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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

Let b be an integer greater than 2, and let

(the sum contains all valid base b numbers up to 100_b). Compute the number of values of b for which the sum of the squares of the base b digits are less than or equal to 512. I understand what N_b's value is, but i don't know about the values. Maybe trying smaller values of b would work.[edited for clarity - bobbym]

Prove that from the set

one can choose 2^k numbers so that none of them can be represented as the arithmetic mean of some pair of distinct chosen numbers.I see you have graph paper.

You must be plotting something

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,395

hi cooljackiec

Compute the number of values of b for which the sum of the squares of the base b digits

Is there a bit missing at the end of this sentence?

In base b the column headings are 1, b, b^2, b^3, ......

so

therefore

but then I don't know what I'm doing with this.

Bob

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

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

the sum of the squares of the base b digits are less than or equal to 512. I understand what N_b's value is, but i don't know about the values. Maybe trying smaller values of b would work.

I see you have graph paper.

You must be plotting something

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,395

If I've understood this problem correctly, there are very few cases, so you could just use trial to get them all.

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: 15,532

But, you need the squares of the digits...

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: 6,395

OK. I don't understand the question then.

Bob

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,786

Compute the number of values of b for which the sum of the squares of the base b digits

Is it my eyes or this browser, is something missing from the end of this line?

**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: 15,532

Hi bobbym

The remainder of the question is in post #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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,786

Bob already asked that question. I am going to edit post #1.

**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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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

We just have to sum the squares of the digits in $N_b$. But finding the digits ain't viable. I don't have any ideas.

I see you have graph paper.

You must be plotting something

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,786

Why is finding the digits not viable?

**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: 15,532

Because you cannot do it in the general case.

Also, I just figured out that Bob posted the expression for the sum. But, that unfortunately tell us nothing about the digits, though.

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: 86,786

What do you do with

add the digits and square? Treat the digts as base 10 number and square? What?

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,395

I have it sorted on paper. Watch this space, while I make a picture.

Bob

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,395

hi

This is my interpretation of the problem.

The picture shows the cases b = 2, b = 3 and b = 4

I have listed all the values from 1 to b^2

The highlighting shows a clear pattern.

With b = 2, we have 2 ones in the first and second column.

With b = 3, we have 3 ones and 3 twos .

With b = 4, we have 4 ones, 4 twos and 4 threes.

Generalising and adding one more for the one in the third column we get this formula for the sum of the squares of the digits:

This checks out with answers 5, 31 and 113 respectively.

Bob

*Last edited by bob bundy (2014-06-15 08:19:56)*

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,532

Hi Bob

I'm not sure what you did, so I think it would be the easiest to explain how you got 5 for b=2. I am getting 2.

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

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,395

hi Stefy,

the sum of the squares of the base b digits

So I added thus

Next one:

Next one:

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,532

But, I think we are supposed to find the sum of the squares of the digits of N_b.

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

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,395

You may be right. The problem is difficult to understand.

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,532

I agree. We should wait for OP to verify which interpretation of the problem is correct.

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

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

Ok, we have N_b. Let's say N_b is in form of

. We wantI see you have graph paper.

You must be plotting something

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,532

cooljackiec wrote:

Ok, we have N_b. Let's say N_b is in form of

. We want

Hm, unfortunately, just as I thought. This will be harder.

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

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

Do you have any ideas for the 2nd question?

I see you have graph paper.

You must be plotting something

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,532

Hi cooljackiec

I found that these numbers satisfy Q2.

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

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,532

anonimnystefy wrote:

Hi cooljackiec

I found that these numbers satisfy Q2.

I have been able to prove that fact, unless I make a mistake somewhere. I will post my proof, if needed.

Also, the formula for Q1. is , I think, so all bases up to 32 have the property.

*Last edited by anonimnystefy (2014-06-16 11:20:47)*

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

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