hard gcse maths question (9-1) (Harder than A Level?) Watch

Flkeavy
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I'm stuck on the last question of my homework paper and even my Chinese friend who's a maths genius can't do it:
22) Here are the first five terms of an arithmetic sequence.
7 13 19 25 31
Prove that the difference between the squares of any two terms of the sequence is always a multiple of 24.
This is what I have written down so far:
6n+1
(v is the closest i can get to a down arrow)
nv1 = 7 nv2 = 13
7^2 = 49
13^2 = 169
169 - 49 = 120
120/24 = 5

xvn+1 = xvn + 6
xvn+2 = xvn+1 + 6
(xvn+2 + 6)^2 - (xvn+1 + 6)^2 = 24m
(xvn+2)^2 + 36 - (xvn+1)^2 = 24m
(xvn+2)^2 - (xvn+1)^2 + 72 = 24m
[(xvn+2)^2 - (xvn+1)^2 + 72]/m = 24

Any help would really be appreciated.
Thank you!
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Sir Cumference
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(Original post by Flkeavy)
I'm stuck on the last question of my homework paper and even my Chinese friend who's a maths genius can't do it:
22) Here are the first five terms of an arithmetic sequence.
7 13 19 25 31
Prove that the difference between the squares of any two terms of the sequence is always a multiple of 24.
This is possibly the hardest GCSE proof question I've seen in an "official" paper. Your working is a bit muddled so I recommend you start again:

SInce the nth term is 6n+1, this means that any terms in this sequence must have this form. So if we're considering any two terms in this sequence then you could represent them as 6n+1 and 6m+1.

So the difference between the squares of these two numbers will be

(6m+1)^2-(6n+1)^2

Can you try to continue from here? If you're still stuck, please post all your working/thoughts.

EDIT : there’s a much quicker way to do this as explained later in the thread.
Last edited by Sir Cumference; 9 months ago
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WarningPlanet02
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i got to 36nsquared + 12n - 36Msquared - 12m
how do u prove this is a multiple of 24
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Sir Cumference
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(Original post by WarningPlanet02)
i got to 36nsquared + 12n - 36Msquared - 12m
how do u prove this is a multiple of 24
You can easily show that this is a multiple of 12. So take out a factor of 12 then to prove it is a factor of 24 you'll need to show that the rest is a multiple of 2 i.e. even.
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WarningPlanet02
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Thanks a lot worked it out but in an exam would never get the multiple of 12 part then prove its even.
Will we get this sort of question in the exam
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Sir Cumference
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(Original post by WarningPlanet02)
Thanks a lot worked it out but in an exam would never get the multiple of 12 part then prove its even.
Will we get this sort of question in the exam
It was in an official specimen paper so something similar to it could be in the real exam!

It's really tough though so if this was in the real thing it would probably be the hardest question in the 3 papers.

By the way, you could have dealt with (6m+1)^2-(6n+1)^2 by using difference of two squares, which saves you having to expand.
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username3059626
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(Original post by danielwinstanley)
One term = 6n+1
Another term = 6m+1

Difference between squares = (36n^2+12n+1)-(36m^2+12m+1)

Simplify this down

36n^2+12n-36m^2+12m
Factorise and take 12 our
12(3n^2-n-3m^2+m)

This is always a multiple of 24 because you times it by 12. A multiple of 24 is always a multiple of 12.

Sorry if the conclusion doesn't make sense, it's late.
It doesn't make sense.
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Sir Cumference
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Can anyone see a nice way to finish this question that is GCSE level? I can do it but the final part of the proof isn't very nice. The question is:

Here are the first five terms of an arithmetic sequence.
7, 13, 19, 25, 31

Prove that the difference between the squares of any two terms of the sequence is always a multiple of 24.

From googling it looks like the question was from a practice paper, but I can't find the actual paper. The JustMaths solution (Q4) is incorrect.

RDKGames _gcx ?
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RDKGames
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(Original post by danielwinstanley)
One term = 6n+1
Another term = 6m+1

Difference between squares = (36n^2+12n+1)-(36m^2+12m+1)

Simplify this down

36n^2+12n-36m^2+12m
Factorise and take 12 our
12(3n^2-n-3m^2+m)

This is always a multiple of 24 because you times it by 12. A multiple of 24 is always a multiple of 12.

Sorry if the conclusion doesn't make sense, it's late.
A number being a multiple of 24 implies the number is a multiple of 12.

A number being a multiple of 12 does not always imply the number is a multiple of 24 - example; 12 is not a multiple of 24.

(Original post by notnek)
Can anyone see a nice way to finish this question that is GCSE level? I can do it but the final part of the proof isn't very nice. The question is:

Here are the first five terms of an arithmetic sequence.
7, 13, 19, 25, 31

Prove that the difference between the squares of any two terms of the sequence is always a multiple of 24.

From googling it looks like the question was from a practice paper, but I can't find the actual paper. The JustMaths solution (Q4) is incorrect.

RDKGames _gcx ?
I can't see it for GCSE level.

You get down to (1+6n)^2-(1+6m)^2=12(n-m)(3n+3m+1), \forall n,m \in \mathbb{N}_0 then you start to consider odd/even parities of n,m to prove that you can always take out a factor of 2 in any case. So if n,m are both even or odd, then n-m is even thus a 2 can be taken out. If n,m are of opposing parities then n-m is odd but (3n+3m+1) is even thus you can take out a factor of 2.

A mere expansion and factorisation won't prove it's always a multiple of 24 due to the reasons I outlined in the reply above - so consideration of parity of brackets would be required.

Of course, you can always say that 12(n-m)(3n+3m+1)= 24[\frac{(n-m)(3n+3m+1)}{2}] but then you'd need to show the numerator is even somehow - and this requires the same approach with parities as above.
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A02
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(Original post by notnek)
Can anyone see a nice way to finish this question that is GCSE level? I can do it but the final part of the proof isn't very nice. The question is:

Here are the first five terms of an arithmetic sequence.
7, 13, 19, 25, 31

Prove that the difference between the squares of any two terms of the sequence is always a multiple of 24.

From googling it looks like the question was from a practice paper, but I can't find the actual paper. The JustMaths solution (Q4) is incorrect.

RDKGames _gcx ?
This is one GCSE level way to finish the proof:

Once you get to 12(3n^2+n-3m^2-m), group terms as follows:
12((3n^2+n)-(3m^2+m)).
The 2 brackets are even, because if n is odd, let n=2k+1. We get 3n^2+n = 12k^2+14k+4 =
2(6k^2+7k+2). If n is even, let n=2k. We get 3n^2+n = 12k^2+2k = 2(6k^2+k)

So 3n^+n and 3m^2+m are even, so let 3n^2 = 2p and 3m^2+m = 2q. Their difference is 2(p-q), which is even.

Thus we have 12(3n^2+n-3m^2-m)= 12(2z)=24z, where z is an element of the set of integers.
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_gcx
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(Original post by danielwinstanley)
One term = 6n+1
Another term = 6m+1

Difference between squares = (36n^2+12n+1)-(36m^2+12m+1)

Simplify this down

36n^2+12n-36m^2+12m
Factorise and take 12 our
12(3n^2-n-3m^2+m)

This is always a multiple of 24 because you times it by 12. A multiple of 24 is always a multiple of 12.

Sorry if the conclusion doesn't make sense, it's late.
(36n^2+12n+1)-(36m^2+12m+1) \neq 36n^2+12n-36m^2+12m. It equals 36n^2 + 12n - 36m^2 - 12m. A multiple of 12 is only a multiple of 24 if and only if the number being multiplied by 12 is even. For example 36 is a multiple of 12 (12*3 = 36), but is not a multiple of 24. I'm not sure if the examiner would expect students to recognise this.
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Sir Cumference
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(Original post by RDKGames)
A number being a multiple of 24 implies the number is a multiple of 12.

A number being a multiple of 12 does not always imply the number is a multiple of 24 - example; 12 is not a multiple of 24.



I can't see it for GCSE level.

You get down to (1+6n)^2-(1+6m)^2=12(n-m)(n+m+1), \forall n,m \in \mathbb{N}_0 then you start to consider odd/even parities of n,m to prove that you can always take out a factor of 2 in any case. So if n,m are both even or odd, then n-m is even thus a 2 can be taken out. If n,m are of opposing parities then n-m is odd but (n+m+1) is even thus you can take out a factor of 2.

A mere expansion and factorisation won't prove it's always a multiple of 24 due to the reasons I outlines in the reply above - so consideration of parity of brackets would be required.

Of course, you can always say that 12(n-m)(n+m+1)= 24[\frac{(n-m)(n+m+1)}{2}] but then you'd need to show the numerator is even somehow - and this requires the same approach with parities as above.
Thanks, yes the parity argument is what I ended up doing. I supose it isn't really beyond GCSE but the percentage of GCSE students that would even consider this method would be tiny.

I'll try to track down the actual question and mark scheme. It's possible that the question I can find online (and the question the OP was given for homework) was copied incorrectly from the actual paper.
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_gcx
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(Original post by notnek)
Thanks, yes the parity argument is what I ended up doing. I supose it isn't really beyond GCSE but the percentage of GCSE students that would even consider this method would be tiny.

I'll try to track down the actual question and mark scheme. It's possible that the question I can find online (and the question the OP was given for homework) was copied incorrectly from the actual paper.
Most candidates would likely prove that the proposition is true in one given case, true for consecutive terms, or skip the question entirely, so the number of full, complete solutions, would be ridiculously small.

Specimen Set 1, Question 22 of the last paper, by the way,

Spoiler:
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I don't see how the concluding statement is worth 1 mark??

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A02
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(Original post by notnek)
Thanks, yes the parity argument is what I ended up doing. I supose it isn't really beyond GCSE but the percentage of GCSE students that would even consider this method would be tiny.

I'll try to track down the actual question and mark scheme. It's possible that the question I can find online (and the question the OP was given for homework) was copied incorrectly from the actual paper.
Would a proof considering quadratic residues mod 24 be accepted at GCSE?
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the bear
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Pearson Edexcel Level 1/Level 2 GCSE (9-1) in Mathematics - Specimen Papers Set 1 - September 2015 © Pearson Education Limited 2015 166 Specimen Papers Set 1 - September 2015 © Pearson Education Limited 2015

Q22
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Sir Cumference
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(Original post by A02)
Would a proof considering quadratic residues mod 24 be accepted at GCSE?
It would be accepted if the marker doesn't immediately mark it incorrect because they don't understand it!

But it isn't taught at GCSE or A Level.
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(Original post by A02)
Would a proof considering quadratic residues mod 24 be accepted at GCSE?
Number theory isn't taught even in current spec A-level further maths. (though it's going to be in the new FP4)
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Sir Cumference
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(Original post by _gcx)
Most candidates would likely prove that the proposition is true in one given case, true for consecutive terms, or skip the question entirely, so the number of full, complete solutions, would be ridiculously small.

Specimen Set 1, Question 22 of the last paper, by the way,

Spoiler:
Show








I don't see how the concluding statement is worth 1 mark??


Thanks for that. I'm sure I looked through this paper twice but didn't see the question

Proof is a grade 8/9 topic so to add in a requirement of an even/odd argument to complete the proof is just unfair. And as you said, 1 mark for it is ridiculous!

It's possible that the question was made and then only afterwards they realised that the solution was harder than they thought (but that's a very cynical view of Edexcel exam writers).
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A02
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(Original post by notnek)
Thanks for that. I'm sure I looked through this paper twice but didn't see the question

Proof is a grade 8/9 topic so to add in a requirement of an even/odd argument to complete the proof is just unfair. And as you said, 1 mark for it is ridiculous!

It's possible that the question was made and then only afterwards they realised that the solution was harder than they thought (but that's a very cynical view of Edexcel exam writers).
I can't imagine more than a handful of students nationwide getting full marks on a question like that. This question is probably more difficult than quite a few of the current further maths proof by induction to show a sequence is divisible by something style questions.
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Sir Cumference
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(Original post by A02)
I can't imagine more than a handful of students nationwide getting full marks on a question like that. This question is probably more difficult than quite a few of the current further maths proof by induction to show a sequence is divisible by something style questions.
I completely agree. I'd be surprised if I saw a parity argument as a requirement for a Further Maths proof question.
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