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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,470

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.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

ganesh wrote:

... but I wanted that to be done without a computer ...

However, if you wanted I could work on some code that does repeated multiplication, but only retains the last "n" digits.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,470

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???????

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

Is it enough to prove that 743740081787109376² ends in ...743740081787109376 ?

If so then I could get a program to run that in a loop of 10 ie:

**0**743740081787109376²**1**743740081787109376²

etc ...

And the results of that (assuming there are any) can be fed in for higher digits.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,470

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.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

Alright, ganesh, this is my plan:

Step A: get a full accuracy multiplier to work using javascript, and make this available on the website for your use.

Step B: make a version of this that hunts for repeated final digits.

I am looking forward to trying this out, but cannot start for a few days, OK?

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,470

Take your time. I am in no hurry.

After you have done it, I'd have to work an hour a day

to add just one or two digits, since that would require checking for all numbers from 0 to 9.

Thank you once again, Administrator.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

OK, ganesh, I have a first version of my "full precision" program for you to test out here

Put your number in the first box, press "x^2" and the square **should** appear in the second box.

Test it out, and tell me what happens!

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

It seems fine to me. I thought I found an error, but I couldn't reproduce it so I think that was just me.

I continued ganesh's thing for a while so:

893380022607743740081787109376^p=(10^30)n+893380022607743740081787109376

Why did the vector cross the road?

It wanted to be normal.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,470

It works very well! Thanks a million!

I wonder how mathsy reached this number so quickly!893380022607743740081787109376.

I knew 43740081787109376,

Just tell me whether you exhausted all the other numbers,

that is, we know 1787109376^2 ends in 1787109376.

But, 2787109376^2 should not end in 2787109376.

mathsy, would you clarify this point?

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

Still experimental software, guys! I know already that I have a problem with negative numbers (not that it directly affects this application)

But this is interesting ... it could be another magical number. It does seem to go on ...

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

No, I just tried numbers from 0 up to 9 until I found one that works, then moved on to the next digit.

If you want to show that that's the only number that does that, you'll have to clean up after me. Sorry!

Why did the vector cross the road?

It wanted to be normal.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

OK, I've gone back and tested everything exhaustively so yes, 893380022607743740081787109376 IS the only number.

Also, I found why I got the error before: if you put in a space, the program reads it as '0'

e.g: (1 2)^2=10404.

I just had a few spaces at the end of my input number and the output had 0000 on the end and I couldn't understand why.

Why did the vector cross the road?

It wanted to be normal.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

(OK, I need to clean up the number entered in to the program - probably just remove anything not 0-9 or ".")

Now with this ...893380022607743740081787109376 number - do you mean it doesn't branch out anywhere, that so far there has always been one and only one digit that it can grow by? If so, that is long odds or something systematic.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

ganesh wrote:

Squares should be sufficient. I'll do the remaining part, by checking for higher powers.

You don't need to! Cubing is just squaring and then multiplying again. Therefore, if you square a number with this property, you get the same ending digits and so in that respect the square is the same as the original, so cubing will always still work. This argument can be carried forward to raising to higher powers as well.

I got thinking a bit more and realised that any number with the last n digits in the form of (n-1) 0's followed by a 1 or another 0 raised to any power will have the same last n digits.

Even better, I realised that 5 raised to any power would end in 5, so I tried to make another chain, but ending in 5.

Not only is there one, it's much easier to find! 5²=**2**5, so 2 is the next number in the sequence. 25²=**6**25, so 6 is the next number. 625² is 39**0**625, so 0 is the next number. I've gone up to 106619977392256259918212890625 successfully using this method, but the sky's the limit!.

MathsIsFun wrote:

Now with this ...893380022607743740081787109376 number - do you mean it doesn't branch out anywhere, that so far there has always been one and only one digit that it can grow by?

Yes. Except for with the very last digit, which I've already mentioned.

*Last edited by mathsyperson (2005-07-09 23:14:32)*

Why did the vector cross the road?

It wanted to be normal.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
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106619977392256259918212890625 has 5 double-digits

893380022607743740081787109376 has 5 double digits

Every number is paired (around 9), except the last digit

Why?

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Isn't maths amazing?

Why did the vector cross the road?

It wanted to be normal.

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**Roraborealis****Member**- Registered: 2005-03-17
- Posts: 1,594

Yes, though all it is really is the imagination.....that's what I find really interesting!

School is practice for the future. Practice makes perfect. But - nobody's perfect, so why practice?

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

Isn't it interesting?

...106619977392256259918212890625

+ ...893380022607743740081787109376

------------------------------------------------

...(1)000000000000000000000000000001

where the (1) is carried

There should be an underlying explanation. Perhaps geometric, or a series ....

What are your ideas on it, ganesh?

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,470

Wow! Great job by you two!

I got to really thank both of you, Admin for putting the calculator on the site,

and Mathsy for making my job pretty easy!

I had 17 digits with me already, now you have given me 30!

I know, Mathsy, you have done a lot of homework....

Testing for all numbers 0-9 for twenty digits is a demanding job....

it requires squaring to be done 200 times! ( after 1787109376)

Regarding my ideas, my explanation is (not too sure),

2 and 5 are numbers that perfectly compliment each other in multiplication,

and I go back to my earlier post....

The last n digits of powers of 2 start repeating at every

2^(4 x 5^n-1). This is a hypothesis, it isn't proven yet.

It works till 2^2500, beyond that, not too sure! 2^2500 ends in 09376.

This is indeed, a great work by the Mathsisfun team!

*Last edited by ganesh (2005-07-10 16:42:16)*

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

Good fun, too.

BTW I have made some minor changes to the program (hopefully fixed the "extra spaces" problem that Mathsy reported) and integrated it with the website style.

So far I have got it to add, subtract and multiply (not divide yet), so if you need it to do other precision work like 2^2500 or something, tell me and I will modify it to suit.

Any ideas on layout would be appreciated, too.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

It would make it a lot easier if someone could prove somehow that there will always be only one number in the sequence.

We've gone up to 30 like that now, so I think it's safe to assume that that's true. If that could be proved though, the of making the number would skyrocket! You just do my shortcut thingy for the 5 one, then take away from 9 to find the 6 one and you wouldn't have to make sure that all the other numbers don't work.

P.S. You only need to prove that for the 6 one. Whatever number you put in front of the 5 one, it always gives you the same digit, which is why it's so easy.

Why did the vector cross the road?

It wanted to be normal.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,470

The layout is fine. Would it be possible to add 2^n? Thats the most important of powers.

And also n! if possible.

And Mathsy, its difficult to prove or disprove that there would always be a number in the sequence. But I strongly feel, the sequence would go on and on.

*Last edited by ganesh (2005-07-10 20:28:46)*

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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?

Why did the vector cross the road?

It wanted to be normal.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

ganesh wrote:

The layout is fine. Would it be possible to add 2^n? Thats the most important of powers.

And also n! if possible.

I will work on it.

Hmmm ... the "brute force" algorithm for 2^n is just to multiply by 2, "n" times, but this could take a long time for high values of n. But I am thinking it may be better to break "n" into factors ...

So for example 2^100 = (2^10)^10 (so 100 multiplies are reduced to 21 multiplies)

Or maybe better 100's prime factors are 2x2x5x5, so 2^100 = (((2^2)^2)^5)^5 (only 17 steps)

In comparison, n! SHOULD be a piece of cake

We will see!

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