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#43426 Re: Jokes » Oh, my achy breaky brain.... » 2005-11-16 00:36:30
This is pure genius! Really! It proves english is complete and utter idiocy and we need to switch to a form of mathematical communication.
110001100110100111001 111010011010001 100010110011101
(Translated in English :You are right!)
#43427 Maths Is Fun - Suggestions and Comments » Scientific Calculator » 2005-11-15 01:11:51
- Jai Ganesh
- Replies: 24
I tried using the scientific calculator on this website.
I wanted the value of 2^2520. I got the result 'infinity'.
I found that the calculator gives a result for 2^1022, but for 2^1023, you get infinity. Can the 'infinity' be substituted by 'error' or 'out of range'?
#43428 Re: Help Me ! » A question on essays cheating » 2005-11-15 01:07:02
May God pardon you!
#43429 Re: Help Me ! » I need your help » 2005-11-15 00:34:56
I'm not good at proofs, but this is my attempt:
Take 109376 as an example.
We need to prove that (y109376)² ends in y109376 for only one value of y.
y109376 can be written as y*10^6+1*10^5+0*10^4+9*10^3+3*10^2+7*10^1+6*10^0.
We are only interested in the 10^6 coefficient of the square of this. That will be equal to 2(y*(6*10^0))+a constant made up of the expansion of the other powers of 10, but that does not involve y.
So, the 10^6 coefficient is 12y+c, which should only be equal to y for one value of y.
We are only interested in the last digit of 12y+c, so it can be called 2y+c without affecting our purposes. However big c is, it can also have all of its digits except the last removed.
We want to solve y=2y+c.
Take away y+c: y=-c
Give back one of c's 10's to make it more sensible: y=10-c
c is a constant, so there is only one value for y.As shown in the expansion of y109376, the important coefficient will always take the form 2(y*6), because 6 is always the last digit. This means that this proof can be carried forward for any point on the chain of the magic number.[/proof]
There's probably a flaw in there somewhere, could someone check it please?
Thanks a million, Mathsyperson! I have checked the proof thoroughly and now, I can declare the proof is precise, elegant, and flawless.
This proof has been very helpful to me.
I thank you, again, MathsIsFun, for the Full Precision Calculator.
#43430 Puzzles and Games » Two easy questions..... » 2005-11-14 18:48:36
- Jai Ganesh
- Replies: 1
1) !, #, %, &, _
2) (a-x)(b-x)(c-x).......(z-x)=?
#43431 Re: Guestbook » Help me please to find a percentage » 2005-11-14 18:35:34
From what I can understand, out of 450 votes, a student got 350 votes.
The percentage of votes he or she got is 350/450 x 100%
i.e. 77.77777777777777....% rounded off to
77.78%.
#43432 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-11-14 18:27:59
Problem # k + 45
Find the sum of all the numbers that can be formed with the digits 1, 2, 3, 4, and 5 taken 4 at a time (without repetition).
#43433 Re: This is Cool » Integral relationship » 2005-11-14 18:20:48
I'm not a mathematician either.
I work taxing manufacturers (collecting taxes).
#43434 Re: Introductions » Hello all! » 2005-11-14 18:14:32
Hi Angelpumpingas,
Trig is quite easy at the preleminary level....
Sinθ = Opposite side/Hypotenuse
Cosθ = Adjacent side/Hypotenuse
Tanθ = Opposite side/Adjacent side = Sinθ /Cosθ
Sinθ = Cos(90 - θ ) in degrees
and Cosθ = Sin(90-θ ) in degrees,
Sin²θ + Cos²θ = 1
#43435 Re: This is Cool » Integral relationship » 2005-11-14 00:06:35
Hi Mr.Keiichi Suzuki,
Welcome to the forum.
Your website is cool, your proof is elgeant and precise.
I do have great respect for Japanese Mathematicians, I had read about Shimura and Taniyama before. 
#43436 Re: Help Me ! » A real tough trig question help will be appreciated » 2005-11-13 22:51:26
Sinx*Sin2x + Cosx*Cos2x
= Sinx[Sin(x+x)] + Cosx[(Cos(x+x)]
= Sinx[2SinxCosx]+ Cosx[Cos²x - Sin²x]
Rearranging the terms,
= Cosx[2Sin²x + Cos²x - Sin²x]
= Cosx[Sin²x + Cos²x]
=Cosx (since Sin²x + Cos²x=1).
#43437 Re: Help Me ! » Compound Interest » 2005-11-13 20:22:58
Present Value = 5,000 + 5,000/(1.06^3) + 5,000/(1.06^4)
= 13,158.46
This is done by using the formula
A = P*(1 + r/100)^n
Since the interest is compounded annually, r =6 and n=0, 3 and 4.
Therefore,
P = A/(1 + r/100)^n
Is that clear?
#43438 Re: Introductions » Hi!!!!!!!! » 2005-11-13 17:40:11
And so would ours be, if it weren't for good moderators
Yay! And where would we be if we didn't have a fantastic administrator...?
Thanks, MathsIsFun!
Thanks, justlookingforthemoment, you are right! A good team requires a good leader!
#43439 Re: Jokes » Thought this was cute... » 2005-11-13 17:27:37
Good ones! Can you think modifications for some of these...
(1) Quo Vadis
(2) Quod Erat Faciendum
(3) Obiter Dictum
(4) Audi Altarem Partem
(5) Persona non grata
#43440 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-11-13 17:22:44
Problem # k + 44
It is believed that James Bernoulli calculated the sum of the tenth powers of the first 1,000 natural numbers in 7½ minutes (without a calculator or computer...they were non-existent during his period!). Given the same time limit, can you tell what the last diigit of the sum of the tenth powers of the first 1,000 natural numbers would be?
#43441 Re: Help Me ! » How do i work out an average? » 2005-11-13 17:20:08
Hi Freelancer_man,
Welcome to the forum.
This is how the average or mean is computed.
Find the sum of all the numbers.
You get 260.
Divide it buy the number of quantities (in this case, numbers).
i.e. 9.
The average is 260/9 or 28.888888888888888.........
rounded off to 28.89.
In general, average or mean of a,b,c,d,e,f is
(a+b+c+d+e+f)/6. There are six quantities a,b,c,d,e,f. Therefore, the sum is divided by 6. Clear?
#43442 Re: Introductions » heya! » 2005-11-10 16:34:46
Hi Make_Me_Yours94
Welcome to the forum!
You can find a lot of friends here, but remember,
this is a forum, chatting isn't possible here, it may take time for replies to your posts!
#43443 Re: Help Me ! » need explaination » 2005-11-10 16:27:49
1) xlog4 + log7 = xlog9+ log5
xlog4 - xlog9 = log5 - log7
xlog9 - xlog4 = log7 - log5
x(log9 - log4) = log7 - log5
x = (log7 - log5)/(log9 - log4)
The second question isn't clear 
#43444 Re: Puzzles and Games » Word puzzle » 2005-11-10 16:17:57
Rora is right!
I had that in my mind, but I don't know if there are any more
#43445 Re: Help Me ! » john wallis's book » 2005-11-09 23:43:36
#43446 Re: This is Cool » Infinity comparisons » 2005-11-09 23:31:52
#43447 Puzzles and Games » Word puzzle » 2005-11-09 16:10:11
- Jai Ganesh
- Replies: 14
Can you think of a pair of words which are both anagrams and homophones?
(Examples of Homophones :- Ice and eyes, Seam and seem, Read and Reed etc.
Examples of Anagrams:- Talent and Latent, Mate and Meat, Slate and Stale etc.)
For an example
#43448 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-11-08 22:44:43
Here........
Problem # k + 43
If x² + 1/x² = 23, what is the value of x^9 + 1/x^9 ?
#43449 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-11-06 16:34:19
A week Diwali break!
(Diwali is the most important festivel for Hindus, the festival of color and lights, sweets and fireworks! The festival celebrated for victory of good over evil!)
#43450 Re: Jokes » Customer Service » 2005-10-29 19:00:43
That was interesting, HiImDan.