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    (Original post by notnek)
    Forget Hannah's sweets, here is the worst answered question in the last 10 years for all the KS4 maths exams:



    This is a treat for all the GCSE students revising hard
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    Not that i recall stats but I got 55.5 minutes
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    (Original post by Kvothe the arcane)
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    Not that i recall stats but I got 55.5 minutes
    That's what I got also and would be accepted by the mark scheme.

    It's a tricky one - I can understand why so few got all the marks.
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    (Original post by Kvothe the arcane)
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    Not that i recall stats but I got 55.5 minutes
    God that question is complete and utter bs waste of time.
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    (Original post by HapaxOromenon)
    Correct. It was from the 2015 Senior Mathematical Challenge.
    Are you allowed any way of answering (i.e. f(5*100))?
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    (Original post by Zacken)
    This 6 mark question came up in my GCSE exam and I quite enjoyed it.

    Integers a and b are such that (a + 3\sqrt{5})^2 + a - b\sqrt{5} = 51. Find the possible values of a and the corresponding values of b.
    hi i've given it a go and for a i got
    a=√36-6a√5-b√5
    2
    i know its probably wrong but any hints would be appreciated
    also which exam board is this from?
    thanks
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    (Original post by OHH_MY_DAYZ:b)
    hi i've given it a go and for a i got
    a=√36-6a√5-b√5
    2
    i know its probably wrong but any hints would be appreciated
    also which exam board is this from?
    thanks
    Nopes, expand them out and compare the rational and irrational parts.

    CIE IGCSE.
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    (Original post by Zacken)
    Nopes, expand them out and compare the rational and irrational parts.

    CIE IGCSE.
    okay so i expanded the brackets and got
    a2 +3a√5 +3a√5 +a -b√5=51
    and then the 3a√5 +3a√5 become 6a√5
    and then i'm lost....
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    (Original post by OHH_MY_DAYZ:b)
    okay so i expanded the brackets and got
    a2 +3a√5 +3a√5 +a -b√5=51
    and then the 3a√5 +3a√5 become 6a√5
    and then i'm lost....
    So you get a^2 + a = 51 and 6a - b = 0
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    Without:

    a) using a calculator or
    b) using long division or
    c) multiplying the factors provided

    show that:

    9,999,999,999,999,999 = 100000001 \times 10001 \times 101 \times 11 \times 9

    Hint:
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    x^2-y^2 = (x+y)(x-y)
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    (Original post by atsruser)
    Without:

    a) using a calculator or
    b) using long division or
    c) multiplying the factors provided

    show that:

    9,999,999,999,999,999 = 100000001 \times 10001 \times 101 \times 11 \times 9

    Hint:
    Spoiler:
    Show
    x^2-y^2 = (x+y)(x-y)
    Spoiler:
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    Note that

    a) x^2-1=x^2-1^2=(x+1)(x-1)
    b) 9,999,999,999,999,999 = 10^{16}-1

    so we have:

    u^{16}-1= (u^8)^2-1^2=(u^8+1)(u^8-1)=(u^8+1)((u^4)^2-1) = \cdots \\ = (u^8+1)(u^4+1)(u^2+1)(u+1)(u-1)

    and now let u=10 to get the quoted result

 
 
 
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