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**zetafunc.****Guest**

We sent 8 e-mails to each other yesterday. I'll find out where she lives... maybe we can walk to school together.

Okay, thanks.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,383

I am coming up with 70 such numbers for 4 digits alone.

**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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**zetafunc.****Guest**

Sorry, should have specified, the number has to be 4-digits long, so 70 is the right answer. How did you do it with GFs?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,383

The answer is the coefficient of x^0, which is 70.

*Last edited by bobbym (2013-03-08 20:49:38)*

**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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**zetafunc.****Guest**

I can see that the second bracket represents the 1s, but what is happening in the others?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,383

The first bracket represents the first digit. It can be from 1 to 4.

The second bracket represents the second digit. It can be from 0 to 4.

Do you follow?

**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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**zetafunc.****Guest**

Okay, that makes sense... but I don't understand why you have negative exponents for the last two. Why isn't it just the second bracket cubed since each of the other numbers can be from 0 to 4?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,383

When you are solving a problem like this you are really solving a linear diophantine equation. Mathematics consists of taking a problem that is in one form and changing into another that has the same number of solutions. One that we can solve. We will get into the "teakettle principle" in just a minute.

The problem really is the number of solutions to

a + b = c + d when 0<a<5, 0 ≤ b,c,d < 5. And a,b,c,d are each a single digit of your 4 digit number.

Follow so far?

*Last edited by bobbym (2013-03-08 21:04:32)*

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Oh, I see... so it becomes

a + b + c + d = 0

where a and b have the same ranges as before, but now c and d can be between 0 and -4.

which is why we need the co-efficient of x^0, and why we have negative powers...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,383

That is one way of looking at it. The last two brackets just represent c and d taking on values of -4 to 0.

When we expand that we get:

We see the constant coefficient ( x^0 ) is 70. What do you

think the 34 x^4 tells us?

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

The number of solutions to a + b + c + d = 4?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,383

Yes! It means that there are 34 ways to write that 4 digit number abcd such that a + b is 4 greater than c + d.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Ah, I see. Thanks for the solution! Now on to the second part of the question.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,383

What second part?

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

It asks for how many 4-digit numbers can be made using the numbers 0 to k. I used the same approach, it did not work...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,383

Hi;

That is more difficult but a tentative experimental answer can be found quickly.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Yes, that's the one -- although, this question is from a public exam paper where only a pen is allowed...

The GF is great for solving the first part, but the second part looks like a systematic approach is needed.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,383

Yes, but the important part is that we have the answer... Do they agree that is correct?

Then I can work on the so called proof.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Yes. It is okay, I found the 'full solution' -- although, I am trying to see if I can do it with a GF.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,383

Can you supply the link?

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

To the question paper, or the solution?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,383

To both if you have them. In case I do not get it.

I would try induction before using the gf approach here.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

w w w . m a t h s h e l p e r . c o . u k / o x b . h t m

Under the heading **STEP Past Papers**, look at Q1 of 2007 Paper 1 for the question, or click on '2007 Solutions' to see the answers.

**zetafunc.****Guest**

How would you use induction?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,383

Sorry, it did not work. Looking at a gf approach.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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