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    Solution 117

    Let \alpha be a solution to \lambda^p+\lambda^q-\lambda =0.

    Since f(x^p)=f(x)^p+\alpha^q and f(y^q)=f(y)^q+\alpha^p we have f(x^p+y^q)=f(x^p)+f(y^q)-\alpha. Set h(z)=f(z)-\alpha gives h(x'+y')=h(x')+h(y') and hence f(x)=ax+\alpha since f,h cont.

    It is now clear that a\in\{-1,0,1\} if p,q are odd and a\in\{0,1\} otherwise. If a\neq 0 then we require \alpha =0, if a=0 it is easily checked that \alpha has three possible values when \max p,q is even and two when \max p,q is odd (both these cases include \alpha=0 of course)



    One of my favourite series questions:

    Problem 119***

    \displaystyle \sum_{n=0}^{\infty} \binom{2n}{n}^{-1}
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    Solution 119

    We use generating functions.

    Observe \displaystyle \dbinom{2n}{n}^{-1}= \frac{n^{2}}{2n(2n-1)} \times \dbinom{2n-2}{n-1}^{-1}
    Hence \displaystyle 2(2n-1) \dbinom{2n}{n}^{-1}= n \dbinom{2n-2}{n-1}^{-1}
    Thus \displaystyle \sum_{n=1}^{\infty} 2(2n-1) \dbinom{2n}{n}^{-1}x^{n} = \sum_{n=1}^{\infty} n \dbinom{2n-2}{n-1}^{-1}x^{n}
    \displaystyle 4\sum_{n=0}^{\infty} n\dbinom{2n}{n}^{-1}x^{n} - 2\sum_{n=0}^{\infty} \dbinom{2n}{n}^{-1}x^{n}+2 = \sum_{n=0}^{\infty} (n+1)\dbinom{2n}{n}^{-1}x^{n+1}
    \displaystyle F(x) = \sum_{n=0}^{\infty} \dbinom{2n}{n}^{-1}x^{n}.
    Further, \displaystyle 4xF'(x)-2F(x)+2=x(xF(x))', which is equivalent to (4x-x^{2})F'(x)=(x+2)F(x)-2.
    Linear differential equation, with variable coefficients. We employ Lagrange's method. After solving this equation (it is too late here, so I am omitting the details (I may add these details in tomorrow)), we obtain \displaystyle F(1)= \frac{4}{3}+\frac{2\pi}{9\sqrt{3  }}.
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    (Original post by Mladenov)
    Solution 119
    Bravo! I took a somewhat different route, if you're interested.

    \displaystyle \sum_{n\geq 0} \binom{2n}{n}^{-1}=\sum_{n\geq 0} \frac{(2n+1)\Gamma^2(n+1)}{ \Gamma (2n+2)}=\sum_{n\geq 0} (2n+1)\text{B}\big (n+1,n+1\big)

    \displaystyle f(\alpha )=\int_0^1 \sum_{n\geq 0}\alpha^{2n+1}x^n(1-x)^n\,dx =\int_0^1 \frac{\alpha\,dx}{1-\alpha^2x(1-x)}

    \displaystyle f'(1)=\int_0^1 \frac{1+x-x^2}{(x^2-x+1)^2}\,dx which is very easy.
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    Oh and here is a quick alternative solution to 108:

    Solution 108

    \alpha, \beta are integer roots to the eq. \alpha+\beta|4pq+1\Rightarrow exactly one of \alpha,\beta is odd/even. Let \alpha be the odd root:

    (4pq+1)\alpha =(p^2+q^2)(\alpha^2+1)

    LHS is odd, RHS is even, hence \alpha cannot exist. So the polynomial is not a quadratic \Rightarrow p=q=0
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    (Original post by Lord of the Flies)
    I took a somewhat different route, if you're interested
    Superb. I would have never come up with this solution, for when I saw the binomial coefficient, I instantly started thinking of combinatorial solution.

    The following problem is inspired by your profile picture.

    Problem 120***

    Evaluate: \displaystyle \int_{-\infty}^{\infty} \frac{\cos x -1}{x^{2}(x^{2}+a)}dx, where a > 1.
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    Solution 120

    \displaystyle\frac{a}{2}\int_{-\infty}^{\infty} \frac{\cos x-1}{x^2(x^2+a)}\,dx=\underbrace{ \int_0^{ \infty} \frac{\cos x-1}{x^2}\,dx}_{(1)}-\underbrace{\int_0^{\infty} \frac{\cos x\,dx}{x^2+a}}_{(2)}+\underbrace  {\int_0^{\infty} \frac{dx}{x^2+a}}_{(3)}

    \displaystyle (1):\;\;\int_0^{ \infty} \frac{\cos x-1}{x^2}\,dx=-\int_0^{\infty}\frac{\sin x}{x}\,dx

    \displaystyle \begin{aligned} w(t)=\int_0^{\infty}\frac{\sin tx}{x}\,dx\Rightarrow\mathcal{L}  \{w(t)\} &=\int_0^{\infty}e^{-st}\int_0^{\infty}\frac{\sin tx}{x}\,dx\,dt\\&=\int_0^{\infty  }\frac{1}{x}\int_0^{\infty}e^{-st}\sin tx\,dt\,dx\\&=\int_0^{\infty} \frac{1}{x}\left( \frac{x}{x^2+s^2} \right)\,dx \\&=\frac{\pi}{2s}

    \displaystyle\mathcal{L}^{-1} \left\{\frac{\pi}{2s}\right\} =\frac{\pi}{2}\Rightarrow \int_0^{ \infty} \frac{\cos x-1}{x^2}\,dx=-\frac{\pi}{2}

    \displaystyle \begin{aligned}(2):\;\; z(t)=\int_0^{\infty} \frac{\cos tx\,dx}{x^2+a}\,dx\Rightarrow \mathcal{L} \{ z(t)\} &=\int_0^{\infty}e^{-st}\int_0^{\infty}\frac{\cos tx\,dx}{x^2+a}\,dx\,dt\\&=\int_0  ^{\infty}\frac{1}{x^2+a}\int_0^{  \infty}e^{-st}\cos tx\,dt\,dx\\&=\int_0^{\infty} \frac{1}{x^2+a} \left(\frac{s}{x^2+s^2} \right)dx\\&=\frac{s}{s^2-a}\left(\int_0^{\infty}\frac{dx}  {x^2+a}-\int_0^{\infty} \frac{dx}{x^2+s^2}\right)\\&= \frac{\pi}{2\sqrt{a}}\left( \frac{1}{s+\sqrt{a}}\right)\end{  aligned}

    \displaystyle\mathcal{L}^{-1} \left\{\frac{1}{s+\sqrt{a}} \right\} = e^{-\sqrt{a}t} \Rightarrow \int_0^{\infty} \frac{\cos x\,dx}{x^2+a}=\frac{\pi }{2\sqrt{a}e^{\sqrt{a}}}

    \displaystyle (3):\;\;\int_0^{\infty}\frac{dx}  {x^2+a}=\frac{\pi}{2\sqrt{a}}

    Therefore \displaystyle\int_{-\infty}^{\infty} \frac{\cos x-1}{x^2(x^2+a)}\,dx=\frac{\pi}{a \sqrt{a}} \left(1-\sqrt{a}-\frac{1}{e^{\sqrt{a}}}\right)
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    (Original post by Lord of the Flies)
    Solution 120
    You would agree that Laplace is indeed quite useful for this problem, would you not?
    I suspect that there is a complex analysis solution, but I am not sure if it is rather laborious.

    Problem 121*** It is not that awful.

    Evaluate \displaystyle \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} {\rm sign}(x){\rm sign}(y)e^{-\frac{x^{2}+y^{2}}{2}}\sin (xy) \,dx \,dy

    Problem 122*** This is tough, unless one has studied analytic number theory.

    Let \displaystyle \sum_{n \equiv 1 \pmod 2} (-1)^{\frac{n-1}{2}}\frac{ \log n}{\sqrt{n}}, and \displaystyle \sum_{n \equiv 1 \pmod 2} (-1)^{\frac{n-1}{2}}\frac{1}{\sqrt{n}}. Find an explicit form for \displaystyle \left(\sum_{n \equiv 1 \pmod 2} (-1)^{\frac{n-1}{2}}\frac{ \log n}{\sqrt{n}} \right)\left(\sum_{n \equiv 1 \pmod 2} (-1)^{\frac{n-1}{2}}\frac{1}{\sqrt{n}} \right)^{-1}.
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    Problem 123***

    Prove that  x^8 \equiv 16 (Mod p) is solvable for every prime p.
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    Solution 123

    For p=2 the problem is trivial. We suppose p \ge 3. Note that x^{n} \equiv a \pmod p is solvable if and only if \displaystyle a^{\frac{p-1}{\gcd (n,p-1)}} \equiv 1 \pmod p. To prove this assertion, we introduce g - primitive root \pmod p. Let a \equiv g^{i} \pmod p and x \equiv g^{j} \pmod p. We then have g^{jn} \equiv g^{i} \pmod p. This is equivalent to jn \equiv i \pmod {p-1}. Let k= \gcd(n,p-1). If i \equiv 0 \pmod k then jn \equiv i \pmod {p-1} is solvable and \displaystyle a^{\frac{p-1}{k}} \equiv g^{\frac{(p-1)i}{k}} \equiv 1 \pmod p. If i \not\equiv 0 \pmod k, then jn \equiv i \pmod {p-1} is not solvable and \displaystyle a^{\frac{p-1}{k}} \equiv g^{\frac{(p-1)i}{k}} \not\equiv 1 \pmod p.
    Hence we have to prove that \displaystyle 2^{4\frac{p-1}{\gcd (8, p-1)}} \equiv 1 \pmod p, which is obvious unless p \equiv 1 \pmod 8. However, in this case \left(\frac{2}{p}\right)=1; thus x^{2} \equiv 2 \pmod p is solvable and we are done.
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    ...alright then.

    Problem 124

    Evaluate  \tau_p = \displaystyle\sum_{k=0}^{p-1}\Big{(}\frac{k}{p}\Big{)} e^{\frac{2\pi i k}{p}}

    Where  \Big{(}\frac{k}{p}\Big{)} is the legendre symbol.
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    Some of these questions are getting a little silly (for the intended audience). Laplace transforms? Legendre symbols? These are things encountered in the second or third year of a good university. Having said that, if people are happy and able to solve them, by all means be my guest
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    Solution 121

    \displaystyle\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \text{sgn}\, x\,\text{sgn}\, y \,\exp \left(-\frac{x^2+y^2}{2}\right)\sin xy\,dx\,dy

    =4\displaystyle\int_0^{\infty} \int_0^{\infty}\exp \left(-\frac{x^2+y^2}{2}\right)\sin xy\,dx\,dy

    =4\displaystyle\int_0^{\frac{\pi  }{2}} \int_0^{\infty} r\exp \left(-\frac{r^2}{2}\right)\sin \left(\frac{r^2}{2}\sin 2\alpha\right)\,dr\,d\alpha

    =4\displaystyle\int_0^{\frac{\pi  }{2}} \left[ -\frac{1}{1+\sin^2 2\alpha}\exp\left(-\frac{r^2}{2}\right)\left(\sin \left(\frac{r^2}{2}\sin 2\alpha\right) +\sin 2\alpha \cos\left(\frac{r^2}{2}\sin 2\alpha\right)  \right)\right]_0^{\infty}d\alpha

    =4\displaystyle\int_0^{\frac{\pi  }{2}} \frac{\sin 2\alpha\,d\alpha}{1+\sin^2 2\alpha }

    =8\displaystyle\int_0^{\infty} \frac{t\,dt}{t^4+6t^2+1}

    =\displaystyle\sqrt{2}\ln \big(3+2\sqrt{2}\big)

    In retrospect I am not sure switching to polar was the best idea, since some of the working is a bit tedious, but whatever.
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    Solution 124

    Denote \displaystyle F(p)= \sum_{k=0}^{p-1} \left(\frac{k}{p}\right)e^{\frac  {2ki\pi}{p}}. Hence we have \displaystyle (F(p))^{2}=F(p)\overline{F(p)}= \left( \sum_{k=0}^{p-1} \left(\frac{k}{p} \right)e^{\frac{2ki\pi}{p}} \right) \left( \sum_{m=0}^{p-1} \left(\frac{m}{p} \right) e^{-\frac{2mi\pi}{p}}\right). Let l be such that k \equiv ml \pmod p. Hence
    \begin{aligned} \displaystyle \left( \sum_{k=0}^{p-1} \left(\frac{k}{p} \right)e^{\frac{2ki\pi}{p}} \right) \left( \sum_{m=0}^{p-1} \left(\frac{m}{p} \right) e^{-\frac{2mi\pi}{p}}\right)&= \sum_{m=1}^{p-1} \sum_{l=1}^{p-1} \left(\frac{l}{p} \right) e^{\frac{2im(l-1)\pi}{p}} \\& = p-1+ \sum_{m=1}^{p-1} \sum_{l=2}^{p-1} \left(\frac{l}{p} \right) e^{\frac{2im(l-1)\pi}{p}} \\&= p-1 + \sum_{l=2}^{p-1} \left(\frac{l}{p} \right) \\&= p \end{aligned}
    Therefore \displaystyle \sum_{k=0}^{p-1} \left(\frac{k}{p}\right)e^{\frac  {2ki\pi}{p}} = \pm \sqrt{p}


    (Original post by Lord of the Flies)
    Solution 121
    In retrospect I am not sure switching to polar was the best idea, since some of the working is a bit tedious, but whatever.
    This integral turned out to be an annoying computation (I have also solved it using polar coordinates). I regret posting it, since most probably there is no concise solution.

    Problem 125**

    Let a_{1}, a_{2},..., a_{n} be n students. Some of these students know each other. Show that we can split these students into two groups such that if given student knows m students from his or her group, then he or she knows at least m students from the other group.

    Problem 126***

    Let f : \left[0,1\right] \to \mathbb{R} be a continuous non-decreasing function. Show that \displaystyle \frac{1}{2}\int_{0}^{1} f(x)dx \le \int_{0}^{1} xf(x)dx \le \int_{\frac{1}{2}}^{1} f(x)dx.
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    This isn't particularly fair, but whatever here goes (they're pretty ):

    Plot r=e^{\cos(\theta)}-2\cos(4\theta)+\sin^7(\theta/12)

    Plot r=\sin(\sin(\theta))-2\sin(4\theta) + \sin^2(\theta/19)
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    (Original post by shamika)
    This isn't particularly fair, but whatever here goes (they're pretty ):

    Plot r=e^{\cos(\theta)}-2\cos(4\theta)+\sin^7(\theta/12)

    Plot r=\sin(\sin(\theta))-2\sin(4\theta) + \sin^2(\theta/19)
    Well, according to Wolfram Alpha:Name:  Polar1.jpg
Views: 130
Size:  20.9 KB and Name:  Polar2.jpg
Views: 132
Size:  19.3 KB
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    (Original post by Zephyr1011)
    Well, according to Wolfram Alpha:Name:  Polar1.jpg
Views: 130
Size:  20.9 KB and Name:  Polar2.jpg
Views: 132
Size:  19.3 KB
    Yep

    The first is called the butterfly curve - the second is my own creation after playing around with it a bit
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    (Original post by shamika)
    Yep

    The first is called the butterfly curve - the second is my own creation after playing around with it a bit
    What do you call this then?
    Spoiler:
    Show

    Let


    \displaystyle w=3\sqrt{1-\left(\frac{x}{7}\right)^{3}}


    \displaystyle l=\frac{1}{2}\left(x+3\right)-\frac{3}{7}\sqrt{10}\sqrt{4-\left(x+1\right)^{2}}+\frac{6}{7  }\sqrt{10}


    \displaystyle h=\frac{1}{2}\left[f\left(\left|x+\frac{1}{2}\right  |+\left|x-\frac{1}{2}\right|+6\right)-11 \left( x+\frac{3}{4}\right)+\left|x-\frac{3}{4} \right| \right]


    \displaystyle r=\frac{1}{2}\left(3-x\right)-\frac{3}{7}\sqrt{10}\sqrt{4-\left(x-1\right)^{2}}+\frac{6}{7}\sqrt{1  0}


    and the functions


    \displaystyle H\left(x\right)=\begin{cases}

\begin{array}{l}

0\\

\frac{1}{2}\\

1

\end{array} & \begin{array}{r}

x<0\\

x=0\\

x>0

\end{array}\end{cases}


    \displaystyle f\left(x\right)=\left(x-l\right)H\left(x+1\right)+\left(  r-h\right)H\left(x-1\right)+\left(l-w\right)H\left(x+3\right)+\left(  w-r\right)H\left(x-3\right)+w


    \displaystyle g\left(x\right)=\frac{1}{2}\left[\left|\frac{x}{2}\right|+\sqrt{1-\left(\left|\left|x\right|-2\right|-1\right)^{2}}-\frac{1}{112}\left(3\sqrt{33}-7\right)x^{2}+w-3\right]\left(\frac{x+4}{\left|x+4\right  |}-\frac{x-4}{\left|x-4\right|}\right)-w


    Plot f and g on the same graph.

    Spoiler:
    Show

    Alternatively, if parametric graphs are more your cup of tea:

    Let

    \displaystyle s\left(z\right)= \sqrt{ \frac{ \left| z \right| }{z}}

    \displaystyle a=\left(\left|x\right|-1\right)\left(\left|x\right|-0.75\right)

    \displaystyle b=\left(0.75-\left|x\right|\right)\left(\left  |x\right|-0.5\right)

    \displaystyle c=\left(0.5-x\right)\left(0.5+x\right)

    Plot where

      \displaystyle  \left[   \left(  \frac { x }  { 7 }   \right)  ^  { 2 }  s  \left(  \left|  x  \right|  - 3  \right)  +  \left( \frac { y }  { 3 }  \right) ^ { 2 } s \left(y+ \frac { 3 \sqrt { 33 }  }  { 7 }  \right) -1 \right] \times   \displaystyle  \left[  \left|  \frac { x }  { 2 }  \right| - \left( \frac { 3 \sqrt { 33 } -7 }  { 112 }  \right) x^ { 2 } -3+ \sqrt { 1- \left( \left|  \left| x \right| -2 \right| -1 \right) ^ { 2 }  } -y \right] \times

    \displaystyle \left[ 9s \left(a \right) -8 \left| x \right| -y \right] \left[ 3 \left| x \right| +0.75s \left(b \right) -y \right]  \left[ 2.25s \left(c \right) -y \right] \times

      \displaystyle  \left[  \frac { 6 \sqrt { 10 }  }  { 7 } + \left(1.5-0.5 \left| x \right|  \right) s \left( \left| x \right| -1 \right) - \frac { 6 \sqrt { 10 }  }  { 14 }  \sqrt { 4- \left( \left| x \right| -1 \right) ^ { 2 }  } -y \right] =0

    Spoiler:
    Show

    You can tell I'm bored. :coffee:


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    (Original post by ukdragon37)
    What do you call this then?
    Spoiler:
    Show

    Let


    \displaystyle w=3\sqrt{1-\left(\frac{x}{7}\right)^{3}}


    \displaystyle l=\frac{1}{2}\left(x+3\right)-\frac{3}{7}\sqrt{10}\sqrt{4-\left(x+1\right)^{2}}+\frac{6}{7  }\sqrt{10}


    \displaystyle h=\frac{1}{2}\left[f\left(\left|x+\frac{1}{2}\right  |+\left|x-\frac{1}{2}\right|+6\right)-11 \left( x+\frac{3}{4}\right)+\left|x-\frac{3}{4} \right| \right]


    \displaystyle r=\frac{1}{2}\left(3-x\right)-\frac{3}{7}\sqrt{10}\sqrt{4-\left(x-1\right)^{2}}+\frac{6}{7}\sqrt{1  0}


    and the functions


    \displaystyle H\left(x\right)=\begin{cases}

\begin{array}{l}

0\\

\frac{1}{2}\\

1

\end{array} & \begin{array}{r}

x<0\\

x=0\\

x>0

\end{array}\end{cases}


    \displaystyle f\left(x\right)=\left(x-l\right)H\left(x+1\right)+\left(  r-h\right)H\left(x-1\right)+\left(l-w\right)H\left(x+3\right)+\left(  w-r\right)H\left(x-3\right)+w


    \displaystyle g\left(x\right)=\frac{1}{2}\left[\left|\frac{x}{2}\right|+\sqrt{1-\left(\left|\left|x\right|-2\right|-1\right)^{2}}-\frac{1}{112}\left(3\sqrt{33}-7\right)x^{2}+w-3\right]\left(\frac{x+4}{\left|x+4\right  |}-\frac{x-4}{\left|x-4\right|}\right)-w


    Plot f and g on the same graph.

    Spoiler:
    Show

    Alternatively, if parametric graphs are more your cup of tea:

    Let

    \displaystyle s\left(z\right)= \sqrt{ \frac{ \left| z \right| }{z}}

    \displaystyle a=\left(\left|x\right|-1\right)\left(\left|x\right|-0.75\right)

    \displaystyle b=\left(0.75-\left|x\right|\right)\left(\left  |x\right|-0.5\right)

    \displaystyle c=\left(0.5-x\right)\left(0.5+x\right)

    Plot where

      \displaystyle  \left[   \left(  \frac { x }  { 7 }   \right)  ^  { 2 }  s  \left(  \left|  x  \right|  - 3  \right)  +  \left( \frac { y }  { 3 }  \right) ^ { 2 } s \left(y+ \frac { 3 \sqrt { 33 }  }  { 7 }  \right) -1 \right] \times   \displaystyle  \left[  \left|  \frac { x }  { 2 }  \right| - \left( \frac { 3 \sqrt { 33 } -7 }  { 112 }  \right) x^ { 2 } -3+ \sqrt { 1- \left( \left|  \left| x \right| -2 \right| -1 \right) ^ { 2 }  } -y \right] \times

    \displaystyle \left[ 9s \left(a \right) -8 \left| x \right| -y \right] \left[ 3 \left| x \right| +0.75s \left(b \right) -y \right]  \left[ 2.25s \left(c \right) -y \right] \times

      \displaystyle  \left[  \frac { 6 \sqrt { 10 }  }  { 7 } + \left(1.5-0.5 \left| x \right|  \right) s \left( \left| x \right| -1 \right) - \frac { 6 \sqrt { 10 }  }  { 14 }  \sqrt { 4- \left( \left| x \right| -1 \right) ^ { 2 }  } -y \right] =0

    Spoiler:
    Show

    You can tell I'm bored. :coffee:


    LOL! At least you could stick my curve into Wolfram Alpha easily (which is what I was intending )

    What is it?
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    (Original post by shamika)
    Some of these questions are getting a little silly (for the intended audience). Laplace transforms? Legendre symbols? These are things encountered in the second or third year of a good university. Having said that, if people are happy and able to solve them, by all means be my guest
    I find it telling how the OP hasn't submitted a solution since about question 50, and we're now past question 120. This thread stopped being interesting (Or even accessible) for the vast majority of people quite a long time ago and now it seems very much to just be a battle of wits between Mladenov and Lord of the Flies.

    (If this comes across as jealous, then that's probably because it is - I'd love to be able to even attempt the questions that are being thrown around lately! :lol: Unfortunately, the questions that are being thrown out are aimed at a very small audience indeed and just looking at the list in the OP it's very easy to see. It reminds me very strongly of what the STEP thread was like before Christmas.)
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    (Original post by DJMayes)
    I find it telling how the OP hasn't submitted a solution since about question 50, and we're now past question 120. This thread stopped being interesting (Or even accessible) for the vast majority of people quite a long time ago and now it seems very much to just be a battle of wits between Mladenov and Lord of the Flies.

    (If this comes across as jealous, then that's probably because it is - I'd love to be able to even attempt the questions that are being thrown around lately! :lol: Unfortunately, the questions that are being thrown out are aimed at a very small audience indeed and just looking at the list in the OP it's very easy to see. It reminds me very strongly of what the STEP thread was like before Christmas.)
    France vs. Bulgaria.
 
 
 
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