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    (Original post by Zacken)
    You did a putnam question? :rofl: - Sure, no problemo (if I can even wrap my head around it... )
    haha well it seemed a little challenging but wasnt too bad, isnt it the Americans version of STEP/ Olympiad sort of stuff?
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    (Original post by EnglishMuon)
    haha well it seemed a little challenging but wasnt too bad, isnt it the Americans version of STEP/ Olympiad sort of stuff?
    It's aimed at American undergraduates. :lol:
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    (Original post by Zacken)
    It's aimed at American undergraduates. :lol:
    oh haha well I thought it was strange they required you to know about Cayley-Hamilton theorem (atleast I needed it for my solution, perhaps missed an easier way)
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    (Original post by EnglishMuon)
    oh haha well I thought it was strange they required you to know about Cayley-Hamilton theorem (atleast I needed it for my solution, perhaps missed an easier way)
    EnglishMuon at King's: Zain, mate - the worst thing happened. I was writing my shopping list and accidentally solved the Riemann Hypothesis. fml why does this happen to me

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    (Original post by Zacken)
    EnglishMuon at King's: Zain, mate - the worst thing happened. I was writing my shopping list and accidentally solved the Riemann Hypothesis. fml why does this happen to me

    PRSOM :rofl: Im literally crying XD
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    (Original post by englishmuon)
    talking about doing maths, please would you look over my solution to a putnam question later?
    wtf dude
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    (Original post by Zacken)
    It's aimed at American undergraduates. :lol:
    Tbh that may as well include you guys, looking at the kind of things you do.
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    (Original post by Student403)
    wtf dude
    I dont think it was one of the hard ones. It helped that I was talking to Zain about some of the theorems I used only a couple days ago
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    (Original post by EnglishMuon)
    I dont think it was one of the hard ones. It helped that I was talking to Zain about some of the theorems I used only a couple days ago
    The name's still Putnam
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    How did you revise the month before?
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    (Original post by EnglishMuon)
    Talking about doing maths, please would you look over my solution to a Putnam question later?
    I will have a look. I like putnam ****


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    (Original post by physicsmaths)
    I will have a look. I like putnam ****


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    Thanks ill write it up in a min
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    (Original post by EnglishMuon)
    Thanks ill write it up in a min
    Wat problem is it.


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    (Original post by physicsmaths)
    Wat problem is it.


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    problem B3 on 2015 paper
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    Well here is some of my working so far so hopefully it isnt completely wrong!
    B3

    Let  \mathbf{M} =  \begin{pmatrix} a & {a+d} \\ {a+2d} & {a+3d}  \end{pmatrix}
    By Cayley-Hamilton theorem, M satisfies its own characteristic polynomial (if thats the correct term for it):
     | \mathbf{M} - \lambda \mathbf{I} | = \lambda ^{2} - (2a+3d) \lambda -2d^{2} on simplification. Hence  \mathbf{M} ^{2} - (2a+3d) \mathbf{M} -2d^{2} \mathbf{I} = \mathbf{0}

    Now this is the part that maybe could be written better but idk.

    Let  d=0 so  \mathbf{M} = \begin{pmatrix} a & {a} \\ {a} & {a} \end{pmatrix} , \mathbf{M}^{2} = 2a \mathbf{M} . Then clearly every power of M also has elements of an arithmetic progression 0, hence  \mathbf{M} = \begin{pmatrix} a & {a} \\ {a} & {a} \end{pmatrix} \in S

    Now let  d= - \frac{2a}{3} :  \mathbf{M} = \begin{pmatrix} a & { \frac{a}{3}} \\ {- \frac{a}{3}} & {-a} \end{pmatrix}, \mathbf{M}^{2} = 2d^{2} \mathbf{I} . Consider M^3:  \mathbf{M}^{3}=   \begin{pmatrix} {2ad^{2}} & {2ad^{2} + 2d^{3}} \\ {2ad^{2} + 4d^{3}} & {2ad^{2} + 6d^{3}} \end{pmatrix} which also has 'arithmetic elements' so   \begin{pmatrix} a & { \frac{a}{3}} \\ {- \frac{a}{3}} & {-a} \end{pmatrix} \in S .

    I think these two types of matrix satisfy all of the elements of our set S, so Im now trying to prove that all our elements take the form. I havent got a solution yet but Ive got a couple of ideas Im looking in to, but not sure if I know enough about vector spaces to do it properly or not. I was thinking something along the lines of writing our general solution as  \mathbf{M}^{k} = ( \begin{pmatrix} a & {a+d} \\ {a+2d} & {a+3d} \end{pmatrix})^{k}= ( a \begin{pmatrix} 1 & {1} \\ {1} & {1} \end{pmatrix} + d \begin{pmatrix} 0 & {1} \\ {2} & {3} \end{pmatrix})^{k} \in S but not convinced of the follow up (if there even is one).
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    (Original post by EnglishMuon)
    I dont think it was one of the hard ones. It helped that I was talking to Zain about some of the theorems I used only a couple days ago
    Yeah there are some not-that-difficult Putnam problems but most are acknowledged to be very difficult. I've done a few of the easier ones since my Calculus text is American . (I've therefore done quite a few AIME problems and the like, which are really fun .)
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    (Original post by IrrationalRoot)
    Yeah there are some not-that-difficult Putnam problems but most are acknowledged to be very difficult. I've done a few of the easier ones since my Calculus text is American . (I've therefore done quite a few AIME problems and the like, which are really fun .)
    ah nice. But yeah that problem still seems hard to get a full refined solution. Ive got the general thing down on paper now but its far from floor proof with 0 assumptions . Which calculus text do you use btw?
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    (Original post by EnglishMuon)
    ah nice. But yeah that problem still seems hard to get a full refined solution. Ive got the general thing down on paper now but its far from floor proof with 0 assumptions . Which calculus text do you use btw?
    AoPS, really nice rigorous treatment of Calculus, but only halfway through since I have done any for a while due to STEP etc.

    Having a look at a page called 'Easy Putnam problems' now since I don't have too much confidence in my ability lol:
    http://www.math.northwestern.edu/~ml...m_problems.pdf
    The last one looks interesting, might try it now .
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    (Original post by EnglishMuon)
    Well here is some of my working so far so hopefully it isnt completely wrong!
    B3

    Let  \mathbf{M} =  \begin{pmatrix} a & {a+d} \\ {a+2d} & {a+3d}  \end{pmatrix}
    By Cayley-Hamilton theorem, M satisfies its own characteristic polynomial (if thats the correct term for it):
     | \mathbf{M} - \lambda \mathbf{I} | = \lambda ^{2} - (2a+3d) \lambda -2d^{2} on simplification. Hence  \mathbf{M} ^{2} - (2a+3d) \mathbf{M} -2d^{2} \mathbf{I} = \mathbf{0}

    Now this is the part that maybe could be written better but idk.

    Let  d=0 so  \mathbf{M} = \begin{pmatrix} a & {a} \\ {a} & {a} \end{pmatrix} , \mathbf{M}^{2} = 2a \mathbf{M} . Then clearly every power of M also has elements of an arithmetic progression 0, hence  \mathbf{M} = \begin{pmatrix} a & {a} \\ {a} & {a} \end{pmatrix} \in S

    Now let  d= - \frac{2a}{3} :  \mathbf{M} = \begin{pmatrix} a & { \frac{a}{3}} \\ {- \frac{a}{3}} & {-a} \end{pmatrix}, \mathbf{M}^{2} = 2d^{2} \mathbf{I} . Consider M^3:  \mathbf{M}^{3}=   \begin{pmatrix} {2ad^{2}} & {2ad^{2} + 2d^{3}} \\ {2ad^{2} + 4d^{3}} & {2ad^{2} + 6d^{3}} \end{pmatrix} which also has 'arithmetic elements' so   \begin{pmatrix} a & { \frac{a}{3}} \\ {- \frac{a}{3}} & {-a} \end{pmatrix} \in S .

    I think these two types of matrix satisfy all of the elements of our set S, so Im now trying to prove that all our elements take the form. I havent got a solution yet but Ive got a couple of ideas Im looking in to, but not sure if I know enough about vector spaces to do it properly or not. I was thinking something along the lines of writing our general solution as  \mathbf{M}^{k} = ( \begin{pmatrix} a & {a+d} \\ {a+2d} & {a+3d} \end{pmatrix})^{k}= ( a \begin{pmatrix} 1 & {1} \\ {1} & {1} \end{pmatrix} + d \begin{pmatrix} 0 & {1} \\ {2} & {3} \end{pmatrix})^{k} \in S but not convinced of the follow up (if there even is one).
    Looking good.

    Ideas
    Spoiler:
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    Explore some more properties of the Characteristic equation, try to find a formula for \ {M}^k then try to use induction.
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    (Original post by EnglishMuon)
    problem B3 on 2015 paper
    Matrices, na boring lol.


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