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    Problem 358*

    Find all positive integers n such that the product of all positive divisors of n is 24^{240}.
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    (Original post by Felix Felicis)
    Problem 385*

    Find all positive integers n such that the product of all positive divisors of n is 24^{240}.
    I am using Dalek's method of periphrastic divergences and Euler-taradiddles to say that;

    We have 24^{240}

    If we take the base,  24 and multiply it by  2 we have  48 .

    Now do the power divided by the base multiplied by two:

     \frac{240}{48} = 5

     5 is now our new exponent after considering the periphrastic divergence.

    Therefore, the new exponent is the value computed above by the Euler-taraddidle method:

     n = 24^5 .
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    (Original post by Zakee)
    I am using Dalek's method of periphrastic divergences and Euler-taradiddles to say that;

    We have 24^{240}

    If we take the base,  24 and multiply it by  2 we have  48 .

    Now do the power divided by the base multiplied by two:

     \frac{240}{48} = 5

     5 is now our new exponent after considering the periphrastic divergence.

    Therefore, the new exponent is the value computed above by the Euler-taraddidle method:

     n = 24^5 .
    :rofl2:

    You've forgotten the lateral solutions though! :shock:

    24^5 = 2\cdot4^5 = 244444.

    Simples.
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    (Original post by Felix Felicis)
    Problem 385*

    Find all positive integers n such that the product of all positive divisors of n is 24^{240}.
    Solution 385*


    The product \pi(n) of all positive divisors of n is \displaystyle n^{\frac{\sigma (n)}{2}}, where \sigma (n) is the divisor function.

    Thus, if n satisfies this condition, then 24^{240} = 2^{720} \cdot 3^{240} must be a power of n.
    Thus, n = 2^{a} \cdot 3^{b}, where ak = 720 and bk = 240, for some integer k.
    Thus, a = 3b, n = 24^b and b|240 = 2^4 \cdot 3 \cdot 5 .
    So b \in \{ 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 40, 48, 60, 80, 120, 240 \}
    Thus, n \in \{ 24, 24^2, 24^3, 24^4, 24^5, 24^6, 24^8, 24^{40}, 24^{48}, 24^{60}, 24^{80}, 24^{120}, 24^{240} \} =: A
    By the multiplicativity of \sigma, since 24^n = 2^{3n} \cdot 3^n, and if m=p^k for some prime p, then \sigma(m) = k + 1, we have that \sigma(24^n) = (3n+1)(n+1).

    If we let f(n) = \pi(24^n) = 24^{\frac{n}{2}(3n+1)(n+1)}, then f is strictly increasing on the naturals, and thus f(n) = 24^{240} has at most one solution. n = 5 is that solution.
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    (Original post by Felix Felicis)
    :rofl2:

    You've forgotten the lateral solutions though! :shock:

    24^5 = 2\cdot4^5 = 244444.

    Simples.

    Oh, I'm truly sorry. I haven't been doing much Mathematics recently, especially Dalekian Mathematics. I'll learn more some day.
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    Solution 355

    Let Q_m \subseteq S be such that |Q_m|=m, the claim is that if m>n then Q_m does not have property P. Let A_k=S\cap \{2^q(2k-1):\;q\in\mathbb{N}_0\} for 1\leq k\leq n (so \bigcup A_k = S), then by the pigeonhole principle there exist p,q\in Q_{m} such that \{p,q\}\in A_k for some k. And of course, Q_n=\{n,n+1,\cdots 2n-1\} has property P

    Solution 356

    Have I completely and utterly missed something here? If for any i,j we have |A_i|\equiv |A_i\cap A_j|\equiv 0\pod{2} we may group the population into pairs and call the set of these T. Then obviously the number of clubs \leq \mathcal{P}(T)=2^n. Equality is only possible if we allow A_i=\varnothing for some i.
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    (Original post by Lord of the Flies)
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    Solution 355

    Let Q_m \subseteq S be such that |Q_m|=m, the claim is that if m>n then Q_m does not have property P. Let A_k=S\cap \{2^q(2k-1):\;q\in\mathbb{N}_0\} for 1\leq k\leq n (so \bigcup A_k = S), then by the pigeonhole principle there exist p,q\in Q_{m} such that \{p,q\}\in A_k for some k. And of course, Q_n=\{n,n+1,\cdots 2n-1\} has property P

    Solution 356

    Have I completely and utterly missed something here? If for any i,j we have |A_i|\equiv |A_i\cap A_j|\equiv 0\pod{2} we may group the population into pairs and call the set of these T. Then obviously the number of clubs \leq \mathcal{P}(T)=2^n. Equality is only possible if we allow A_i=\varnothing for some i.
    It's fine.

    Problem 359**

    In a town there are n people and m clubs - A_{1},\cdots, A_{m}. If |A_{i}| is odd for all i and |A_{i} \cap A_{j}| is even for all i,j \in \{1,\cdots,m\}, then m \le n. Is the bound attainable?

    Problem 360**

    Let A_{i} \subseteq \{1,\cdots,n \}, i \in \{1,\cdots,m \} be such that for i \not= j the inequality |A_{i} \cap A_{j}| \le 1 holds. Show that \displaystyle m \le n+1+ \dbinom{n}{2}.

    Problem 361**

    Let a,A,B be integers, M - a positive integer. Prove that the numbers Aa^{m}+Bm \pmod M form a complete residue system \pmod M if and only if \gcd(B,M)=1.
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    Did anyone ever get an answer to that horrible integral a while back? I don't recall seeing one.
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    Problem 362*

    This is actually very similar to an A level question but without the added guidance that they usually provide.

    A sports association is planning to construct a running track in the shape of a rectangle surmounted by a semicircle. Let the letter x represents the length of the rectangular section and r represents the radius of the semicircle.
    Determine the length, x, that maximises the area enclosed by the track.

    EDIT

    The perimeter of the track must be 600 metres.

    I completely forgot to put that in. Sorry

    Problem 363*

    Given that the sum of the angles A, B and C of a triangle is \pi radians, show that \sin(A) = \sin(B+C)

    Problem 364***

    Determine the extreme values of

    f(x,y,z,u,v,w) = \frac{1}{1+x+u} + \frac{1}{1+y+v} + \frac{1}{1+z+w}


    where xyz = a^3, uvw = b^3, and x,y,z,u,v,w>0.
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    (Original post by Arieisit)
    Problem 363*

    Given that the sum of the angles A, B and C of a triangle is \pi radians, show that \sin(A) = \sin(B+C)
    Spoiler:
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    Triangle, so c+B=180-A, for any x, \sin(x)=\sin(180-x)
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    (Original post by The nameless one)
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    Triangle, so c+B=180-A, for any x, \sin(x)=\sin(180-x)
    Is that all? :holmes:

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    (Original post by Arieisit)
    Is that all? :holmes:

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    well, almost
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    I am known for being lazy, and that is why i just skipped the previous question, cannot be bothered to read it
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    (Original post by Arieisit)

    Problem 364***

    Determine the extreme values of

    f(x,y,z,u,v,w) = \frac{1}{1+x+u} + \frac{1}{1+y+v} + \frac{1}{1+z+w}


    where xyz = a^3, uvw = b^3, and x,y,z,u,v,w>0.
    Spoiler:
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    Is Lagrange the way to go here or am I barking up the wrong tree?
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    (Original post by james22)
    Did anyone ever get an answer to that horrible integral a while back? I don't recall seeing one.
    Do you mean the cos(tan(x))/x one? (or something like that)

    If so, no solution was posted. I feel I made a tiny bit of progress though hit a wall pretty quickly. I can post my "progress" if you want to try and get the ball rolling again?
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    (Original post by CD315)
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    Is Lagrange the way to go here or am I barking up the wrong tree?
    You may be on to something.

    I don't like giving hints if I didn't see any progress.

    Posted from TSR Mobile
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    (Original post by Arieisit)
    Problem 362*

    This is actually very similar to an A level question but without the added guidance that they usually provide.

    A sports association is planning to construct a running track in the shape of a rectangle surmounted by a semicircle. Let the letter x represents the length of the rectangular section and r represents the radius of the semicircle.
    Determine the length, x, that maximises the area enclosed by the track.

    I must have misunderstood the question.

    The area enclosed is A = 2rx+\pi r^2 this is maximised with respect to x when you differentiate with respect to x and equate it to 0.
    0 = 2r As this only gives a value of the minimum, this implies the maximum will increase as x increases.


    Problem 363*

    Given that the sum of the angles A, B and C of a triangle is \pi radians, show that \sin(A) = \sin(B+C)

    This is very simple I think.
    \sin(A) = Im(e^{iA}) eq 1
    \sin(B+C) = Im(e^{i(B+C)}) eq 2
    A+B+C=\pi
    so:
    A=\pi - B - C eq 3
    sub eq 3 into eq 1:
    \sin(A) = Im(e^{i(\pi -B - C)})
    Therefore:
    \sin(A) = Im(e^{i\pi}e^{-i(B + C)})
    e^{i\pi} = -1
    So:
    \sin(A) = Im(-e^{-i(B + C)})
    Sin is odd so:
    \sin(A) = Im(e^{i(B + C)})
    We have written this equivalently to eq 2.
    Thus:
    \sin(A) = \sin(B+C)


    Problem 364***

    Determine the extreme values of

    f(x,y,z,u,v,w) = \frac{1}{1+x+u} + \frac{1}{1+y+v} + \frac{1}{1+z+w}


    where xyz = a^3, uvw = b^3, and x,y,z,u,v,w>0.
    I liked that middle question can't be bothered with the last one.
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    I've been generalising random expressions and thought you lot might like this one.

    Problem 365
    **/***

    Find a closed form expression for \displaystyle \int_0^{\infty} \frac{x^a}{(b+x^c)^d} \ dx in terms of elementary functions where d \in \mathbb{Z}.

    Be sure to consider the different cases that arise according to the values of a, b and c.
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    (Original post by Arieisit)
    Problem 363*

    Given that the sum of the angles A, B and C of a triangle is \pi radians, show that \sin(A) = \sin(B+C)
    Solution 363

    A + B + C = \pi

    \Rightarrow B + C = \pi - A

    \begin{aligned} \Rightarrow \sin (B+C) & = \sin (\pi - A) \\ & = \sin \pi \cos A - \sin A \cos \pi \\ & = - \sin (A) \times (-1) \\ & = \sin (A) \end{aligned}

    My first solution of the thread! :woo:
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    (Original post by Jkn)
    Do you mean the cos(tan(x))/x one? (or something like that)

    If so, no solution was posted. I feel I made a tiny bit of progress though hit a wall pretty quickly. I can post my "progress" if you want to try and get the ball rolling again?
    That's the one I meant, I'm curious what the solution is because I could not find any way to make simpler. I managed to convert it to several different integrals but they didn't look any nicer.
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    (Original post by k9markiii)
    I liked that middle question can't be bothered with the last one.
    I absolutely love your solution for question 363! When I did it, it was in no way similar to your version. Nice perspective! :yy:

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