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**gyanshrestha****Member**- Registered: 2007-11-06
- Posts: 41

what is the last digit of 3^555555?

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Try the first few powers of 3, find the pattern, and prove it. Then take 55555 modulo the size of this pattern.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,590

Hi gyanshrestha;

It is a 7.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,600

And the first one is a 1.

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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: 88,590

Hoooo, that is a good one.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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How could you say that its 1?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,590

You could also!

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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Hmm....

and without Mathmatica?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Very much so.

1) You could use Alpha.

2) You could use a program whose abilities dwarf Mathematica when it comes to numbers...

**In mathematics, you don't understand things. You just get used to them.**

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and with paper and pencil and my brain?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Some problems are not for your brain. Would you calculate √ 2 to 1000 digits with pencil and paper even though theoretically you could.

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,600

I used bobbym's method from another thread.

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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I was wondering if there's a shortcut

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,853

anonimnystefy wrote:

And the first one is a 1.

And the middle two digits of the even-length answer are {9,0}.

"3^555555" contains six fives; and so, using these figures a little differently, we get:

3(5+5+5+5+5+5) = 90...which verifies the correctness of the middle two digits of 3^555555.

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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I was wondering if there's a shortcut

The front digit is a strange beast. Because of carries it sort of depends on all the others!

3(5+5+5+5+5+5) = 90...which verifies the correctness of the middle two digits of 3^555555.

Okay, what is 12345567890987654321! Just the first 50 digits will do!

**In mathematics, you don't understand things. You just get used to them.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,853

As far as I know (after having done just one test like this), my example is unique and doesn't extend to other sums.

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,590

You mean of the type

3^(xxxxxx)?

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,853

bobbym wrote:

You mean of the type

3^(xxxxxx)?

Nearly. It is y^(xxxxxx). x and y are single-digit integers >0, and y may = x.

So the test is this:

For a=y^(xxxxxx) and b=y(x+x+x+x+x+x), the middle digit for Length[a]=odd (or the middle two digits for Length[a]=even) = b.

So far, after not looking any further than my example in post #14, all I've found is just that one solution.

*Last edited by phrontister (2013-06-17 12:13:44)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,590

How many have you looked at?

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,853

No others.

It was only something completely frivolous where the numbers just happened to fall into place, but now I've set it up like this I might see if there are other solutions...if only to exercise my M.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,590

Hi;

Okay, let me know if you find one more.

**In mathematics, you don't understand things. You just get used to them.**

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 215

How did you calculate all this?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,590

The last digit can be done by mods. The first digit is usually just raw computation except in specific cases. The middle digits are like the first digit.

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,600

phrontister wrote:

bobbym wrote:You mean of the type

3^(xxxxxx)?

Nearly. It is y^(xxxxxx). x and y are single-digit integers >0, and y may = x.

So the test is this:

For a=y^(xxxxxx) and b=y(x+x+x+x+x+x), the middle digit for Length[a]=odd (or the middle two digits for Length[a]=even) = b.So far, after not looking any further than my example in post #14, all I've found is just that one solution.

There are no such numbers besides x=5 and y=3.

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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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,853

Hi stefy,

I couldn't find any others either.

*Last edited by phrontister (2013-06-17 12:12:15)*

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