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    (Original post by j.alexanderh)
    Solution 7:

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     2^a + p^b = 19^a  \Rightarrow p^b = 19^a - 2^a

\Rightarrow p^b = (19-2)(19^{a-1}+ \cdots + 2^{a-1})

\Rightarrow p=17
    since p is prime.

    Now

     2^a + 17^b = (2+17)^a

    If a=1 we have the solution (1,1,17)

    It is easily verified that there is no solution if a=2

    If a \geq 3,
    (2+17)^a = 17(17k+2^{a-1})+2^a

\Rightarrow 17^b = 17(17k + 2^j)

    which clearly has no solutions.

    Hence the only solution is (1,1,17)
    Haven't seen you around in a while! :hi: How have you been?
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    Problem 9**

    Show that there is no function f:\;\mathbb{N}\to\mathbb{N} satisfying f(f(n))=n+2013
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    (Original post by und)
    Haven't seen you around in a while! :hi: How have you been?
    I'm pretty good, thank you. Et toi?

    Problem 10*:

    Evaluate \displaystyle\int^\frac{\pi}{2}_  0 \frac {\sin^{17}x}{ \sin^{17}x + \cos^{17}x}\ dx
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    Solution 10

    \displaystyle x\to \frac{\pi}{2}-x\Rightarrow \int_0^{\frac{\pi}{2}} \frac{\sin^{17} x\,dx}{\sin^{17} x+\cos^{17} x}= \int_0^{\frac{\pi}{2}} \frac{\cos^{17} x\,dx}{\sin^{17} x+\cos^{17} x}

    Hence \displaystyle I=\frac{1}{2}\int_0^{\frac{\pi}{  2}} dx=\frac{\pi}{4}
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    (Original post by Lord of the Flies)
    Solution 10

    \displaystyle x\to \frac{\pi}{2}-x\Rightarrow \int_0^{\frac{\pi}{2}} \frac{\sin^{17} x\,dx}{\sin^{17} x+\cos^{17} x}= \int_0^{\frac{\pi}{2}} \frac{\cos^{17} x\,dx}{\sin^{17} x+\cos^{17} x}

    Hence \displaystyle I=\frac{1}{2}\int_0^{\frac{\pi}{  2}} dx=\frac{\pi}{4}
    Dat substitution :sexface:
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    Solution 8

    Let m run from 2 to n+1 and p_i is the i^{\text{th}} prime number (i<p_i).

    Consider N_m=\big(p_1\cdots p_{n}\big)^{q+1}+m where q is the highest power occuring in the prime factorisations of all possible values of m

    When m is not a prime power N_m has a factor of the form p_i\cdots p_j where p_i,\cdots p_j are the primes in the prime fact. of m. When m is a prime power we need to check that N_m=p_k^a(P+1) is not a power of p_k. This is clearly true since P is a multiple of p_k and thus P+1 is not
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    (Original post by Lord of the Flies)
    Problem 9**

    Show that there is no function f:\;\mathbb{N}\to\mathbb{N} satisfying f(f(n))=n+2013
    Solution 9

    Looking at the sets S = \{f(n) : n \in \mathbb{N}\} and T = f(f(n)) = \{n+2013 : n \in \mathbb{N}\} clearly T \subset S \subset \mathbb{N}. If such an f exists, it is injective and is also bijective between the sets \mathbb{N} \backslash S and S \backslash T so (\mathbb{N} \backslash S) \cup (S \backslash T) = \mathbb{N} \backslash T contains an even number of elements, which is contradicted by the fact the cardinality of \mathbb{N} \backslash T is 2013
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    (Original post by Mladenov)
    Let me try again.

    Solution 6

    Firstly, \displaystyle f(F) = \sqcup\left\{f^{n+1}(1)|n \in \mathbb{N}\cup\{0\}\right\}, and f(1)\ge 1 lead to f(F)=F
    Secondly, by induction, we obtain \displaystyle f^{n}(1)\le f^{n+1}(1). Consequently, \displaystyle \left\{f^n(1)|n \in \mathbb{N}\cup\{0\}\right\} is a directed complete partial order, and its supremum is obviously F.
    Excellent. However the steps after this point can also be completed (more cleanly I think) without the use of contradiction.

    Consider any N \in \mathbb{N}_\omega satisfying f(N) = N. Since 1 \leq N we have  f(1) \leq f(N) = N. Hence by induction  f^n(1) \leq N for any natural n and therefore any N satisfying f(N) = N is an upper bound for the CPO \displaystyle \left\{f^n(1)|n \in \mathbb{N}\cup\{0\}\right\}. However out of all of them F is the supremum of the CPO, so by definition the smallest.

    The point of the question was to get you to prove an instance of Kleene's fixed-point theorem.
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    (Original post by Lord of the Flies)
    Solution 10

    \displaystyle x\to \frac{\pi}{2}-x\Rightarrow \int_0^{\frac{\pi}{2}} \frac{\sin^{17} x\,dx}{\sin^{17} x+\cos^{17} x}= \int_0^{\frac{\pi}{2}} \frac{\cos^{17} x\,dx}{\sin^{17} x+\cos^{17} x}

    Hence \displaystyle I=\frac{1}{2}\int_0^{\frac{\pi}{  2}} dx=\frac{\pi}{4}
    Concise, as usual.
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    (Original post by Noble.)
    ...
    Nail on the head!
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    Am I allowed to bombard this thread with problems?
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    Problem 11* (if you have never seen substitution before, **)

    Find the general solution, by a suitable substitution or otherwise of this differential equation.

    \displaystyle \dfrac{dy}{dx}=\dfrac{sinx(cosx+  y)}{cosx-y}
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    (Original post by shamika)
    Am I allowed to bombard this thread with problems?
    Yes, go right ahead!

    (Original post by j.alexanderh)
    I'm pretty good, thank you. Et toi?
    I'm good. Applying for maths at university I take it?
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    Problem 12*

    I have a cube and I want to paint each side a different colour. How many different ways can I paint the cube with n colours, n \in \mathbb{N} , n\geq 6 ?
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    (Original post by Star-girl)
    Problem 12*

    I have a cube and I want to paint each side a different colour. How many different ways can I paint the cube with n colours?
    Solution 12

    First we choose the six colours, so there are n choose 6 possibilities =\frac{n!}{6!(n-6)!}. Without considering identical cases we have 6! possibilities for colouring the cube, but we divide by 4*6=24 where 6 is the number of possible anchor faces and 4 is due to rotation about the anchor face, so we get \frac{n!}{24(n-6)!}
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    (Original post by Star-girl)
    Problem 12*

    I have a cube and I want to paint each side a different colour. How many different ways can I paint the cube with n colours?
    A cube has 6 sides and there are 24 ways to orientate a cube (each one of six faces can be placed and each placement can be rotated about 0, 90, 180 or 270 degrees). So the number of distinct cubes that are possible if we consider each side to be different is 6!/24 = 30. Given n colours, we must choose 6 (one for each face). So overall we have 30*nC6

    (I assumed that n>6).

    Problem 13*

    Find the exact value of:
    \sqrt{42+\sqrt{42+\sqrt{42+...}}  }
    As an extention, try the above with 6 instead of 42. Also try 2 and then try 8. Can you spot a pattern and/or find a way to spot more numbers that "work"?
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    Solution 13

    x=\sqrt{n+\sqrt{n+\cdots }}\Rightarrow x=\sqrt{n+x}\Rightarrow x^2-x-n=0 which has solutions for all positive n (pick the positive one)

    The required solution is an integer when n=k(k-1) (the solution is then k).
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    Problem 14**

    Prove that 2^{60}-1 is divisible by 61.
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    (Original post by Star-girl)
    Problem 14**

    Prove that 2^{60}-1 is divisible by 61.
    Solution 14

    As 61 is prime and 2 is not divisible by 61

    2^{60}\equiv 1 \pmod{61} by Fermat's Little Theorem. Hence:

    2^{60} - 1\equiv 0 \pmod{61}
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    Problem 15*/**

    Evaluate \displaystyle \lim_{n\to\infty} \int_1^{n} \frac{n}{1+x^n}\,dx

    What if the upper limit in the integral were fixed, say \pi?
 
 
 
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