, and what are they?

The Dyson number for 2 is indeed 18 digits long, as Mr. Dyson indicated. Some of the others are even longer, one being over 50 digits long. Such large numbers look very daunting. Even computers, conventionally used, are of no help in working with them since most common computer math libraries do not deal with integers up to so many significant digits. But you don’t need a computer to calculate them.

2. Let’s define a reverse Dyson number as one that results from the reverse procedure: moving the first digit to the end. Now there exists a number that is the Dyson number for one of the digits and the reverse Dyson number for another digit. What number is that?

3. Does every digit have a Dyson number? Explain why, or why not.

[yittz]

Cedric, the wording of your problem is seriously unclear. What the heck is the 'balance of $924'? Balance as in saving minus loan? Or balance as a redundant term refering to the number $924 (like bank balance)?

And $Y is such an arbitrary term that can be anything, it could be anything from $0 to $924.

Based on the wording, there is nothing wrong with Icy's solution.

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.

[IcyFox]

My initiation problem works based on, solving one to get to another. Rightfully, in that order I've asked.

Should take quite a while. I have a concept of how it can be solved. Not now though , til my side of load is lighten, I'll be in full focus to get this done! Hasta la vista!

OK. Can't wait for it...

P/s: BTW, Icyfoxy, you seem to have a passion in Maths. Do you know about that theory of Maths, Matter, Mind? Under psychics.

Do you frequent your profile visitor messages, BTW?

NO I DO NOT have a passion for math. I'm a major in physics, and as a result I get to do lots of math - LOTS and LOTS of math...


What's on my profile? And I think psychics is a bunch of baloney. So there.

[IcyFox]

As for the Dyson numbers problem, I dun think I can solve it. Oh well...


BTW guys, Freeman Dyson made very important contributions to physics and math, in particular, showing that Feynman's path integral formulation of quantum electrodynamics is equivalent to Schwinger's and Tomonaga's formal mathematical method.

The history of quantum electrodynamics is very interesting in itself and is told from Feynman's perspective here:

http://nobelprize.org/nobel_prizes/p...n-lecture.html

[Han Solo]

As for the Dyson numbers problem, I dun think I can solve it. Oh well... [/url]

If you don't try, you will never solve it. In fact, it's not that hard.

I'll post the answer after the weekend is over.

Han Solo

[Felix]

Dyson number...

click to show/hide spoilers

105263157894736842

[IcyFox]

If you don't try, you will never solve it. In fact, it's not that hard.

I'll post the answer after the weekend is over.

OK then at least wait for me to attempt it before posting the solutions.

But I think posting the answers are OK. I've seen them already.

[Han Solo]

Dyson number...

click to show/hide spoilers

105263157894736842

Correct!!.

Can you post your solution?

Han Solo

[IcyFox]

OK I managed to solve Dyson Number 4 (sort of).


(For the trivial cases of 0 and 1, their Dyson numbers are just themselves.)


Here are the Dyson numbers from 2 to 9:


2: 105,263,157,894,736,842 — 18 digits
3: 1,034,482,758,620,689,655,172,413,793 — 28 digits
4: 102,564 — 6 digits
5: 142,857 — 6 digits
6: 1,016,949,152,542,372,881,355,932,203,389,830,508, 474,576,271,186,440,677,966 — 58 digits
7: 1,014,492,753,623,188,405,797 — 22 digits
8: 1,012,658,227,848 — 13 digits
9: 10,112,359,550,561,797,752,808,988,764,044,943,820 ,224,719 — 44 digits




Link : http://tierneylab.blogs.nytimes.com/...r-dyson-puzzle

[IcyFox]

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!

[IcyFox]

OK here are the obstacles I've met while working on the Dyson numbers.


First off, computing the number of digits can be tedious, especially for 6 and 9, where there are 58 and 44 digits respectively.

But this is easily fixed by recognizing the patterns. For example, in the case of 4, the numbers are always in the form:


99...6a = 39...f + 39...e + etc ...

So I just had to 'add' another 9 in front and see if it's divisible by 39.


Second, I found that this method is a bit buggy for Dyson Number 5. Either it's a unique case or I was careless again.

But, I noticed, among other things, that Dyson Number 5 doesn't end in 5. It ends in 7, while the rest of the Dyson numbers end in 2, 3, 4 etc... respectively.

And, as pointed out by others, Dyson Number 5 is also the reverse Dyson Number for 3, which means
142,857 x 3 = 428571.


Third, using my method doesn't tell you what the last digit of the number should be, and I believe it's only coincidental that they end in 2, 3, 4 etc...

However, neither does the 'official' method I found online, which is different from mine, by the way, provide any solution or proof for that.

So save guessing or trying all the numbers systematically, I don't see any way around this problem currently.


I'd be glad to hear of any possible solutions from other users here.

[IcyFox]

Anyways, here's the official solution.

click to show/hide 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.