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    (Original post by physicsmaths)
    Wait till 9am 15th.


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    Ah okay, it's because I never wanna see the answers after the finishing the exam.
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    (Original post by jjsnyder)
    Ohh okay, thanks


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    It's arcsinhx, so you'd expect that the same trick you use for arcsinx might work.
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    Reminder that no discussion about the exam should take place until 24 hours after the exam has been sat.
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    Q1, STEP I 2014,

    The last part, I don't see how it's linked to the previous parts, Part v talks about two prime numbers pq but 675=5^3*3^4 and 5^3 and 3^4 are not prime numbers. So how can the previous parts be applied? So why there are factorising 675 into that form gives 6 factors pairs?
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    (Original post by Geraer100)
    Q1, STEP I 2014,

    The last part, I don't see how it's linked to the previous parts, Part v talks about two prime numbers pq but 675=5^3*3^4 and 5^3 and 3^4 are not prime numbers. So how can the previous parts be applied? So why there are factorising 675 into that form gives 6 factors pairs?
    There are quite a few unrelated parts in that question, it's kinda silly yeah.
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    how u feeling Zacken
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    (Original post by IrrationalRoot)
    There are quite a few unrelated parts in that question, it's kinda silly yeah.
    Yeah, but I don't really understand how to work out the last part. Don't know why they worked out the numbers of ways to factorize and then deduce that it's the same as the numbers of ways that it could be written as a difference of 2 squares.. Hope tomorrow the first question is nice
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    (Original post by Geraer100)
    Yeah, but I don't really understand how to work out the last part. Don't know why they worked out the numbers of ways to factorize and then deduce that it's the same as the numbers of ways that it could be written as a difference of 2 squares.. Hope tomorrow the first question is nice
    Because you know that x^2-y^2=(x+y)(x-y) so you're finding all the ways to factorise and you can see that all these factor pairs can be written as x+y and x-y for some integers x and y.
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    (Original post by IrrationalRoot)
    Because you know that x^2-y^2=(x+y)(x-y) so you're finding all the ways to factorise and you can see that all these factor pairs can be written as x+y and x-y for some integers x and y.
    Oh I see, as one way of factorizing means one way of written that number in a difference of 2 squares. I was trying to link this part to the previous part.
    Thanks!!
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    Im a lost cause when it comes to STEP, even for tomorrow's paper my ultimate goal is a 2 but even that has a slim chance of happening. I do however want to thank everyone on here for the advice and especially solutions (and of course curse all of you for making me feel like a village idiot :ahee:) and I would like to wish the best of luck to those who are actually potent at Mathematics and actually need STEP for their unis. Maybe one day I will STEP my game up (signed my death sentence with that "first in the history of the exam" pun) and finally understand how it works.
    It has been quite a journey. See y'all on the other side.
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    Remember someone(dfranklin?) made a really good post about necessary and sufficient conditions a while back, can any one link it to me?


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    (Original post by drandy76)
    Remember someone(dfranklin?) made a really good post about necessary and sufficient conditions a while back, can any one link it to me?


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    I can't find it, but if it helps this is how I always think about it:

    Sufficient condition: if the condition is true, then the result must be true, but not necessarily the other way around. In other words, knowing only the condition is true is sufficient to assert that the result is true.

    Necessary condition: Result is true implies that the condition is true. In other words, the result cannot possibly hold without the condition. However, note that this says nothing about whether or not the result is actually true - all it says is that if it is true, then the condition must be also.

    If we then have a condition that is both sufficient and necessary, we deduce that the condition and the result are logically equivalent, since without the condition the result cannot be true, and with the condition the result must be true.

    If you want me to go through a specific example I'd be happy to do so
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    are there any nice step 1 esc. qs in ur AOPS book(s)? (I plan to get my own as soon as exams finish so sorry for stealing )
    IrrationalRoot
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    (Original post by Alex_Aits)
    I can't find it, but if it helps this is how I always think about it:

    Sufficient condition: if the condition is true, then the result must be true, but not necessarily the other way around. In other words, knowing only the condition is true is sufficient to assert that the result is true.

    Necessary condition: Result is true implies that the condition is true. In other words, the result cannot possibly hold without the condition. However, note that this says nothing about whether or not the result is actually true - all it says is that if it is true, then the condition must be also.

    If we then have a condition that is both sufficient and necessary, we deduce that the condition and the result are logically equivalent.

    If you want me to go through a specific example I'd be happy to do so
    I know it well enough to work in out in the context of the question usually, but wanna just go over it to make sure I understand it properly, thanks


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    (Original post by EnglishMuon)
    are there any nice step 1 esc. qs in ur AOPS book(s)? (I plan to get my own as soon as exams finish so sorry for stealing )
    IrrationalRoot
    Haha dw, you're not stealing, I'm sharing .
    I'll try and find one now, but I'm not sure how STEP-like they'll be.
    In the mean time if you haven't tried STEP III 1991 Q7 that's a bit STEP I -y and it's really nice.
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    (Original post by IrrationalRoot)
    Haha dw, you're not stealing, I'm sharing .
    I'll try and find one now, but I'm not sure how STEP-like they'll be.
    In the mean time if you haven't tried STEP III 1991 Q7 that's a bit STEP I -y and it's really nice.
    ah yes I have tried that one before, another nice q also thanks!
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    (Original post by EnglishMuon)
    ah yes I have tried that one before, another nice q also thanks!
    I've got an infinite series and an infinite product for you, which one? (The hard ones all seem to be infinite series and products lol.)
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    (Original post by IrrationalRoot)
    I've got an infinite series and an infinite product for you, which one? (The hard ones all seem to be infinite series and products lol.)
    ooh any thanks
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    (Original post by EnglishMuon)
    ooh any thanks
    Latex not working for some reason but
    Evaluate the sum of (Fn)/3^n from n=1 to infinity where Fn is the nth Fibonacci number with F0=0.
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    (Original post by IrrationalRoot)
    Latex not working for some reason but
    Evaluate the sum of (Fn)/3^n from n=1 to infinity where Fn is the nth Fibonacci number with F0=0.
    Lol thats basicaly a step question with a 2 replaced by 3.


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