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Maths Help (for the know alls :P)
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RE: Maths Help (for the know alls :P)
quote:
Originally posted by John Anderton
Where did i go wrong
quote:
Originally posted by CookieRevised
This is not yet the answer! Now it is time to read the question again. They asked what position the number is in. And positions useually start at 1. To make no confusion say this also in your answer :

= 411 (with first position being 1)

04-03-2005 03:25 PM
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RE: RE: Maths Help (for the know alls :P)
quote:
Originally posted by John Anderton
sorry that should have been 4 letters .... a,b,c,d and e
a,b,c,d and e are still 5 letters. Make up your mind :P

quote:
Originally posted by John Anderton
I dunno that formula that u r using .... can u do the same question with permutation and combination .....
perhaps, but that is a very very confusing and long long way to do it.

quote:
Originally posted by John Anderton
1st word aaaaa
2nd        aaaab
3           aaaac
4           aaaad
5           aaaba
So now you're saying only 4 letters are possible??? a, b, c, d? In that case:

adbca = 03120

To calculate from a base 4 number to the equivalent base ten number:
= 0 * 4^4   +   3 * 4^3   +   1 * 4^2   +   2 * 4^1   +   0 * 4^0[/b]
= 216

or 217, considering the first position is position 1.

I'm not going to repeat the whole thing again though, see my previous reply in how you calculate this and the simple logic behind it. And follow the method with a piece of paper and create the tables on paper.

The important part is (for 5 possible letters): a, b, c, d, e
  dictionary | base 5 number | base 10 number (aka position-1)
  aaaaa          00000               0
  aaaab          00001               1
  aaaac          00002               2
  aaaad          00003               3
  aaaae          00004               4
  aaaba          00010               5
  aaabb          00011               6
  aaabc          00012               7
  aaabd          00013               8
  aaabe          00014               9
  aaaca          00020               10
  (...)            (...)                (...)

The important part is (for 4 possible letters): a, b, c, d
  dictionary | base 4 number | base 10 number (aka position-1)
  aaaaa          00000               0
  aaaab          00001               1
  aaaac          00002               2
  aaaad          00003               3
  aaaba          00010               4
  aaabb          00011               5
  aaabc          00012               6
  aaabd          00013               7
  aaaca          00020               8
  (...)            (...)                (...)


quote:
Originally posted by John Anderton
Ok so 4 changes r possible in the 1st letter (from the right) (like units place in numbers) ....
Wrong... looking at your dictionary you just posted:
1st word aaaaa => no change!
2nd        aaaab => 1st change
3           aaaac => 2nd change
4           aaaad => 3rd change
5           aaaba => 4th change, we go to the next unit

I see only 3 possible changes for the first unit!!! a->b  b->c  c->d

As I say, don't use the method your using, it is very very long and very very confusing and you bound to make (logic) errors.

Look at "adbca" as a code, a series of numbers, thus convert it to numbers first. Not as some string where you calculate the "changes"...


quote:
Originally posted by John Anderton
5th change (note: this is wrong if there are only 4 possible letters... it's the 4th change!!! see above) will make the 2nd letter from the right change once. So When the 1st letter changes 25 times (5*5) then the 3rd letter from the right will change. When the 1st letter changes 125 times then the 4th letter changes once .....
the 4th letter from the right changes 3 times (a -> b -> c ->d) thus ..... 125*3 = 375
Revise what you just told here and put it in simple maths:

for each unit: "number of changes" * "total possible numbers" ^ "position of unit"

note: in maths we start counting at 0, not 1!! 0 is also a number, the first number.

eg 1:
- total numbers = 5 possible letters: a, b, c, d, e
- number of changes (value of the letter) for unit 0 in "adbca" = 0  => 0 * 5 ^ 0
- number of changes (value of the letter) for unit 1 in "adbca" = 2  => 2 * 5 ^ 1
- number of changes (value of the letter) for unit 2 in "adbca" = 1  => 1 * 5 ^ 2
- number of changes (value of the letter) for unit 3 in "adbca" = 3  => 3 * 5 ^ 3
- number of changes (value of the letter) for unit 4 in "adbca" = 0  => 0 * 5 ^ 4
- 0*5^4 + 3*5^3 + 1*5^2 + 2*5^1 + 0*5^0 = 410


eg 2:
- total numbers = 4 possible letters: a, b, c, d
- number of changes (value of the letter) for unit 0 in "adbca" = 0  => 0 * 4 ^ 0
- number of changes (value of the letter) for unit 1 in "adbca" = 2  => 2 * 4 ^ 1
- number of changes (value of the letter) for unit 2 in "adbca" = 1  => 1 * 4 ^ 2
- number of changes (value of the letter) for unit 3 in "adbca" = 3  => 3 * 4 ^ 3
- number of changes (value of the letter) for unit 4 in "adbca" = 0  => 0 * 4 ^ 4
- 0*4^4 + 3*4^3 + 1*4^2 + 2*4^1 + 0*4^0 = 216



eg 3 ("special" case: 10 possible numbers, in other words, convert from base 10 (=a decimal number) to base 10 (=a decimal number), aka there is no change at all):
- total numbers = 10 possible letters: a, b, c, d, e, f, g, h, i, j
- number of changes (value of the letter) for unit 0 in "adbca" = 0  => 0 * 10 ^ 0
- number of changes (value of the letter) for unit 1 in "adbca" = 2  => 2 * 10 ^ 1
- number of changes (value of the letter) for unit 2 in "adbca" = 1  => 1 * 10 ^ 2
- number of changes (value of the letter) for unit 3 in "adbca" = 3  => 3 * 10 ^ 3
- number of changes (value of the letter) for unit 4 in "adbca" = 0  => 0 * 10 ^ 4
- 0*10^4 + 3*10^3 + 1*10^2 + 2*10^1 + 0*10^0 = 3120


eg 4 ("special" case: 2 possible numbers, in other words, convert from base 2 to base 10, aka convert a binary number to a decimal number):
- total numbers = 2 possible letters: a, b       so lets take the string "ababa" as an example
- number of changes (value of the letter) for unit 0 in "ababa" = 0  => 0 * 2 ^ 0
- number of changes (value of the letter) for unit 1 in "ababa" = 1  => 1 * 2 ^ 1
- number of changes (value of the letter) for unit 2 in "ababa" = 0  => 0 * 2 ^ 2
- number of changes (value of the letter) for unit 3 in "ababa" = 1  => 1 * 2 ^ 3
- number of changes (value of the letter) for unit 4 in "ababa" = 0  => 0 * 2 ^ 4
- 0*2^4 + 1*2^3 + 0*2^2 + 1*2^1 + 0*2^0 = 10

=> this last example can easly be done totaly on paper (only 11 rows to be made in the table), so do this to understand the whole thing:
  dictionary | base 2 number | base 10 number (aka position-1)
  aaaaa          00000               0
  aaaab          00001               1
  aaaba          00010               2
  aaabb          00011               3
  aabaa          00100               4
  aabab          00101               5
  aabba          00110               6
  aabbb          00111               7
  abaaa          01000               8
  abaab          01001               9
  ababa         01010              10   < =row/position 11
  (...)            (...)                (...)



quote:
Originally posted by John Anderton
so a total of 410 words have passed and we have reached adbca. So shouldnt that be the answer :S
Read what you just said ;) we have passed 410 words, so we are at position 411...
(provided there are 5 possible letters (a, b, c, d, e)! Which is apparently not what you've showed in your last dictionary in your last post, so make up your mind on the number of possible letters.)





This post was edited on 04-03-2005 at 04:39 PM by CookieRevised.
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04-03-2005 04:04 PM
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RE: Maths Help (for the know alls :P)
quote:
Originally posted by CookieRevised
(provided there are 5 possible letters (a, b, c, d, e)! Which is apparently not what you've showed in your last dictionary in your last post, so make up your mind on the number of possible letters.)
Damn dodgy typos :dodgy:
Sorry ....
4 letters .... a, b, c and d :P
Now the answer is ^o)
Ohh ... i give up :P
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04-03-2005 05:35 PM
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RE: Maths Help (for the know alls :P)
the answer is:

- total numbers = 4 possible letters: a, b, c, d
- number of changes (value of the letter) for unit 0 in "adbca" = 0  => 0 * 4 ^ 0
- number of changes (value of the letter) for unit 1 in "adbca" = 2  => 2 * 4 ^ 1
- number of changes (value of the letter) for unit 2 in "adbca" = 1  => 1 * 4 ^ 2
- number of changes (value of the letter) for unit 3 in "adbca" = 3  => 3 * 4 ^ 3
- number of changes (value of the letter) for unit 4 in "adbca" = 0  => 0 * 4 ^ 4
- 0*4^4 + 3*4^3 + 1*4^2 + 2*4^1 + 0*4^0 = 216

=> position or row number 217
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04-04-2005 04:04 AM
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RE: Maths Help (for the know alls :P)
Cookie, explain how you can have the square root of minus 1?!? I mean, I thought/think that the square root is like the opposite of ^2, as + is opposite to -. Since x^2 = x*x, the answer is always positive ( -3 * -3 = 9 and 3 * 3 = 9). So you can't have [something]^2 and then get -1, so you can't do sqrt(-1) either. That's what I learned anyway.. :dodgy: Dutch schools :P (or :dodgy: Cookie)
04-04-2005 05:49 PM
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RE: Maths Help (for the know alls :P)
Excellent and interesting question though. But, you will learn that in the futur. It's the same as you didn't learn about integrals and deviations yet. Everything will come in time, and you first need to understand other things before you can understand this ;)

Anyways, here is a shot (and some interesting history of maths)...

No, taking a square root of a negative number is, as you pointed out, impossible in the real world. However, that doesn't mean you can't do maths with the "impossible". It's something like the number "infinity". Infinity doesn't exist in real life, everything is limited. But you still need something to do maths with and to represent "something" beyond our scope of thinking.

This is the same for taking a square root of a negative number. In some calculations like quadratic and cubic equations (like used ealier in this thread) you will see that you can have a result which actually can't exist, but still it _is_ a result.

In the past, people stopped there, they couldn't go further. But one day, somebody said, what if we just invent a new number, and lets call it i (from "imaginary number"), i would be equal to the square root to -1. From that time on people could do maths and work out solutions further where they were stuck in the past.

note: such things happened in the past also (and still are happening now). It is only since the 13th century that negative numbers were slowly considered real numbers in the Western world. Before that, anything lower then 0 simpy didn't exist.

So, a whole new group of numbers emerged, the "complex numbers" written as "(x, y)" where x is a real number and y the imaginary number. So the complex number (5, 4) means 5+4i or 5+4*SQR(-1)...

So you can do maths with them too of course:
(5, 4) + (5, 4) = 5+4i + 5+4i = 10 + 8i

And you can even solve this strange question: divide the number 10 into two parts so that their product is 40! Impossible you would think, well not with complex numbers, the solution: 5+SQR(–15) and 5–SQR(–15). The proof:

[5+SQR(–15)]  +  [5–SQR(–15)]
=>  5 + 5 + SQR(–15) - SQR(–15)
=>  5 + 5
=>  10, hey presto!

[5+SQR(–15)]  *  [5–SQR(–15)]
=>  5²  -  5*SQR(–15)  +  5*SQR(–15)  -  SQR(–15)²
=>  25  -  -15
=>  25  +  15
=>  40, hey presto!

Note: this is actually a "famous" calculation/question from Cardano (16th century). He didn't actualy invented the imaginary number i and also didn't do anything further with this calculation or understood what he just found. This was actually the result of a competition to solve the cubic equation. Which was thought to be impossible. Cardona solved it by using negative numbers (a great controversy in that time) and this "complex" number thing was only a small byproduct of it.

Cardano did not go further into this, and it is only later that they were called complex numbers. A few years later Bombelli gave several examples involving these new type of numbers.

eg:

One of Cardano's cubic formulas gives the solution to the equation:
x³ = cx + d
as
x = SQR³[d/2 + SQR(e)] + SQR³[d/2 – SQR(e)]    where e = (d/2)² – (c/3)³
I've used "SQR³(x)" to indicate the third root of x

The 'problem' is that e could be negative, but there isn't any real number which you can multipy by itself and get a negative number, so Cardano was stuck there. Nevertheless Bombelli used this formula to solve the equation:
x³ = 15x + 4

according to Cardano's formula:
c = 15
d = 4
thus e = (4/2)² – (15/3)³  =>  2² - 5³  =>  -121  (a negative number no less!!!)
and thus x = SQR³[2 + SQR(–121)] + SQR³[2 – SQR(–121)]

Now, the square root of –121 is not a real number; it's neither positive, negative, nor zero. Nevertheless, Bombelli continued to work with this expression until he found equations that lead him to the solution 4, a real number!
So there _is_ a real solution eventhough SQR(–121) can't exist.

This shows nicely how complex numbers do have their use to solve things, even though they can't exist. Prior to the discovery of this, people said there was no solution. Although 4³ is equal to 15 * 4 + 4....


There is more detail to this story though:

When cubic formulas were a big hush-hush in the 16th century, Cardano noted that the sum of the three solutions to a cubic equation (x³ + bx² + cx + d = 0) is –b, the negation of the coefficient of x².

By the 17th century the theory of equations had developed so far so that someone called Girard developped a principle of algebra out of this. He also said that an nth degree equation always had n solutions. This is what we now call "the fundamental theorem of algebra". But He never was able to proof this principle though, just because some solutions couldn't "exist".

Another guy also studied this relation between solutions and coefficients, and was able to show that Cardano and Girard were right if you disregarded negative solutions as "false" and those other impossible solutions as "imaginary".

And slowly negative numbers became a common good and raised in status (by the results and efforts of other people), but it was not until the 18th century that complex numbers became in real use, roughly 200 years after their first "discovery" . They weren't considered to be real numbers though, but they were useful in the theories and formulas back in that time.

But still nobody was able to proof the fundamental theorem of algebra which stated that there are n solutions of an nth degree equation. So, although the complex numbers were not fully understood, the square root of -1 was being used more and more.

By the end of the 18th century numbers of the form x + yi were used pretty often by research mathematicians. For example Euler, a very influencial mathematician used it often in his trigonometric functions. And because of this and his influences, it was Gauss, and others like Wessel and Argand, who began to study complex numbers in function of xy-planes (as coordinates of planes) and that complex numbers were fully understood. And thus in 1799, Gauss published the first proof to the Girard's principle of algebra.

(You might know Gauss from the "Gauss-curve", it's the same one. And Euler is also a name which you certainly will hear in the next years in school, if you already didn't. And this whole deal with complex numbers will certainly be on the agenda and you'll notice that it will all start with talk about coördinates of planes and the polar coördinates system ;))

This post was edited on 04-05-2005 at 03:33 AM by CookieRevised.
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04-05-2005 02:07 AM
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RE: Maths Help (for the know alls :P)
holy shit... [Image: train.gif]



im suuuuuuuch an asshole for not understanding a  thing of this ! *cries*

This post was edited on 04-05-2005 at 11:27 AM by Fredzz.
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04-05-2005 11:26 AM
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RE: Maths Help (for the know alls :P)
a quote from one of my maths proffessors...

"life is complex....
...it has real and imaginary parts"
04-05-2005 11:33 AM
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RE: Maths Help (for the know alls :P)
quote:
Originally posted by Fredzz
im suuuuuuuch an asshole for not understanding a  thing of this ! *cries*
Its an imaginary number :D
quote:
Originally posted by CookieRevised
- 0*4^4 + 3*4^3 + 1*4^2 + 2*4^1 + 0*4^0 = 216
=> position or row number 217
So i got the 1st one correct ... my ans at 1st was correct ... but the question i typed wasnt :P
But by the method i use .... i count only the changes ... oh i get it ...
i got the no of changes as 216 so the position is 217 ;)
thanks cookie :)
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04-05-2005 11:37 AM
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RE: Maths Help (for the know alls :P)
Thanks aswell Cookie.
04-05-2005 02:00 PM
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