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**sulley****Member**- Registered: 2012-11-26
- Posts: 4

Hello all,

I was hoping you would be so kind as to help me with a problem that I'm having. I would like to calculate 141^(162^164), but the result is rather large and exceeds the capabilities of any software I have yet encountered.

I've tried using BC (see en.wikipedia.org/wiki/Bc_programming_language), you probably know that it's a very capable arbitrary precision calculator, but it can't manage it, it tells me that the 'exponent is too large in raise'.

Does anyone have any ideas? Do I have any chance of producing an actual result?

It's might be worth telling you that I know C, so can write some code to help, but the issue I have is that the types available in C aren't big enough, so I wouldn't even know where to start.

Thanks,

Rob

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

Hi sulley;

Welcome to the forum!

The front part 16870151161094535473499554809767331400820946875159,,, called the mantissa is the first 50 digits of the number if you need more let me know.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**sulley****Member**- Registered: 2012-11-26
- Posts: 4

Wow, impressive! Thank you very much. Can you share how you did it please?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

There are a couple of ways of tackling tower problems as these are called.

There are math programs that can handle this large a number directly.

Both Maxima using the BFloat class and Derive 6.1 are capable of getting this answer and I used Derive 6.1 to check.

There is a math way but it is a bit complicated if you have never seen it before.

Are you interested?

If you think that number is massive then think again:

that's massive^massive. 10 or more years ago I went after the front digit of that number. I was a mere lad of only 82 years of age so I figured I would bring it to its knees before I died...

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

You never told which digit was actually the first one...

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

For the tower problem? I do not know which one it is.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**sulley****Member**- Registered: 2012-11-26
- Posts: 4

bobbym wrote:

There is a math way but it is a bit complicated if you have never seen it before.

Are you interested?

Yes it would interest me, but realistically I will have very little use for the knowledge, so I won't take any more of your time.

bobbym wrote:

that's massive^massive. 10 or more years ago I went after the front digit of that number. I was a mere lad of only 82 years of age so I figured I would bring it to its knees before I died...

Doesn't that make you 92? Or am I missing something.

Thanks again for you help,

Rob

*Last edited by sulley (2012-11-28 02:37:10)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,391

Or am I missing something.

Yes! bobbym regularly lies about his age.

As far as I know only three people know for sure, and two of those are doubtful.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Well, he has given us some info about his age. He was born on the 3rd of July and on Sunday...

*Last edited by anonimnystefy (2012-11-28 04:10:48)*

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,391

Is that reliable information though?

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

I think so.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

Hi sulley and all;

I was not born in July. Records of my birth have been destroyed. Hospital burned down and so did the rectory.

Yes it would interest me, but realistically I will have very little use for the knowledge, so I won't take any more of your time.

Then it will die with me. It does appear I am 92 but no one believes that so sometimes I am younger.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**sulley****Member**- Registered: 2012-11-26
- Posts: 4

I'm back again! Is that 92 in hex?

I have an even bigger challenge for you, would you be so kind as to calculate 141!^(162!^164!) for me please? to as large an accuracy as you dare! (ridiculous, I know!)

Thanks,

Rob

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

Hi;

That is 92 in decimal. Your question boils down to this:

which means it is larger than my current limit of 9^(9^(9^5)). As a matter of fact

it is larger than 9^(9^(9^(9))) which I have struggled with for more than 10 years.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Hi bobbym

Is there a way to get the front digits of a number other than the one you showed me in the big Oh thread?

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

Hi;

That is the only one that I use consistently. Of course there are other methods but they are all experimental.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Do you know any?

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

This suggests another method but in this case it fails. As a matter of fact it fails for

9^(9^(9^9))) too! His number is just too large.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

I meant-do you have another method for evaluating those kinds of numbers in general, for example, less than 9^9^9^5?

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

Hi;

For 9^9^9^5 ( I leave out the bracketing from now on ) is a major undertaking and I used a

recurrence.

And you missed my point, post #18 does suggest another way.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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