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    (Original post by DJMayes)
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    Why does x have to be a non-negative 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.

    Just got rid of that part, can't be asked to redo tbh.
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    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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    (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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    At the beginning and towards the end, the star rating is fairly close to the level of knowledge required to solve these problems, in the middle the number of stars is not an indicator of what knowledge is required.
    Hypothetically, some of these could be solved with A-Level knowledge I suspect, but to spot the trick you'd have to be some kind of genius
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    (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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    To be honest, no, not all of them can.
    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.
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    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) \land (p \land r)} V {p \land (~q V r)}
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    (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) \land (p \land r)} V {p \land (~q V r)}
    What do you mean "evaluate"? What do you want people to do exactly? :confused:
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    (Original post by ukdragon37)
    What do you mean "evaluate"? What do you want people to do exactly? :confused:
    Simplify, probably.
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    (Original post by FireGarden)
    Simplify, probably.
    Yes, exactly.
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    (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) \land (p \land r)} V {p \land (~q V r)}
    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)}.
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    (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)}.
    This is NOT the algebra of propositions (Boolean Algebra)

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    (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) \land (p \land r)} V {p \land (~q V r)}
    Solution 328

    \left[\left(\lnot p \lor \lnot q \lor r \right) \land \left (p \land r\right)\right] \lor  \left[ p \land \left( \lnot q \lor r \right) \right]

    \iff p \land \left\{ \left[\left(\lnot p \lor \lnot q \lor r \right) \land r \right] \lor \lnot q \lor r\right\} by associativity and distributivity of \land over \lor

    \iff p \land \left( \lnot q \lor r \right) by absorption on r.

    (Original post by Arieisit)
    This is NOT the algebra of propositions (Boolean Algebra)

    Posted from TSR Mobile
    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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    (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.
    I suppose that you are right :facepalm:



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    (Original post by ukdragon37)
    Solution 328

    \left[\left(\lnot p \lor \lnot q \lor r \right) \land \left (p \land r\right)\right] \lor  \left[ p \land \left( \lnot q \lor r \right) \right]

    \iff p \land \left\{ \left[\left(\lnot p \lor \lnot q \lor r \right) \land r \right] \lor \lnot q \lor r\right\} by associativity and distributivity of \land over \lor

    \iff p \land \left( \lnot q \lor r \right) by absorption on r.
    Ah many commendations my mathematical friend. There was me thinking that you were just going through a mundane mathematical question via conventional methods... but then you just shock your dedicated audience tremendously by unveiling the beautiful esoteric technique that is absorption... ABSORPTION! I ask, WHO IN THE RIGHT MIND WOULD HAVE DARED TO HAVE UTILIZED SUCH AN IMMENSELY DIFFICULT TECHNIQUE?! and what an elegant sponge you chose to use for the job too! Wonderful. Keep up the good work. :pierre: I will be greatly anticipating and anxiously awaiting your next ground-breaking mathematical display.
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    (Original post by Arieisit)
    This is NOT the algebra of propositions (Boolean Algebra)
    It is similar, though - I'm just walking through the derivation of the Boolean algebra rules in this specific case. In a similar way, I could say "solve x^2+2x+5 = 0 by the quadratic formula", and solve it by completing the square - in that case, I would have done the same thing, really, since completing the square is how we derive the quadratic formula.
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    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?
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    (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?
    Solution:
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    In 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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    (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?
    Solution: (it's nice - have a go before you look)
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    We 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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    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?
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    Solution 330

    Straightforward, \text{lcm} (4,5,9,11)=1980 minutes.

    Problem 331*

    Show that if \frac{1}{x}f(x) tends to a non-zero limit as x\to 0 then \displaystyle \sum_{k=1}^{\infty} f\left(\frac{1}{k}\right) cannot converge.

    (that one is roughly the same level as the most recent questions on this thread)

    Here is a nice one:

    Problem 332*

    x,y,z>0 are positive integers satisfying x^2+y^2+1=xyz. Show that z=3
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    (Original post by Lord of the Flies)
    Problem 332*

    x,y,z>0 are positive integers satisfying x^2+y^2+1=xyz. Show that z=3
    This was on one of our example sheets
 
 
 
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