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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,812

(1) 1.444667861...............

(a) This is the maximum value of xth root of x for any value of x.

(b) This number is also the maximum value such that

x^x^x^x^x...............ad infinitum is finite.

The number is eth root of e, the Natural logarithm base.

e is approximately 2.7182818284.

2. 1787109376

It is known that the square or any higher power of a number ending in 6 is always 6.

This holds good for 76, 376, 9376, etc.

The ten digit number given also has this property.

This can be continued indefinitely, and you get more and more digits.

PS:- Try searching with Google for these numbers. You'd be surprised by the result

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

I remember that 2nd number! There was a topic about it a while ago.

Then further investigation revealed that there was a similar number to that, but it ended in 5.

8212890625 was the last bit. And there was the very peculiar property that if you added together the last n digits of both numbers, you always got 1000...0001, with (n-1) zeroes.

Why did the vector cross the road?

It wanted to be normal.

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**namealreadychosen****Member**- Registered: 2011-07-23
- Posts: 16

I agree with (a) but I thought that x^x^x... would be infinite if x>1 because x^x^x>x^x>x, and time ^x is added the amount by which it increases also increases.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,154

Hi;

Which is approximately

1.111782011041844...

Welcome to the forum!

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,018

hi

actualy the value of x^x^x^x^...:

Let s=x^x^x^...

s=x^(x^x^x^...)

s=x^s

So once we can sole this equation in terms of x then we get the value of s.

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