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    S U T = set of odd integers
    S ∩ T = empty set (cannot leave two different remainders)

    Proving the closure under multiplication of S is trivial, just express numbers in terms of 4k + 1 for non-negative integer k
    As for disproving close under multiplication of T, just take 3 in T, then 3^2 = 9 which is not in T

    For part iii, if none of its prime factors are in T, then they are all in S (since S U T = set of odd integers, and all prime factors must be odd as otherwise we'll have an even number in T). But then the product is in S, not in T, contradiction as T is a product of powers of its prime factors

    for iv a), you can go the "hand-wavey" route/intuitive route but I dunno how permissible that would be. Basically you can first inductively show using modular arithmetic or whatever you want that any product of an odd number of integers in T is itself in T, while any product of an even number of integers in T is in S. an integer in T that is not T-prime can be expressed as the product of two or more integers in T, so wee see it must be expressible as the product of an odd number of integers in T. Then we apply the same logic to each of the factors in such a product, and eventually this process stops as it is finite (and the final product must be of an odd number of the fellas since it is in T). I thought I had something more rigorous but realised it assumed a little bit much, maybe come back to this if I care enough but that's the gist..
    Wordy mostly done solution to 7 (haven't looked for an example but whatever, screw examples)
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    how many marks for doing 1 full, 2 full but missing the constant, 3 full, 4 and getting a circle center 0,0 and radius 1/k by missing the constant again but otherwise full, 10 did all up to finding the value of lamda, where i got bogged down in algebra and but said how to ding lamda then sub back in into the equation for lamba and e to find e, then 11 i did all except find the value of cos2a where i got bogged down again but i showed how to find the max distance by pythag that was like hsec2a or something then subbed my value of cos2a...
    thanks i really dont have much idea how they are marked especially thepartials
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    (Original post by 13 1 20 8 42)
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    S U T = set of odd integers
    S ∩ T = empty set (cannot leave two different remainders)

    Proving the closure under multiplication of S is trivial, just express numbers in terms of 4k + 1 for non-negative integer k
    As for disproving close under multiplication of T, just take 3 in T, then 3^2 = 9 which is not in T

    For part iii, if none of its prime factors are in T, then they are all in S (since S U T = set of odd integers, and all prime factors must be odd as otherwise we'll have an even number in T). But then the product is in S, not in T, contradiction as T is a product of powers of its prime factors

    for iv a), you can go the "hand-wavey" route/intuitive route but I dunno how permissible that would be. Basically you can first inductively show using modular arithmetic or whatever you want that any product of an odd number of integers in T is itself in T, while any product of an even number of integers in T is in S. an integer in T that is not T-prime can be expressed as the product of two or more integers in T, so wee see it must be expressible as the product of an odd number of integers in T. Then we apply the same logic to each of the factors in such a product, and eventually this process stops as it is finite (and the final product must be of an odd number of the fellas since it is in T). I thought I had something more rigorous but realised it assumed a little bit much, maybe come back to this if I care enough but that's the gist..
    Wordy mostly done solution to 7 (haven't looked for an example but whatever, screw examples)
    Pretty much what I did, except I didn't know what the notation meant for the first part


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    (Original post by drandy76)
    Pretty much what I did, except I didn't know what the notation meant for the first part


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    Ah fair enough, probs only worth like 2-3 marks anyway
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    (Original post by johnthedog)
    would quite like one of them fancy polls to gauge how hard the paper was compared to previous years.
    Shamika and the other experienced hands will give a more valuable opinion in due course when they get hold of the whole paper.

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    (Original post by Duke Glacia)
    ur opinions on this paper ?
    Did badly but nice paper. No idea how to do the second part of the circles Q I'll admit.
    Everything else I got wrong/didn't do was just being being 5x dumber than usual.
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    (Original post by IrrationalRoot)
    Did badly but nice paper. No idea how to do the second part of the circles Q I'll admit.
    Everything else I got wrong/didn't do was just being being 5x dumber than usual.
    predicted grade boundries ?
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    (Original post by Duke Glacia)
    predicted grade boundries ?
    Probably fairly high . But I'm not completely sure, depends how other people found it.
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    (Original post by 13 1 20 8 42)
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    S U T = set of odd integers
    S ∩ T = empty set (cannot leave two different remainders)

    Proving the closure under multiplication of S is trivial, just express numbers in terms of 4k + 1 for non-negative integer k
    As for disproving close under multiplication of T, just take 3 in T, then 3^2 = 9 which is not in T

    For part iii, if none of its prime factors are in T, then they are all in S (since S U T = set of odd integers, and all prime factors must be odd as otherwise we'll have an even number in T). But then the product is in S, not in T, contradiction as T is a product of powers of its prime factors

    for iv a), you can go the "hand-wavey" route/intuitive route but I dunno how permissible that would be. Basically you can first inductively show using modular arithmetic or whatever you want that any product of an odd number of integers in T is itself in T, while any product of an even number of integers in T is in S. an integer in T that is not T-prime can be expressed as the product of two or more integers in T, so wee see it must be expressible as the product of an odd number of integers in T. Then we apply the same logic to each of the factors in such a product, and eventually this process stops as it is finite (and the final product must be of an odd number of the fellas since it is in T). I thought I had something more rigorous but realised it assumed a little bit much, maybe come back to this if I care enough but that's the gist..
    Wordy mostly done solution to 7 (haven't looked for an example but whatever, screw examples)
    For iv)a), you could say this. Suppose t (not 1) in T is not T-prime. Then express t as a product of integers in T (excluding 1). Decompose each integer in the product until t is written as a product of T-prime integers (cannot be decomposed further). In this state let t = t1 * t2 * ... * tm. Now, the product of two 3 mod 4 integers is congruent to 1 mod 4. So if m is even, then t = 1 mod 4, a contradiction. Therefore m is odd.




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    (Original post by LukaVendov)
    how many marks for doing 1 full, 2 full but missing the constant, 3 full, 4 and getting a circle center 0,0 and radius 1/k by missing the constant again but otherwise full, 10 did all up to finding the value of lamda, where i got bogged down in algebra and but said how to ding lamda then sub back in into the equation for lamba and e to find e, then 11 i did all except find the value of cos2a where i got bogged down again but i showed how to find the max distance by pythag that was like hsec2a or something then subbed my value of cos2a...
    thanks i really dont have much idea how they are marked especially thepartials
    20, 19, 20, 16/17, 16/17, 14/15
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    (Original post by Ecasx)
    For iv)a), you could say this. Suppose t (not 1) in T is not T-prime. Then express t as a product of integers in T (excluding 1). Decompose each integer in the product until t is written as a product of T-prime integers (cannot be decomposed further). In this state let t = t1 * t2 * ... * tm. Now, the product of two 3 mod 4 integers is congruent to 1 mod 4. So if m is even, then t = 1 mod 4, a contradiction. Therefore m is odd.


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    Yes, that is in essence very similar just less wordy. I was just thinking one could do something rigorous/that doesn't rely on just saying "keep doing this until you stop" but considering STEPpers aren't supposed to have done set theory should probs be full marks for that kind of argument
    (A well-ordering principle method for instance like the one used for integers in general)
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    (Original post by shamika)
    EDIT: if someone can let me know when the paper has been uploaded that'd be great
    Solution thread's been made and I'll give you a tag/quote when the paper gets released (shouldn't be too long now, just waiting for the person who's uploading it to come out of his M4 exam).

    I'm afraid that I didn't do very well but the boundaries are going to be really high.
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    Can we only use pencil ?
    step central page : They should write their answers in hard pencil or pen.
    Can we only use pencil only in graphs ?
    Also is HB2 (regular pencils) considered to be hard ?
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    Don't care about step I bring on ii tomorrow
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    Looks like it was a nice paper*
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    * = Dun kill me if it wasn't!
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    (Original post by Vesniep)
    Can we only use pencil ?
    step central page : They should write their answers in hard pencil or pen.
    Can we only use pencil only in graphs ?
    Also is HB2 (regular pencils) considered to be hard ?
    They aren't going to chemically analyse the pencils everyone used and fail you if it's not 4H.

    It's just to ensure your answers are (and remain) legible.

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    (Original post by johnthedog)
    oh for **** ******** ****** ************ sake
    oh i know that feeling. Did exactly the same as you
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    shamika whats ur opinion on the paper ?
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    (Original post by 13 1 20 8 42)
    ...
    How do you have the paper? How do you even have the time for STEP (or are Warwick exams over)?
 
 
 
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