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#41876 Re: This is Cool » 0.9999....(recurring) = 1? » 2005-08-09 16:20:35

I thought of it in a different way. I used the summation of a Geometric Progression and this is the result I got:-
0.9999999999999999999999999999.....................
= 0.9(1 + 1/10 + 1/100 + 1/1000 + 1/10,000 + 1/100,000 + 1/1,000,000.......)
The ∑ of the series within the brackets would be 1/(1-1/10) = 1/(9/10)
= 10/9
10/9 x (0.9) = 9/9 = 1
I know this is an approximation for an infinite series.
But 0.999999999999999999999999999999999.................... is a devilish number. You never encounter this number in any mathematical problem. I have never come across recurring 9s after the decimal in any division. It can be proved that there can be no recurring 9s in any division.
Let me prove this by a counter-solution.
Let there be a recurring decimal 0.2999999999999999999.......
When we try to convert this into a fraction, this is what happens.
x = 0.2999999999999999999............
10x = 2.999999999999999999999......
100x = 29.9999999999999999999999......
90x = 27
x = 27/90 = 3/10, which is 0.3 !
Therefore, I think 0.9999999999999..........exists only theoretically. Only in the imagination.

#41877 Re: This is Cool » Remembering sin and cos values » 2005-08-09 00:08:29

kylekatarn wrote:

I think mnemonics is a disrespect to math.

You are right, but sometimes, it is important to remember things. We all know that the square root of 2 can be calculated when required, but remembering 1.4142 helps when we are required to solve problems quickly. The mnemonic 'All Silver Ta Cups' was for the benefit of students who may need to know whether Cosθ or Sinθ or Tanθ  is positve or negative for a certain value of θ while solving problems, when time is crucial.  smile

#41878 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-08-08 22:58:38

I get a different answer.
There are 9 numbers ending with zero up to 99. 10,20,30,40,50,60,70,80,90.
100 has two zeros. Then, in each set of ten consecutive numbers (1-10, 11-20, 21-30 etc.),
one ends in 2 and one ends in 5, which gives us another 0. There are ten tens in a hundred.
Therefore, I thought 100! would end with 21 consecuitive zeros.
But when I calculated 100! using the full precision calculator on this website, I got
93,326,215,443,944,152,681,699,238,856,
266,700,490,715,968,264,381,621,468,592,
963,895,217,599,993,229,915,608,941,463,
976,156,518,286,253,697,920,827,223,758,
251,185,210,916,864,000,000,000,000,000,
000,000,000
That is 24 zeros! I wonder where those three extra zeros came from roll
Okay, one additional zero 40, 50 since 40x50 = 2000, the other by using 4 (instead of 2)for 25 and 8 for 75?

Did I miss something, Mathsy?

#41879 Re: Dark Discussions at Cafe Infinity » Wink Murder 2 » 2005-08-08 22:51:12

Robert Recorde, who gave the world the '=' symbol, would be fuming!
What more do I do?
Multiply both sides by n (n ≠ 0,1), we get n=n !
n IS EQUAL TO n. How do I prove that?
Taking Log on both sides wouldn't help, because we would than get 0=0, which gives us absurd results.
When both the sides are raised to the power n, we still get 1=1 !
When both sides are divided by n, we bget 1/n = 1/n which is the same as n=n !
How about some practical methods? Draw a straight line of 1 m. Draw another of 1 m. Both of them would be exactly equal. Or, 1 hour is the same everywhere on the planet!
I give up cool

#41880 Re: Dark Discussions at Cafe Infinity » Wink Murder 2 » 2005-08-08 21:39:02

MathsIsFun wrote:

I have just had a puzzle sent to me ... it seems very easy.

... a simple note that says "Prove that 1=1"

Wellll... doesn't 1=1? but how do we PROVE it?

You have shaken the foundations of Mathematics.
If someone asxked me to do it,
I'd simply say take cube roots on both sides.
If the cube roots are equal, the numbers are equal big_smile

#41881 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-08-08 21:03:36

mathsyperson wrote:

About the rectangles in your previous problem, think about it a bit more.

Got it, Mathsy! By cutting 4 rectangles from the four corners, a cuboidal box cannot be made as the heights of the two adjacent sides would differ! smile

#41882 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-08-08 20:31:41

Problme #n+5
How many zeros would 1,000! end in?

#41883 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-08-08 20:24:49

There are n vertices (or points) in a polygon of n sides.
The n points can be connected to any other point in nC2 ways.
That is we get nC2 straight lines. But, of these n straight lines form the sides of the polygon.
Therefore, the number of diagonals is equal to
nC2 - n
= n!/(n-2)!(2!) - n
= n(n-1)/2 - n
= [n(n-1)-2n]/2
= [n² -n-2n]/2 = (n²-3n)/2 = n(n-3)/2

The sum of intenral angles of a polygon of n sides is 180(n-2) degrees as rightly replied by Mathsy!

#41884 Re: This is Cool » Nice graphs » 2005-08-08 19:32:09

NIH wrote:

However, there is a very simple proof that no primes of the form 4n-1 can be written as the sum of two squares.

4n-1 = a² + b²
4n    = a² + b² + 1
  n    = (a² + b² + 1)/4
Since 4n-1 is an odd number, one of a and b is odd and the other is even.
When an even number is squared, the resultant is always a multiple of 4.
An odd number when squared is always of the form 2n+1.
Any number of the form 2n+2 would never be divisible by 4 when n is even! When n is odd, it is of the form(2n+1). When this is squared, we get
4n² +4n+1. i.e. 4(n² +n)+1. When 1 is added to this number, we get
4(n² +n)+2. This number is NOT divisible by 4.
Therefore,
a² + b² ≠ 4n-1
q.e.d smile

#41885 Re: Help Me ! » coordinates of a point on an ellipse » 2005-08-08 17:22:11

The equation of an ellipse is given by
x²/a² + y²/b² = 1
The height and the width of the rectangle would be 2b and 2a.
Now, from this information, get the value of ba and a.
Substitute in the equation for the ellipse.
You get the equation of the ellipse.
Thereafter, give values for x like -2, -1, 0, 1, 2 etc. and find the corresponding values of y.

#41886 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-08-08 16:35:07

Problem # n+4
(1) How many diagonals would a polygon of 'n' sides contain?
(2) What would be the sum of the internal angles of a polygon of 'n' sides?

#41887 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-08-08 16:33:22

Absolutely right, Mathsy!
What happens if we're allowed to cut rectangles?
We get two variables, x and y.
Volume = (30-2x)(14-2y)(2x)
How do we proceed? roll Any clues?

#41888 Re: Puzzles and Games » Who owns the fish? » 2005-08-08 16:28:39

Well done, tt! You got it right! smile

#41889 Re: Help Me ! » Square root of 0 » 2005-08-08 00:10:18

Square root of a number A is a number B which multiplied by itself gives the number A.
eg
square root of 25 = ±5 (that is +5 or -5, because both +5 x +5 = 25 and -5 x -5 = 25)
square root of 2401 = ±49
square root of 65536 = ±256
square root of 0 = 0
square root of 1 = ±1
square root of 2 = 1.4142135623..........................
square root of 10 = 3.1622776601........................
Those decimals appear because 2, 10 etc are NOT perfect squares.

#41890 Re: Help Me ! » Can Someone Please Tell me a Good Way to Remember Square Route? =-S » 2005-08-07 17:32:08

Here are the squares and square roots of all squares up to 31.
1² = 1
2² = 4
3² = 9
4² = 16
5² = 25
6² = 36
7² = 49
8² = 64
9² = 81
10² = 100
11² = 121
12² = 144
13² = 169
14² = 256
15² = 225
16² = 256
17² = 289
18² = 324
19² = 361
20² = 400
21² = 441
22² = 484
23² = 529
24² = 576
25² = 625
26² = 676
27² = 729
28² = 784
29² = 841
30² = 900
31² = 961

Remember,
(1) The last digit of a perfect square is always 0, 1, 5, or 6.
(2) Finding square roots is easier by prime factorisation.
eg. 784 = 2 x 2 x 2 x 7 x 7 x 7
Therefore, the square root of 784 is 2 x 2 x 7, i.e. 28

#41891 Re: Help Me ! » Computer Nerd needing Math Nerd Help - SciNotation? » 2005-08-07 17:14:41

wcy, i viewed the base converter. It works well.
I tried converting 759 to base 20 and got the right answer smile

#41892 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-08-07 17:09:29

Problem #n+3
A rectangular metal sheet has dimensions 30m, 14m.
Four equal squares are cut off from the four corners of the rectangle.
Then, the sheet is folded to form a cuboidal box with top open.
What is the maximum volume of the cuboid you can get?

#41893 Re: Puzzles and Games » Who owns the fish? » 2005-08-07 16:56:20

If you all are interested in logical puzzles, here's one:-

A Captain has three smart soldiers under him. The three soldiers are informed that their intelligence is being tested and that the captain has three blue and two red caps. They are made to stand in ascending order of their heights. They are blindfolded and the captain puts a cap, chosen at random, on the heads of the three soldiers. The blindfold is then removed. The captain asks the third soldier whats the color of the cap on his head. The third soldier sees the color of the cap on the heads of the first two soldiers, but he says he is unable to tell the color of the cap on his head.
The Captain asks the same question to the second solider. The second solider takes into consideration the third soldier's reply and the color of the cap on the first soldier, which he is able to see. But, he too is unable to tell the color of the cap on his head.
The Captain then asks the same question to the first soldier. The first soldier has only the replies of the third and second soldiers to draw any conclusion. But he tells the color of the cap on his head. He gives the right answer.

What's the color of the cap on the first soldier's head and how did he answer correctly?

#41894 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-08-07 16:44:16

You are right, Mathsy, but the number of bags required is 10!
They would contain $1, $2, $4, $8, $16, $32, $64, $128, $256 and $489. For any amount less than or equal to $511, the first 9 bags can be used. For amount greater than $511, one bag of $489 and remaining from one or more of the first nine bags can be used.

#41895 Re: Puzzles and Games » Paradox » 2005-08-07 16:39:12

wcy wrote:

(i²/1)= (1/i²)
√(i²)/√1= (√1)/(√i²)

Normally, in such paradoxes, this is where the mistake lies.
Since a² = b²,
it cannot be inferred that a=b.
Taking square root on both sides of an equation should always have the
'±' sign, after the square root is worked out.

#41896 Re: Help Me ! » sequence » 2005-08-05 00:06:53

Good post, and Welcome, Robin!
Subsequent differences, yes, sometimes they do help in find the next number in sequences.
But, what you have pointed out is, I feel, not acceptable.
You have found the difference, and then added that to the previous term.
In mathematical terms, you have found the difference between 2 terms, a and b, then added the difference to a, which would have to be b!

#41897 Re: This is Cool » Digit-perfect Numbers!! » 2005-08-04 23:58:09

That is, it is to be proved that
(abcd..........)^n = a^n+b^n+c^n+d^n..................
Here, both a,b,c,d...etc and n go on and on.
It is easy to say there would always exist a n digit number whose sum of digits raised to the power n is equal to the number. To prove that may be quite difficult. As n becomes higher, the combinations available increase, thereby increasing the possibility of such a number existing. smile

#41898 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-08-04 19:04:54

Problem #n+2

Divide $1,000 (in whole $ increments) into a number of bags so that I can ask for any amount between $1 and $1,000 and you can give me the proper amount by giving me a certain number of these bags without opening them. What is the minimum number of bags you will require?

#41899 Re: Guestbook » maths home » 2005-08-04 18:32:39

Good, you liked the website, Tara Witney!
You are wlecome to join the forum!

#41900 Re: Help Me ! » three today three more tomorrow.... » 2005-08-04 18:24:48

This is an Arithmetic Progression (AP).
An Arithmetic Progression is something like
1,2,3,4,5,6,7,8 or
2,4,6,8,10,12, or
9,12,15,18,21,24 or
16, 13, 10, 7...........
where the difference between two successive terms is the same.
This is called 'd' or common difference.
If 'a' is the first term of an AP and 'd' is the common difference,
the nth term would be a+(n-1)d
In your case, d is 3 minutes. n is 23, and a is also 3.
Therefore, on the 23rd day, you have trusted me for
3 + (23-1)3 = 3 + (22)3 = 3 + 66 = 69 minutes.


If you require the sum of all the terms, the formula is
n/2 [2a+(n-1)d]
In the example given by you, it would be
23/2 [6 + 22(3)] = 23/2[6+66] = 23/2[72]=23*36 = 828 minutes
Is that clear?

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