Right, I believe that I have come up with a proof. But I don't think it is what you expected, as it is quite long winded. I think I have proved it anyway. I have basically proved that the difference in the differences of the rise of the output when n is increased by 1 each time, is always equal to 6. That is quite complicated and difficult to understand.(Original post by ian.slater)
Sadly you've been given one of the methods already. If you are given a problem like this, a good way to start is to check that it works for small cases of n. And then you try to look for why it's true. Is there some pattern? Then you try to turn the pattern into a proof.
You are probably already familiar with two kinds of maths problem  those you know how to do and those you don't. There is a third kind  those you can figure out how to do if you work on them for long enough.
There is a more direct method than modular arithmetic.
E.g. When n=4&5, your outputs are 39 and 93. 9339=54.
When n=3&4, '' '' '' 9 and 39. 399=30
Now, 5430=24.
When you repeat this for the next set of numbers (4&5with5&6), the final number you get is 30, which is exactly 6 more than 26. Then this continually increases by 6 as you increase the values of n like I just did. This is the pattern I found. Then I proved that you will always get that number 6 no matter what with some long winded basic algebra. Considering that n^3 7n + 3 gives a multiple of 3 for a few tested values of n, and then proving what I have stated that I have proved, does this then prove that the output will always be a multiple of 3 (Because 6 is a multiple of three too)?
I told you it was long winded /:
EDIT: This is the Algebra I did...
Cancels to...
Cancels to...
Cancels to...
Which basically is
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CharlieBoardman
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 05042012 00:31
Last edited by CharlieBoardman; 05042012 at 01:23. 
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 05042012 01:12
I wonder if you could get away with a proof that is almost identical to induction. If you can prove that for n = 1 is true, and then consider what happens between the differences  that is, when you have n^3  7n + 3 and (n+1)^3  7(n+1) + 3  is a multiple of three, then n^3  7n + 3 must be a multiple of three because you are adding multiple of threes to your first term...it's not strictly induction if you don't put on the extra comments and assuming it's true for n = k, no?

CharlieBoardman
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 05042012 01:24
(Original post by Blazy)
I wonder if you could get away with a proof that is almost identical to induction. If you can prove that for n = 1 is true, and then consider what happens between the differences  that is, when you have n^3  7n + 3 and (n+1)^3  7(n+1) + 3  is a multiple of three, then n^3  7n + 3 must be a multiple of three because you are adding multiple of threes to your first term...it's not strictly induction if you don't put on the extra comments and assuming it's true for n = k, no? 
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 1484
 05042012 01:26
(Original post by CharlieBoardman)
Right, I believe that I have come up with a proof. But I don't think it is what you expected, as it is quite long winded. I think I have proved it anyway. I have basically proved that the difference in the differences of the rise of the output when n is increased by 1 each time, is always equal to 6. That is quite complicated and difficult to understand.
E.g. When n=4&5, your outputs are 39 and 93. 9339=54.
When n=3&4, '' '' '' 9 and 39. 399=30
Now, 5430=24.
When you repeat this for the next set of numbers (4&5with5&6), the final number you get is 30, which is exactly 6 more than 26. Then this continually increases by 6 as you increase the values of n like I just did. This is the pattern I found. Then I proved that you will always get that number 6 no matter what with some long winded basic algebra. Considering that n^3 7n + 3 gives a multiple of 3 for a few tested values of n, and then proving what I have stated that I have proved, does this then prove that the output will always be a multiple of 3 (Because 6 is a multiple of three too)?
I told you it was long winded /:Last edited by Zuzuzu; 05042012 at 01:28. 
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 1485
 05042012 01:31
(Original post by Blazy)
I wonder if you could get away with a proof that is almost identical to induction. If you can prove that for n = 1 is true, and then consider what happens between the differences  that is, when you have n^3  7n + 3 and (n+1)^3  7(n+1) + 3  is a multiple of three, then n^3  7n + 3 must be a multiple of three because you are adding multiple of threes to your first term...it's not strictly induction if you don't put on the extra comments and assuming it's true for n = k, no?
x_(n+1) = x_n + 3y_n where y_n is an integer (*)
As x_0 is divisible by 3, and (*) clearly means if x_n is divisible by 3, so is x_(n+1), x_n is divisible by 3 for all by induction.Last edited by matt2k8; 05042012 at 01:32. 
CharlieBoardman
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 05042012 01:35
(Original post by Zuzuzu)
You're almost there. Instead of subbing specific values of n into your expression, consider the difference between any two consecutive values of it. ie. and see what this is. 
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 1487
 05042012 01:39
(Original post by CharlieBoardman)
Right, I believe that I have come up with a proof. But I don't think it is what you expected, as it is quite long winded. I think I have proved it anyway. I have basically proved that the difference in the differences of the rise of the output when n is increased by 1 each time, is always equal to 6. That is quite complicated and difficult to understand.
E.g. When n=4&5, your outputs are 39 and 93. 9339=54.
When n=3&4, '' '' '' 9 and 39. 399=30
Now, 5430=24.
When you repeat this for the next set of numbers (4&5with5&6), the final number you get is 30, which is exactly 6 more than 26. Then this continually increases by 6 as you increase the values of n like I just did. This is the pattern I found. Then I proved that you will always get that number 6 no matter what with some long winded basic algebra. Considering that n^3 7n + 3 gives a multiple of 3 for a few tested values of n, and then proving what I have stated that I have proved, does this then prove that the output will always be a multiple of 3 (Because 6 is a multiple of three too)?
I told you it was long winded /:
EDIT: This is the Algebra I did...
Cancels to...
Cancels to...
Cancels to...
Which basically is
WTS (x+1)^2 + (x1)^2 = 2x^2 + 2.
LHS = (x+1)^2 + (x1)^2 = (x^2 + 2x + 1) + (x^2 2x +1)
= 2x^2 + 2 = RHS. 
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 1488
 05042012 01:44
(Original post by CharlieBoardman)
I have added the algebra that I did onto my last comment... Can you make sense out of it? Or do I need to explain why I did things to make it understandable?

CharlieBoardman
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 1489
 05042012 01:45
(Original post by matt2k8)
In this working, you've basically assumed the identity you want is true, then shown you don't get a contradiction  it's better to state what you want to show, then start with one side, and get to the other. e.g. for proving (x+1)^2 + (x1)^2 = 2x^2 + 2, I'd write:
WTS (x+1)^2 + (x1)^2 = 2x^2 + 2.
LHS = (x+1)^2 + (x1)^2 = (x^2 + 2x + 1) + (x^2 2x +1)
= 2x^2 + 2 = RHS.
Also, I don't know how I would do what you suggested to my equation anyway... 
CharlieBoardman
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 05042012 01:48
(Original post by twig)
The intended method I think (someone has probably posted this already, I have not checked):

Dirac Spinor
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 05042012 01:50
(Original post by CharlieBoardman)
I can see how you've got the RHS, but I'm not sure as to how that proves it? 
CharlieBoardman
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 05042012 02:00
(Original post by twig)
The intended method I think (someone has probably posted this already, I have not checked):
(Original post by bensmith)
how often do multiples of 3 repeat?
And that 3(12n) is obviously divisible by 3? 
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 1493
 05042012 02:01
(Original post by twig)
The intended method I think (someone has probably posted this already, I have not checked):
Spoiler:Show. At least one of n(n7)(n+1) will be a multiple of 3 for whatever the value of n i.e.
n is a multiple of three when n= 3, 6, 9, 12...3k for some integer k
(n7) is a multiple of three when n = 1, 4, 7, 10...3k2
(n+1) is a multiple of three when n = 2, 5, 8...3k1
This should cover all cases for whatever the value of n.
I guess it's that's what's nice about maths  you can have many different solutions, ugly and beautiful... 
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 1494
 05042012 02:15
(Original post by Blazy)
Yeah I was gonna post a version that works along the lines of taking things and adding things (except mine was a lot worse...)
Spoiler:Show. At least one of n(n7)(n+1) will be a multiple of 3 for whatever the value of n i.e.
n is a multiple of three when n= 3, 6, 9, 12...3k for some integer k
(n7) is a multiple of three when n = 1, 4, 7, 10...3k2
(n+1) is a multiple of three when n = 2, 5, 8...3k1
This should cover all cases for whatever the value of n.
I guess it's that's what's nice about maths  you can have many different solutions, ugly and beautiful... 
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 1495
 05042012 02:18
(Original post by CharlieBoardman)
But by proving that I didn't get a contradiction, doesn't that mean than the proof is correct that all values of n will get the answer 6? Therefore proving the statement?
Also, I don't know how I would do what you suggested to my equation anyway...
WTS .
LHS =
= RHS. 
CharlieBoardman
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 05042012 02:30

GreenLantern1
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 05042012 09:46

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 1498
 05042012 15:30
(Original post by GreenLantern1)
Wouldn't it be shorter and quicker just using modular arithmetic? 
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 1499
 05042012 15:44
I want to prepare for STEP 1 only...is it worth going through 'Advanced Problems in Core Mathematics' by Stephen Siklos? Or is the focus there primarily on STEP 2 and 3?
Also how is everyone preparing for the applied questions? In particular, the probability ones? 
GreenLantern1
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 05042012 15:48
(Original post by matt2k8)
Yes, as I said in one of my earlier posts  I was just showing a better way to write out the proof of the identity used in his proof.
EDIT: And can you use modular arithmetic in STEP, since it isn't on the A Level course?Last edited by GreenLantern1; 05042012 at 15:56.
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