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    Enjoy

    Problem 1

    Let

    p and q be positive integers (not necessarily distinct)

    Prove that

    p^q - 1 is divisible by (p-1)

    without using the remainder theorem




    Problem 2

    Find

    \displaystyle \int \dfrac{\sin t - \cos t}{(\sin t + \cos t)\sqrt{\sin^2 t \cos^2 t + \sin t \cos t}} \ dt




    Problem 3

    Let f be a function satisfying

    f + f' = |x|

    for real x

    You are given that f(-1)=0

    Find the value of f(1)




    Problem 4

    Find

    \displaystyle \int \dfrac{x\cos x + 1}{\sqrt{x^2 + 2x^3 e^{\sin x}}} \ dx




    Problem 5

    A rational function g(x) is defined as follows

    g(x) = \displaystyle \int \dfrac{f(x)}{x^2(x+1)^3}\ dx

    where f(x) is a quadratic satisfying f(0)=1

    Find the value of f'(0)

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    Will demotivate myself with these after the Arsenal game
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    (Original post by drandy76)
    Will demotivate myself with these after the Arsenal game
    Me too after the rivals game ....
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    (Original post by Indeterminate)
    Enjoy

    Problem 1

    Let

    p and q be positive integers (not necessarily distinct)

    Prove that

    p^q - 1 is divisible by (p-1)

    I'm tired and it's time to hit this with an ugly hammer, I think there's a nice geometrical series proof?

    Claim: x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \cdots + 1)

    The base case (n=2) holds, since x^2 - 1 = (x-1)(x+1).

    Assuming x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \cdots + 1), we then multiply by x on both sides to get:

    x^{n+1} - x = (x-1)(x^n + x^{n-1} + \cdots + x)

    Cleverly adding zero:

    x^{n+1} - 1 - (x-1) = (x-1)(x^n + x^{n-1} + \cdots + x)

    So, adding x-1 to both sides, we have:

    x^{n+1} - 1 = (x-1)(x^n + x^{n-1} + \cdots +x + 1) and we are done!

    Hence, p^q - 1 = (p-1)(p^{q-1} + \cdots + 1) and is hence definitely divisible by p-1.
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    Simple? What level would you call this?
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    Surely this is trivial, as you can just set p=1 -> p^q-1=0, so by the Remainder Theorem it divides (p-1) as required.

    (Original post by Indeterminate)
    Enjoy

    Let

    p and q be positive integers (not necessarily distinct)

    Prove that

    p^q - 1 is divisible by (p-1)
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    (Original post by TeeEm)
    Me too after the rivals game ....
    I fear a positive result for tottenham will mean a postpone of my attempts until tomorrow
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    (Original post by Zacken)
    I'm tired and it's time to hit this with an ugly hammer, I think there's a nice geometrical series proof?

    Claim: x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \cdots + 1)

    The base case (n=2) holds, since x^2 - 1 = (x-1)(x+1).

    Assuming x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \cdots + 1), we then multiply by x on both sides to get:

    x^{n+1} - x = (x-1)(x^n + x^{n-1} + \cdots + x)

    Cleverly adding zero:

    x^{n+1} - 1 - (x-1) = (x-1)(x^n + x^{n-1} + \cdots + x)

    So, adding x-1 to both sides, we have:

    x^{n+1} - 1 = (x-1)(x^n + x^{n-1} + \cdots +x + 1) and we are done!

    Hence, p^q - 1 = (p-1)(p^{q-1} + \cdots + 1) and is hence definitely divisible by p-1.
    No need for all that - see my post above.
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    (Original post by drandy76)
    I fear a positive result for tottenham will mean a postpone of my attempts until tomorrow
    West ham has been a boggie team for us for years ...
    The odds are in your favour tonight
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    (Original post by constellarknight)
    No need for all that - see my post above.
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    (Original post by constellarknight)
    Surely this is trivial, as you can just set p=1 -> p^q-1=0, so by the Remainder Theorem it divides (p-1) as required.
    I initially thought of that as well - but then I figured that it was too trivial, so I figured assuming the remainder theorem wasn't allowed.
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    (Original post by TeeEm)
    West ham has been a boggie team for us for years ...
    The odds are in your favour tonight
    hopefully they chip away at your shiny goal difference as well
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    (Original post by drandy76)
    hopefully they chip away at your shiny goal difference as well
    we are losing already
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    we just took the lead , do i dare hope?

    (Original post by TeeEm)
    we are losing already
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    • Political Ambassador
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    (Original post by undercxver)
    Simple? What level would you call this?
    A-level :yep:
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    (Original post by Indeterminate)
    This proves that p^q - 1 is divisible by (p-1) for all q when p =1

    What happens when p \neq 1 ?
    x=1 is a root of x^q-1, regardless of the value of q. Thus (x-1) is a factor, by the Remainder Theorem. I don't really understand your point??
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    (Original post by Indeterminate)
    This proves that p^q - 1 is divisible by (p-1) for all q when p =1

    What happens when p \neq 1 ?
    The remainder theorem implies the other factor is always a polynomial in p, no?
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    (Original post by constellarknight)
    x=1 is a root of x^q-1, regardless of the value of q. Thus (x-1) is a factor, by the Remainder Theorem. I don't really understand your point??
    (Original post by EricPiphany)
    The remainder theorem implies the other factor is always a polynomial in p, no?
    Ignore me :facepalm:

    I'll edit the OP
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    (Original post by EricPiphany)
    The remainder theorem implies the other factor is always a polynomial in p, no?
    Exactly. Like I said in my previous post, Indeterminate is not making much sense, which is apt given his username.
 
 
 
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