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## #1 Re: Help Me ! » Proof - Right Triangle » 2006-01-23 22:14:59

As I have to go to school now I will post the proof tonight.

## #2 Re: Help Me ! » Looks simple but actually quite hard » 2006-01-15 05:25:35

Ricky,

Can you show us the algorithm that is being used or write here how  x^4 - x^3 - 10x^2 - 5x + 7 = 0 is being factored as it is very important for me to understand it. Thanks.

## #3 Help Me ! » Looks simple but actually quite hard » 2006-01-14 23:56:45

krisper
Replies: 7

Hi guys,
I have a problem with an equation. Here it is: (x^2 + 4x + 5)/(x+1) = 3√(x+2). I have tried everything but still not getting the roots?! Please help. Thanks.

## #4 Re: Help Me ! » Please Help Me With A Permutation Question!!! » 2006-01-14 23:11:17

Ah, reading again the M.ANAND KUMAR post I understand that he wanted to use any digit in his 8 digits phone number. Sorry, I was wrong, I thought that the 8 digits are constant. Your answer is correct.

## #5 Re: Help Me ! » Please Help Me With A Permutation Question!!! » 2006-01-14 23:05:06

Ricky,

I checked again and it is correct - the formula is n!/k! where n is the number of all digits and k is the number of the repeated one. And here we are talking about exactly 8 digits that are used: for example we have the number
98347836 - and we want to keep the first 2 digits constants and we can permute the other ones: 3,4,7,8,3,6.
and here we have 2 threes, so the formula is 6!/(2*1).
98 347836
98 427836
98 428736
....
You can count them to see if I am wrong.

## #6 Re: This is Cool » Really cool thing with the numbers like a^x » 2006-01-14 08:20:36

When is 0 you will get Z=1 and 0! equals 1, so n can be 0 And you are right again, it is only for integers because n! formula can only be used for integers that are >=0.

P.S. I have edited my first post telling that x must be integer

## #7 Re: Help Me ! » Please Help Me With A Permutation Question!!! » 2006-01-14 07:42:55

Frankly,

OK, if you have 8 digits number and you want to keep the first 2 digits constants here is what you do:

You have left 6 digits to permute. If the 6 left digits are unique your formula is n! where n equals 6. So you have 6*5*4*3*2*1 different 8 digits numbers. BUT when you have reapeated digits and their number is k, your formula is n!/k!

I hope it makes sence.

## #8 Re: This is Cool » Really cool thing with the numbers like a^x » 2006-01-14 07:12:52

And yes, you are correct and  that why I said that x should be >=0 By the way I found this on my own a couple of years ago when I was playing of numbers hoping I will find something significant (everyone's dream ). It is very nice though, because when repeating the steps you will always get to the 0 value.

## #9 This is Cool » Really cool thing with the numbers like a^x » 2006-01-14 05:37:40

krisper
Replies: 6

Here is what we do:
1. We have the numbers from 1 to 10 a∈{1,2,3,4,5,6,7,8,9,10}.
2. x is the power of these numbers.
3. Take any random integer number x that is between 0 and +∞.
4. Calculate
1^x = b1
2^x = b2
3^x = b3
4^x = b4
5^x = b5
6^x = b6
7^x = b7
8^x = b8
9^x = b9
10^x = b10
5. Now the fun part We calculate this:
c1 = b2 - b1;
c2 = b3 - b2;
c3 = b4 - b3;
c4 = b5 - b4;
c5 = b6 - b5;
c6 = b7 - b6;
c7 = b8 - b7;
c8 = b9 - b8;
c9 = b10 - b9;
6. Now we have the set of numbers c1..c9. We do the same and get the set d1..d8 (d1 = c2 - c1, d2 = c3 - c2 and etc.). We do this until we get a set of numbers that has equal numbers and if we decide to subtract again we will get 0. These equal numbers we will call Z.
7. Write here what value of x have you used and what Z have you got as answer?

P.S.
If all your calculations are correct you will have a Z that equals x!

## #10 Re: Help Me ! » help help » 2006-01-13 23:21:22

kempos you are correct
It should be:

cos144 + cos72 = 2cos(144+72)/2*cos(144-72)/2=2cos108*cos36=
2cos(180-72)*cos36=-2cos72*cos36=(-cos72*2cos36*sin36)/sin36=
(-cos72*sin72)/sin36=(-1/2)*(sin144)/sin36=(-1/2)*(sin144)/(sin(180-36))=
(-1/2)(sin144)/(sin144)=-1/2

## #11 Introductions » And Hello From Me Too :) » 2006-01-13 10:15:10

krisper
Replies: 3

Hi all My name is Ivan and I am from Bulgaria. I live in a small city called Botevgrad which is situated near our capital city Sofia. I am 18 years old, and the things I like doing most are - maths, programming and basketball I found this site accidently and in the next 15 to 20 seconds I loved it. You can learn great things here. I applaud everyone responsible for this site to happen - great work guys! And to all others - keep up the good work.

## #12 Re: Help Me ! » help help » 2006-01-12 22:33:31

cos144 + cos72 = 2cos(144+72)/2*cos(144-72)/2=2cos108*cos36=
2cos(180-72)*cos36=-2cos72*cos36=(-cos72*2cos36*sin36)/sin36=
(-cos72*sin72)/sin36=(-1/2)*(sin144)/sin36=(-1/2)*(sin144)/(sin(180-36))=
(-1/2)(sin144)/(sin144)=-1/2

## #13 Re: Help Me ! » A cubic... » 2006-01-04 22:16:31

I am sure you could find this without using the cubic formula.
x^3 - x - 1 = 0
x(x^2-1) = 1
if x = 0, the equasion has no meaning, so x must differ from 0 (x<>0); Now we can devide by x.
x^2 - 1 = 1/x.
We draw the graphics of these two expressions - x^2 - 1 and 1/x. After that we check where they cross eachother. This happens only in I quadrant which means that x^3 - x - 1 = 0 has only one real root. Afterwards doing some calculas we find that the root is somewhere between 1 and √2. Now we have to use the tangents and make some calculations and I am sure we will get the exact value of this root.

## #14 Re: Help Me ! » inequality » 2006-01-04 06:19:13

This is exactly where I am, but I just cant figure out how to use that. I guess I have to express x with y but just dont know how.

## #15 Help Me ! » inequality » 2006-01-03 22:32:52

krisper
Replies: 4

For y and x we have this : x(x-15) + y(y-10) + 78 <= 0; Find the max and the min value of the expression : 3x + 2y.

I dont know if this is the right method for this problem but here is where I am:

I. x^2 - 15x + y^2 - 10y + 78 <= 0
5(3x + 2y) => x^2 + y^2 + 78
And here I dont know what to do so I try another method:

II.  x^2 - 15x + y^2 - 10y + 78 <= 78
x^2 - 15x + 225/4 <= 10y - y^2 - 25 + 13/4
(x-7,5)(x+7,5) <= -(y-5)(y-5) + 13/4
(x-7,5)(x+7,5) <= (√13/2 - y + 5)(√13/2 + y - 5)
(x-7,5)(x+7,5) <= 1/4(√13 - 2y + 10)(√13 + 2y - 10)

But here I am stuck again. I find the roots of these 2 equasions for x and y, draw their parabolas and compare them but the only thing I find is an interval of solutions not min and max values of the 3x + 2y. Please help Thanks.

## #16 Re: Help Me ! » lg problem » 2005-12-25 23:11:55

ganesh wrote:

We know that for any number greater than 25, n!>10^n.

Is this a theorem or ? Thanks very much for your explanation. Happy holidays.

## #17 Re: Help Me ! » lg problem » 2005-12-25 22:36:15

ganesh wrote:

2^10,000 contain 3011 digits.
The factorial of this number would be of the order of 10^10^3011 (actually, much greater!).

Sorry again for asking but can you explain how did you got to this - (2^10 000)! = 10^10^3011 I tried the Stirling formula but nothing like this Thanks again.

## #18 Re: Help Me ! » lg problem » 2005-12-23 01:31:51

"2^10,000 contain 3011 digits" - Can you explain me please how did you find that out Thanks.

## #19 Help Me ! » lg problem » 2005-12-20 20:56:04

krisper
Replies: 11

This is a bit of chalange for me and that is why I am posting it here The problem says: Prove that
lg(lgX) > 2500, where X equals (2^10000)! . The ! sign means factoriel.  Please if you dont know how to solve it atleast write where you got after trying so we could solve this problem together. Thanks.