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    There was no need to try 10 different permutations of A and B to find c. In the first part you showed that the composition of functions A and B is commutative. Function composition is also associative, so any composition with 2 As and 3 Bs would be equivalent to AABBB. You could then check it for this value and be satisfied it was the only solution. Sadly I must have made an arithmetic blunder because I don't think I got 107 when I did that - I can't believe I made such a mistake, but I gave all my reasoning so if they're reasonable marking it maybe they'll only take off a mark for the arithmetic slip, given I explained my answer in full.
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    (Original post by 11234)
    did u get n has to be even
    I messed that one up too. Everyone else is saying n has to be even
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    (Original post by furmat)
    There was no need to try 10 different permutations of A and B to find c. In the first part you showed that the composition of functions A and B is commutative. Function composition is also associative, so any composition with 2 As and 3 Bs would be equivalent to AABBB. You could then check it for this value and be satisfied it was the only solution. Sadly I must have made an arithmetic blunder because I don't think I got 107 when I did that - I can't believe I made such a mistake, but I gave all my reasoning so if they're reasonable marking it maybe they'll only take off a mark for the arithmetic slip, given I explained my answer in full.
    How did you know that about function composition?
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    How did everyone find q3 and q4
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    (Original post by RuairiMorrissey)
    How did you know that about function composition?
    Yeah it's not the most obscure knowledge but it's certainly not in C1 or C2. I think there must be a C1/C2 way of figuring out there's only 1 value of c
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    (Original post by furmat)
    There was no need to try 10 different permutations of A and B to find c. In the first part you showed that the composition of functions A and B is commutative. Function composition is also associative, so any composition with 2 As and 3 Bs would be equivalent to AABBB. You could then check it for this value and be satisfied it was the only solution. Sadly I must have made an arithmetic blunder because I don't think I got 107 when I did that - I can't believe I made such a mistake, but I gave all my reasoning so if they're reasonable marking it maybe they'll only take off a mark for the arithmetic slip, given I explained my answer in full.
    Could you explain this because the way i see it, I dont think proving that A(B(x)) = B(A(x)) implies commutativity.

    I may be wrong but how do you know that A(B(B(x))) = B(A(A(x))) etc..
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    when do they start givin out interviews
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    (Original post by 11234)
    How did everyone find q3 and q4
    I found q3 strange but easy. I didnt do q4 cuz i did q6 instead
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    (Original post by m0.4444)
    I found q3 strange but easy. I didnt do q4 cuz i did q6 instead
    would u say this paper was harder than last years?
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    (Original post by sillygoose123)
    could you explain this because the way i see it, i dont think proving that a(b(x)) = b(a(x)) implies commutativity.

    I may be wrong but how do you know that a(b(b(x))) = b(a(a(x))) etc..
    ab = ba

    abb = bab

    bab = baa
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    (Original post by m0.4444)
    I messed that one up too. Everyone else is saying n has to be even
    I got no solutions : (
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    (Original post by collmr)
    ab = ba

    abb = bab

    bab = baa
    Ahhh i see now i feel like a sillygoose
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    (Original post by RuairiMorrissey)
    I got no solutions : (
    I'm pretty sure it's no solutions...
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    (Original post by sillygoose123)
    Could you explain this because the way i see it, I dont think proving that A(B(x)) = B(A(x)) implies commutativity.

    I may be wrong but how do you know that A(B(B(x))) = B(A(A(x))) etc..
    A(B(x)) = B(A(x)) is kind of the definition of commutativity for these two functions, if you treat the functions as elements and function composition as a binary operation that is associative (which is the case). The way I'm saying it makes it sound more obscure than it really is. I still made an error in the actual calculation for c
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    (Original post by RuairiMorrissey)
    I got no solutions : (
    Thats what i put
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    (Original post by RuairiMorrissey)
    I got no solutions : (
    Basically you show that if n is even, you can do difference of squares and play around with the algebra until you get a factor of x^2 + 1 which therefore proves works for all even n. To prove that it doesnt work for odd n, I just tested it with random values of odd n and found counter examples.
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    (Original post by sillygoose123)
    Basically you show that if n is even, you can do difference of squares and play around with the algebra until you get a factor of x^2 + 1 which therefore proves works for all even n. To prove that it doesnt work for odd n, I just tested it with random values of odd n and found counter examples.
    I just subbed in x = i

    : I
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    (Original post by RuairiMorrissey)
    How did you know that about function composition?
    if AB = BA...
    AAB = A(BA) = (BA)A
    AABB = A(BA)B = (BA)AB = AB(BA) = (BA)(BA) = B(BA)A
    etc.
    basically for any two combinations of A and B, as long as the number of A and number of B are constant, they come out as the same answer
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    (Original post by RuairiMorrissey)
    I just subbed in x = i

    : I
    Yeah I think that's the simplest way to do it. You can even avoid using i by just subbing in x^2 = -1 but this requires extreme reliance on the factor theorem if you haven't been taught about imaginary numbers lol
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    (Original post by boombox111)
    if AB = BA...
    AAB = A(BA) = (BA)A
    AABB = A(BA)B = (BA)AB = AB(BA) = (BA)(BA) = B(BA)A
    etc.
    basically for any two combinations of A and B, as long as the number of A and number of B are constant, they come out as the same answer
    Ah I see, well done.

    What do you reckon the shortlisted/succesful average will be?
 
 
 
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