# A Maths Question

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#21

(Original post by

Here's a solution here.

**Fermat**)Here's a solution here.

I see a slight problem:

1+r^2+r^4+r^6 =/= (1+r)(1+r^4).

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#22

(Original post by

I see a slight problem:

1+r^2+r^4+r^6 =/= (1+r)(1+r^4).

**Ralfskini**)I see a slight problem:

1+r^2+r^4+r^6 =/= (1+r)(1+r^4).

Also Fermat I think the problem is that without knowledge or a hint to use the substitution u = r + 1/r the polynomial has no obvious roots so is hard to factorise and solve for r, so Katie thinks there must be an alternative method that doesn't require this change of variable

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#23

(Original post by

Its just a typo - it should be 1 + r² + r³ + r^6 = (1 + r²)(1 + r^4)

Also Fermat I think the problem is that without knowledge or a hint to use the substitution u = r + 1/r the polynomial has no obvious roots so is hard to factorise and solve for r, so Katie thinks there must be an alternative method that doesn't require this change of variable

**It'sPhil...**)Its just a typo - it should be 1 + r² + r³ + r^6 = (1 + r²)(1 + r^4)

Also Fermat I think the problem is that without knowledge or a hint to use the substitution u = r + 1/r the polynomial has no obvious roots so is hard to factorise and solve for r, so Katie thinks there must be an alternative method that doesn't require this change of variable

Using that substitution is about the only way I can see of solving it?

High power polynomials are usually solved by some method which will reduce the power.

Spotting a factor in the polynonmial is about the only other way I can think it might be solved.

Edit: In an analytic way that is.

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(Original post by

Yes, that is a problem. Unfortuanetely I have little idea of what maths level people are at!

Using that substitution is about the only way I can see of solving it?

High power polynomials are usually solved by some method which will reduce the power.

Spotting a factor in the polynonmial is about the only other way I can think it might be solved.

**Fermat**)Yes, that is a problem. Unfortuanetely I have little idea of what maths level people are at!

Using that substitution is about the only way I can see of solving it?

High power polynomials are usually solved by some method which will reduce the power.

Spotting a factor in the polynonmial is about the only other way I can think it might be solved.

The solutions I have seen all contain that substitution. But, as I keep saying, I can't imagine that substituion would be obvious to a GCSE student so that's why I thought, considering its meant to be accessible to that age group, there must be an alternative method. But no-one I know has yet found it. Perhaps it doesn't exist.

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#25

I think I know how this problem could be solved using very basic maths. It does involve using all possible combinations, but there aren't many.

A] We want a sequence of four whole numbers, n, nr, nrr, nrrr {I've not worked out how to enter maths notation on this thing, so apologies], with the highest number below 36.

B]Either one or three of the numbers must be odd.

C] (nrrr+nr)-(nrr+n)=13.

n and r must both be negative, but that doesn't matter for calculation.

There can't be many ratios that produce four whole numbers below 36, so apply the obvious, r=2. This gives n=1 or n=3, and neither fulfils condition C.

The next simplest ratio is r=3/2.

This gives the sequence n, 3n/2, 9n/4, 27n/8, which in turn gives 8, 12, 18, 27. Squares are 64, 144, 324, 729

This fulfils condition C and the squares of these numbers equal 1261.

Both n and r are negative, so the sequence is -8, 12,-18, 27 and r is -3/2.

In fact, the only geometrical progression ratios that can produce four whole numbers below 36 are 2 or 3/2.

I didn't time myself on this, but it took a lot less time than typing this did- I think that while younger students go for a formula or a calculator first, people of my age who were drilled through mental arithmetic, probably produce the answer without much thought: I "knew" almost at once that the sequence above was right: i had to spend much longer showing it.

A] We want a sequence of four whole numbers, n, nr, nrr, nrrr {I've not worked out how to enter maths notation on this thing, so apologies], with the highest number below 36.

B]Either one or three of the numbers must be odd.

C] (nrrr+nr)-(nrr+n)=13.

n and r must both be negative, but that doesn't matter for calculation.

There can't be many ratios that produce four whole numbers below 36, so apply the obvious, r=2. This gives n=1 or n=3, and neither fulfils condition C.

The next simplest ratio is r=3/2.

This gives the sequence n, 3n/2, 9n/4, 27n/8, which in turn gives 8, 12, 18, 27. Squares are 64, 144, 324, 729

This fulfils condition C and the squares of these numbers equal 1261.

Both n and r are negative, so the sequence is -8, 12,-18, 27 and r is -3/2.

In fact, the only geometrical progression ratios that can produce four whole numbers below 36 are 2 or 3/2.

I didn't time myself on this, but it took a lot less time than typing this did- I think that while younger students go for a formula or a calculator first, people of my age who were drilled through mental arithmetic, probably produce the answer without much thought: I "knew" almost at once that the sequence above was right: i had to spend much longer showing it.

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#26

Easy crap,

xsquare + ysquare = 1256

x + y = 13

Solve the simultaneous equation, This is under GCSE stuff

xsquare + ysquare = 1256

x + y = 13

Solve the simultaneous equation, This is under GCSE stuff

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#27

(Original post by

note the fact that the method they used was Laurent poly's.

**Katie Heskins**)note the fact that the method they used was Laurent poly's.

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#28

(Original post by

Easy crap,

xsquare + ysquare = 1256

x + y = 13

Solve the simultaneous equation, This is under GCSE stuff

**Godsize**)Easy crap,

xsquare + ysquare = 1256

x + y = 13

Solve the simultaneous equation, This is under GCSE stuff

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(Original post by

Did you read my solution? It may have used somthing called laurent polynomials, but that would have been unintentional, I do not have a clue what they are!

**mikesgt2**)Did you read my solution? It may have used somthing called laurent polynomials, but that would have been unintentional, I do not have a clue what they are!

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**Godsize**)

Easy crap,

xsquare + ysquare = 1256

x + y = 13

Solve the simultaneous equation, This is under GCSE stuff

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#31

Just because you bloody pieces of **** cant uynderstand this it isnt my problem. Look up your maths books you rancorous coiffured old sows

Solve this and you'll get the answer.

xsquare + ysquare = 1261

x + y = 13

Now I have no time for games.

Solve this and you'll get the answer.

xsquare + ysquare = 1261

x + y = 13

Now I have no time for games.

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#32

(Original post by

Just because you bloody pieces of **** cant uynderstand this it isnt my problem. Look up your maths books you rancorous coiffured old sows

Solve this and you'll get the answer.

xsquare + ysquare = 1261

x + y = 13

Now I have no time for games.

**Godsize**)Just because you bloody pieces of **** cant uynderstand this it isnt my problem. Look up your maths books you rancorous coiffured old sows

Solve this and you'll get the answer.

xsquare + ysquare = 1261

x + y = 13

Now I have no time for games.

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#33

**Godsize**)

Just because you bloody pieces of **** cant uynderstand this it isnt my problem. Look up your maths books you rancorous coiffured old sows

Solve this and you'll get the answer.

xsquare + ysquare = 1261

x + y = 13

Now I have no time for games.

have you actually read the question?

surely you can see that your answer is wrong

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#34

Solve the simultaneous equations first. You'll get two quadtratic equations solve those and you get FOUR FOUR FOUR FOUR FOUR answers...

And stop giving me those bad rep points, I had 8 now I have 4....

And stop giving me those bad rep points, I had 8 now I have 4....

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#35

(Original post by

Solve the simultaneous equations first. You'll get two quadtratic equations solve those and you get FOUR FOUR FOUR FOUR FOUR answers...

And stop giving me those bad rep points, I had 8 now I have 4....

**Godsize**)Solve the simultaneous equations first. You'll get two quadtratic equations solve those and you get FOUR FOUR FOUR FOUR FOUR answers...

And stop giving me those bad rep points, I had 8 now I have 4....

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#36

I've got them, And whoever is giving me the bad reputation:

May the air make you sick and thuder strike your bones, May you fall into the deepest of the abyss and burn in thy purest form. Make food make you weak and breathing cause you death.

Mustaine sez,

Bye

May the air make you sick and thuder strike your bones, May you fall into the deepest of the abyss and burn in thy purest form. Make food make you weak and breathing cause you death.

Mustaine sez,

Bye

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#38

I cant Its too long a method, it also uses the quadratic equation anyhow,

You get a quadratic equation from subtituting equation x, You solve it and then place the values in the other equation (You have two from the quad eq.) You get two more values which are the answers, I made my brother (who's a engineer) solve it he said it was correct. Now give me back those repuatations~!

You get a quadratic equation from subtituting equation x, You solve it and then place the values in the other equation (You have two from the quad eq.) You get two more values which are the answers, I made my brother (who's a engineer) solve it he said it was correct. Now give me back those repuatations~!

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