Just got rid of that part, can't be asked to redo tbh.(Original post by DJMayes)
Spoiler:Show
Why does x have to be a nonnegative integer? sinx has negative roots as well. This is also behind the fact that the factorised expression contains quadratics (Not linear factors, which is what the factor theorem states you can take out as your quadratic can contain extra roots you haven't proved it has). As it is your explanation of this is insufficient. Secondly, you haven't proved that the constant on the outside has to be pi, which is necessary.
On a final note (although I think this was a silly mistake) you should be matching up cubic terms, not quadratic terms which sinx doesn't have in its expansion.

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RoyalBlue7
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 29082013 11:32
Good day to all of you.
I've been going through this thread and found most of the questions to be exceptionally difficult.
Can all of these questions in this thread be solved with A level knowledge; are they based on the Further Pure maths modules or not?
Thanks...
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 29082013 12:06
(Original post by StUdEnTIGCSE)
Good day to all of you.
I've been going through this thread and found most of the questions to be exceptionally difficult.
Can all of these questions in this thread be solved with A level knowledge; are they based on the Further Pure maths modules or not?
Thanks...
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Hypothetically, some of these could be solved with ALevel knowledge I suspect, but to spot the trick you'd have to be some kind of genius 
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 29082013 12:32
(Original post by StUdEnTIGCSE)
Good day to all of you.
I've been going through this thread and found most of the questions to be exceptionally difficult.
Can all of these questions in this thread be solved with A level knowledge; are they based on the Further Pure maths modules or not?
Thanks...
Posted from TSR Mobile
It depends on the star rating of the problem really.
But then again, the knowledge required to solve a problem is no indicator of its difficulty.
In fact, oxymoronically, I hazard that its precisely the opposite. The really advanced problems are often just a simple or otherwise not too difficult application of higher material, whereas some of the 1 or 2 star problems honge on some resourceful and percipient thinking.Last edited by MW24595; 29082013 at 12:33. 
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 02092013 04:18
I noticed the thread's been quiet lately so I think that I would give it a bump with this question.
Problem 328**
Using the algebra of propositions evaluate {(~p V ~q V r) (p r)} V {p (~q V r)}Last edited by Arieisit; 02092013 at 23:30. 
ukdragon37
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 02092013 08:02
(Original post by Arieisit)
I noticed the thread's been quiet lately so I think that I would give it a bump with this question.
Problem 326**
Using the algebra of propositions evaluate {(~p V ~q V r) (p r)} V {p (~q V r)} 
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 02092013 11:44
(Original post by ukdragon37)
What do you mean "evaluate"? What do you want people to do exactly? 
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 02092013 11:56
(Original post by FireGarden)
Simplify, probably. 
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 02092013 20:38
(Original post by Arieisit)
I noticed the thread's been quiet lately so I think that I would give it a bump with this question.
Problem 326**
Using the algebra of propositions evaluate {(~p V ~q V r) (p r)} V {p (~q V r)}
If p is true, then the expression becomes {(~q V r) and r} or {~q or r}}. Considering the case r false, the first bracket is then false, the second is simply ~q; so p true, q false, r false makes the expression true. Considering the case r true, the first bracket becomes true (as does the second) so p true, r true makes the expression true.
Hence the expression is {p and (r or ~q)}. 
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 02092013 22:29
(Original post by Smaug123)
If p is false, then the expression is false. (since then the first bracket is "true and (false and r)" which is "true and false" which is false; and the second bracket is "false and (something)" which is false.)
If p is true, then the expression becomes {(~q V r) and r} or {~q or r}}. Considering the case r false, the first bracket is then false, the second is simply ~q; so p true, q false, r false makes the expression true. Considering the case r true, the first bracket becomes true (as does the second) so p true, r true makes the expression true.
Hence the expression is {p and (r or ~q)}.
Posted from TSR MobileLast edited by Arieisit; 02092013 at 22:30. 
ukdragon37
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 03092013 00:39
(Original post by Arieisit)
I noticed the thread's been quiet lately so I think that I would give it a bump with this question.
Problem 328**
Using the algebra of propositions evaluate {(~p V ~q V r) (p r)} V {p (~q V r)}
by associativity and distributivity of over
by absorption on .
(Original post by Arieisit)
This is NOT the algebra of propositions (Boolean Algebra)
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 03092013 01:00
(Original post by ukdragon37)
Solution 328
To be honest I don't see the value in restricting what method people may solve a problem by. The whole point of this thread is to find interesting methods. If you take that away then it just reduces to a computation that everyone could do if you've learned the textbook method.
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 03092013 10:52
(Original post by ukdragon37)
Solution 328
by associativity and distributivity of over
by absorption on .
Last edited by The Sorcerer; 03092013 at 10:59. 
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 03092013 22:44
(Original post by Arieisit)
This is NOT the algebra of propositions (Boolean Algebra) 
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 03092013 22:52
Problem 329*
a,b,c,d are positive real numbers.
5 of the set ab,ac,ad,bc,bd,cd are the numbers 2,3,4,5,6 in no particular order. Can you find what the other is? 
The nameless one
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 03092013 23:16
(Original post by james22)
Problem 329*
a,b,c,d are positive real numbers.
5 of the set ab,ac,ad,bc,bd,cd are the numbers 2,3,4,5,6 in no particular order. Can you find what the other is?
Spoiler:Show2.4
Spoiler:ShowIn the whole set, there ought to be 3 sets of 2 numbers[(ab,cd),(ac,bd),(ad,bc)], where the product of the 2 elements in each set are the same.Clearly 2 sets are given, (3,4),and(2,6), which both multiplies to 12, so the last set should be(5,2.4) 
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 03092013 23:23
(Original post by james22)
Problem 329*
a,b,c,d are positive real numbers.
5 of the set ab,ac,ad,bc,bd,cd are the numbers 2,3,4,5,6 in no particular order. Can you find what the other is?
Spoiler:ShowWe can suppose (because there's so much symmetry) that we are missing bd.
Then {ab,ac,ad,bc,cd} = {2,3,4,5,6}. But then two pairs of the LHS (namely, {ab,cd} and {ad,bc}) multiply to the same thing. And by unique factorisation, that means ac=5 and abcd = 12.
But then abcd/(ac) is exactly what we're looking for, so the last number is 12/5.
(Very nice puzzle!) 
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 03092013 23:47
Problem 330* (this could be done with gcse knowledge in fact)
Consider the sentence:
THIS IS ONE GREAT CHALLENGE IN MATHEMATICS
Every minute, the first letter of each word is moved to the other end of the word. After how many minutes will the original sentence first reappear? 
Lord of the Flies
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 03092013 23:51
Solution 330
Straightforward, minutes.
Problem 331*
Show that if tends to a nonzero limit as then cannot converge.
(that one is roughly the same level as the most recent questions on this thread)
Here is a nice one:
Problem 332*
are positive integers satisfying . Show thatLast edited by Lord of the Flies; 04092013 at 00:20. Reason: Added two problems 
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