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    What field are we working in?

    For example, 9+10 in integers modulo 3 is 9mod3 + 10mod3 = 0mod3 + 1mod3 = 1.

    Integers modulo 13 on the other hand, 9+10 = 6.
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    (Original post by SeanFM)
    What field are we working in?

    For example, 9+10 in integers modulo 3 is 9mod3 + 10mod3 = 0mod3 + 1mod3 = 1.

    Integers modulo 13 on the other hand, 9+10 = 6.
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    If you work for the Tories, whatever you want it to be.
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    (Original post by TheCe)
    Year 2 maths exam? Genuinely curious
    Nooo it was in year 6 in SATs...i still remember it it was so hard and i was so confused omygod
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    Vaginal Lubrication Cream? :dontknow:
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    (Original post by SeanFM)
    What field are we working in?

    For example, 9+10 in integers modulo 3 is 9mod3 + 10mod3 = 0mod3 + 1mod3 = 1.

    Integers modulo 13 on the other hand, 9+10 = 6.
    You smartass you... :^_^:
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    9+10=18
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    Hmm, let's try a trig sub.

    9 + 10 = 9*1 + 10*1 = 9*(cos^2(x) + sin^2(x)) + 10*(cos^2(x) + sin^2(x))
    = 9*((cos(2x) + 1)/2 + sin^2(x)) + 10*(cos^2(x) - (cos(2x) -1)/2)
    = 9/2(cos(2x) + 1) + 9sin^2(x) + 10cos^2(x) - 5(cos(2x) - 1)
    = 9/2(cos(2x) + 1) + 9sin^2(x) + 10cos^2*(x) -5(cos(2x) + 1) + 10
    = 9sin^2(x) - (1/2)(cos(2x) + 1) + 10cos^2(x) + 10
    = 9sin^2(x) - cos^2(x) + 10cos^2*(x) + 10
    = 9sin^2(x) + 9cos^2(x) + 10
    = 9sin^2(x) + 9cos^2(x) + 9 + 1
    = 9(sin^2(x) + cos^2(x) + 1) + 1
    = 9(1 + 1) + 1
    = 18 + 1
    = 19
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    It depends what group you're working in. If you're working in a group containing the integers you'd just add them normally.

    (Original post by XOR_)
    19! = 1.216451e+17
    I make it 121645100408832000. It's not equal to what you said.
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    (Original post by serah.exe)
    I thought that was an A2 Maths question.
    The answer is 420 by the way.
    STEP III question
 
 
 
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