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The three values of the sides of a right angled triangle are also called a Pythagorean Triple.
A Pythagorean triple is a set of three whole numbers , such that one number squared added to another number squared equals the third number squared. Euclid could prove that there are an infinite number of such Pythagorean triples.
Euclid's proof begins with the observation that the difference between successive square numbers is always an odd number.
4 - 1 = 3, 9 - 4 = 5, 16 - 9 = 7, 25 - 16 = 9, 36 - 25 = 11, 49 - 36 = 13 etc.
Mathematicians are notorious for being sticklers when it comes to requiring absolute proof before accepting any statement. Their reputation is clearly expressed in a story told by Ian Stewart in 'Concepts of Modern Mathematics':-
An astronomer, a physicist, and a mathematician (it is said) were holidaying in Scotland. Glancing from a train window, they observed a black sheep in the middle of a field.
'How interesting' observed the astronomer, 'all Scottish sheep are black!' To which the
physicist responded, 'No, no! Some Scottish sheep are black!' The mathematician gazed heavenward in supplication, and then intoned, 'In Scotland there exists at least one field, containing at least one sheep, at least one side of which is black.'
Taken from 'Fermat's Last Theorem' by Simon Singh
Mathsy is right! This is a puzzle and crept into 'Help me' by mistake.
Anyway, I shall give the solution.
One dice is marked 0,1,2,3,4,5.
The other is marked 0,1,2,7,8,9.
Don't ask me how you get 6!
Similalry, how do you tell whether sin or cos or tan of an angle is positive or negative?
Just remember the mnemonic 'All Silver Tea Cups'.
In the first quadrant, i.e. 0 to 90 degrees, All the angles are positive.
In the second quadrant, 90-180 degrees, only Sine is positive (remember S of Silver!)
In the third quadrant, 180-270 degrees, only Tan is positive. (remember T for Tea)
In the fourth quadrant, i.e. 270 to 360 degrees, only Cos is positive. (C for Cups)
Nevermind getting a silly combination like 38,
but how would you display, say 22, 23, 24 or 25???
I clarify...
On each of the six sides of the two dices, any six numbers
0, 1, 2, 3, 4, 5, 6, 7, 8 or 9
are marked. It could be any of these ten numbers.
The condition is that it should be possible to display all the dates from 01 to 31.
Further, dates with single digits should be displayed as 01, 02, 03 etc.
What markings would make this possible, thats the question.
There's only one solution to this problem.
Two dices are marked with numbers from zero to nine on each face.
The two dices are meant to display all the dates in a month from 01 to 31.
Remember, single digit dates would have to be shown as 01, 02, 03 etc.
That is, one dice should show zero and the other 1 or 2 or 3 and so on.
The two dices can be interchanged i.e. placed in any order.
What would be the numbers written on the six faces of the two dices???
Clue No.1 :- First, figure out what time the clocks were set
oh yes, we did....Maybe we started the same time, but you finished earlier.
Squares should be sufficient. I'll do the remaining part, by checking for higher powers.
I am too not sure if this can be done much further, because there's a possibility it may end somewhere.
10 raised to the power 6 is 1,000,000.
We say Logarithm of 1,000,000 to the base 10 is 6.
if a^b=c
(read as a raised to the power b equals c)
then, Logarithm of c to the base a is b.
In short, we say log(to the base a) c = b
Common logarithm uses the base 10,
Natural logarithm uses the base 'e'.
Logarithms, discovered by John Napier, makes calculations far easier.
Multiplying two numbers is simply adding their logarithms (to the same base).
For example, logarithm value (to base 10) of a few numbers are given.
log 2 = 0.30103
log 3 = 0.47712
log 7 = 0.8451
log 10 = 1
and so on.
No, Thank you, Administrator...
That was only for 'Im really bored'.
What I meant was that question was intended to be answered without use of computers.
Anyhow, what would be of help to me is.......
I discovered that the higher power of any number ending in
743740081787109376 would end in the same number.
If you can add a few more digits ahead of that???????
You maybe right, I'm not too sure,
but I wanted that to be done without a computer.
And the last digit is 7.
Moreoever, I wanted only the last digit.
There's a pattern which makes it easy to tell.
Anyhow, than you, dear member, for first helping out with the
last 5 digits of 2^500 and 2^2500.
When I moved from step(2) to step(3),
I was taking logarithm of 1/2.
Logarithm of 1/2 is a negative number.
(For example, logarithm(base10) of 1/2 is -0.3010 approximately)
When an inequation is mutliplied by a negative number,
the inequality reverses direction.
Thats it!
eg. 5>2
But when this inequation is multiplied by -2,
-10<-4
Higher powers of numbers ending in 0,1,5, and 6 end in themselves.
Of these numbers, I was most interested in 6 because numbers of the form 2^4n end in 6.
For that matter, the last digits of higher powers rotate in cycles of 4, thus, they can easily be known.
For example, try to figure out what the last digit of 23^611 would be. I shall help you with that if there are no answers.
You are right. But all numbers upto 20 only, not 22.
At some point,
2 raised to the power of
4 x 5^n
would end
1787109376.
Thereafter, every higher power of that number would end
in 1787109376.
This is because any power of a
number having its last 10 digits as
1787109376
ends in 1787109376 !!!
I really don't need to bother what the higher dgitis are.
The last 10 digits would do.
Thanks again, for the trouble taken and the time spent.
I have 4 marked weights with me and they all add up to 40 kg.
With a pair of scales, I can distribute any weighment from 1 to 40 kg(Whole numbers).
What are the four weights????
Thank you, Gentleman!
Yes, I just want to generate the output of
2^500, 2^2500 and 2^12500.
If possible, even 2^62500 and 2^312500
The last number of these would be about 100,000 digits.
Yes, you are right. Infinity is diffuclt to comprehend for others, not for mathematicians.
The calculator on my PC (in the scientific mode) gives the complete output of a function of resultant about 40 digits. I wonder how you did that with a calculator. Tell us how.
I read this interesting information in 'The Book of Nothing' by John D Barrow.
With the discovery of negative numbers, new jokes became possible.
Like that of the individual whose personality was so negative that when he walked into a party, the guests would look around each other and ask each other 'Who left?'
Or the scientist whose return to the country was said to have added to the brain drain.
I am pretty serious, Administrator.
I discovered that 2^20 ends in 76.
Thereafter, any number of the form 2^20n ends in 76.
Similarly, any number of the form 2^100n ends in 376.
Next, I had to know the last few digits of 2^500 and 2^2500.
When I learnt that they are 9376 and 09376, I was excited.
Because, any number of the form 2^500n would then have to end in 9376
and every 2^2500n would have to end in 09376.
But I wanted to be sure about it, thats the reason I sought to know how it was obtained.
Are you sure?
I am happy if they are 89376 and 09376.
Please also tell me how you got those numbers.