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    (Original post by Smaug123)
    Kind of, but more flexible - I'm visualising something like the roots of a tree, stretching down, but sometimes meeting up and joining. Not just a straight binary tree, no meeting-and-joining.
    So you want a dag?
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    (Original post by ukdragon37)
    So you want a dag?
    For just a moment there, I thought you were talking about:

    http://www.youtube.com/watch?v=zH64dlgyydM
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    (Original post by ukdragon37)
    So you want a dag?
    That's the one - I've never done any formal graph theory, and my only graph-theoretic knowledge is "that which can easily be done in Mathematica" :P I'd forgotten the term "directed acyclic", although actually it comes up fairly often.
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    Solution 280

    \displaystyle \begin{aligned} \frac{d}{dx} \text{arsinh} x = (1+x^2)^{- \frac{1}{2}} & = 1 + \left( - \tfrac{1}{2} \right) x^2 + \frac{ \left( - \frac{1}{2} \right) \left( - \frac{3}{2} \right) (x^2)^2}{2!} + \cdots \quad (\text{binomial expansion}) \\ & = 1 + \sum_{r=1}^{\infty} \left( \frac{(-1)^{r} x^{2r}}{r! \cdot 2^{r}} \prod_{i=0}^{r-1} (2i + 1) \right) \\ & = 1 + \sum_{r=1}^{\infty} \left( \frac{(-1)^{r} x^{2r}}{r! \cdot 2^{r}} \cdot \frac{(2r-1)!}{(r-1)! \cdot 2^{r-1}} \right) \\ & = 1+ \sum_{r=1}^{\infty} \left( \frac{(-1)^{r} x^{2r}}{4^{r}} \cdot \frac{2 \cdot (2r)!}{2r \cdot r! \cdot (r-1)!} \right)\\ & = \sum_{r=0}^{\infty} \frac{(-1)^{r} x^{2r}}{4^{r}} \cdot \frac{(2r)!}{(r!)^{2}}  \\ \implies \text{arsinh} x & = \int \sum_{r=0}^{\infty} \frac{(-1)^{r} x^{2r}}{4^{r}} \cdot \frac{(2r)!}{(r!)^{2}}  \ dx \\ & = \boxed{\sum_{r=0}^{\infty} \frac{(-1)^{r} (2r)! x^{2r+1}}{4^{r}(2r+1)(r!)^2}  + \mathcal{C}} \end{aligned}

    Subbing in x = 0 gives \mathcal{C} = 0, which yields the desired series.

    I can't quite justify the interval of convergence, though :confused: I can show that it converges if |x| < 1 by the ratio test but not sure how to when |x| = 1
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    (Original post by Felix Felicis)
    Solution 280
    Nice stuff, bro!
    I can't quite justify the interval of convergence, though :confused: I can show that it converges if |x| < 1 by the ratio test but not sure how to when |x| = 1
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    Clearly it is sufficient to prove convergence in the case x=-1. Try arranging the series into a different form and then comparing it to a known series.
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    Problem 296**

    Find
    \int_0^{\pi} e^{\cos(x)}\cos(\sin(x)) dx

    Quite a pretty one.

    Spoiler:
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    I wonder if \int_{-\pi}^{\pi} e^{e^{e^{ix}}} dx is doable.
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    (Original post by henpen)
    Problem 296**

    Find
    \int_0^{\pi} e^{\cos(x)}\cos(\sin(x)) dx

    Quite a pretty one.

    Spoiler:
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    I wonder if \int_{-\pi}^{\pi} e^{e^{e^{ix}}} dx is doable.
    Solution 296:
    I=\displaystyle\int_0^{\pi} e^{\cos(x)}\cos(\sin(x)) dx = \Re \left(\displaystyle\int_0^{\pi} e^{e^{ix}} \ dx \right)

\Rightarrow I= \Re \left(\displaystyle\int_0^{\pi} \displaystyle\sum_{r=0}^{\infty} \dfrac{(e^{ix})^r}{r!} \ dx \right)=\Re \left(\displaystyle\int_0^{\pi} 1+\dfrac{e^{ix}}{1!}+\dfrac{e^{2  ix}}{2!}+. . . +\dfrac{e^{ikx}}{k!}+. . . \right) \ dx

\Rightarrow I =\Re \left[x+\dfrac{e^{ix}}{i \times 1!}+\dfrac{e^{2ix}}{2i \times 2!}+. .  .+ \dfrac{e^{ikx}}{ki\times k!}+ . . .\right]^{\pi}_0

\Rightarrow I=\Re \left[x+\dfrac{1}{i}\displaystyle\sum_  {r=1}^{\infty}\dfrac{(e^{ix})^r}  {r\times r!}\right]^{\pi}_0

\Rightarrow I=\Re \left( \left( \pi-i \displaystyle\sum_{r=1}^{\infty} \dfrac{(e^{i\pi})^r}{r\times r!}\right)-\left(0-i\displaystyle\sum_{r=1}^{\infty  } \dfrac{(e^{i0})^r}{r\times r!}\right) \right)

\Rightarrow I =\Re \left( \pi- i \left( \displaystyle\sum_{r=1}^{\infty} \dfrac{(-1)^r}{r\times r!} -  \displaystyle\sum_{r=1}^{\infty} \dfrac{(1)^r}{r\times r!}\right) \right)

\Rightarrow I=\pi
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    Problem 297***

    Calculate the volume of the ellipsoid defined by

     V: \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} \leq 1

    Where a,b,c are positive constants.
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    (Original post by henpen)
    I wonder if \int_{-\pi}^{\pi} e^{e^{e^{ix}}} dx is doable.
    (Original post by joostan)
    x
    Interesting stuff guys!

    I have been able to do the first few cases of the generalisation and a pattern has emerged. I thought I had been able to prove the result below but, whilst Wolfram agrees with me for n=2,3,4 it doesn't for n=5.

    Essentially \displaystyle \Re \int_0^{\pi} e^{e^{e^{ix}}} \ dx = \pi e and \displaystyle \Re \int_0^{\pi} e^{e^{e^{e^{ix}}}} \ dx = \pi e^e.

    I thought I had proven the generalisation though clearly I had made some mistake. I even tried n=5 separately and still got \pi e^e^e. Back to the drawing board:

    Problem 298*

    Evaluate \displaystyle \Re \int_0^{\pi} \underbrace{{e^{e^{\cdots}^{ix}}  }}_n \ dx for n \ge 2 where the underbrace represents the number of times e appears in the exponentiation.
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    (Original post by FireGarden)
    Problem 297***

    Calculate the volume of the ellipsoid defined by

     V: \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} \leq 1

    Where a,b,c are positive constants.
    Instead of using 'triple integration', spherical coordinates and/or complicated parametrization, I will derive the result in a way that could have been done by an A-level student.

    Solution 297
    *

    We know that the area enclosed by an ellipse \frac{x^2}{a^2}+\frac{y^2}{b^2}=  1 is \pi ab. The proof is part of the A-level specification (and is of little interest).

    Rearranging the equation of the ellipsoid bounding our volume, we get \frac{x^2}{\left( a \sqrt{1-\frac{z^2}{c^2}} \right)^2}+\frac{y^2}{\left( ab\sqrt{1-\frac{z^2}{c^2}} \right)^2}=1.

    As the volume is a dimesional extension to area as area is to length, it can be easy inferred that \displaystyle V=\int_a^b f(z) \ dz where f(z) is the area of the ellipse contained by the surface generated at each z coordinate and a and b are the boundaries on the volume enclosed by the ellipsoid. These boundaries correspond to z= \pm c, as can be seen by setting x=y=0 in the original equation and, using the above result, we get that f(z) = \pi \left( a \sqrt{1-\frac{z^2}{c^2}} \right) \left( b \sqrt{1-\frac{z^2}{c^2}} \right)=\pi ab \left(1-\frac{z^2}{c^2} \right). This gives:

    \displaystyle V = \pi ab \int_{-c}^c 1-\frac{z^2}{c^2} \ dz =\frac{4}{3} \pi abc \ \square

    Note that this equation has the required a-b-c symmetry as well as the property that it reduces to the volume of a sphere when a=b=c.

    Generalization:


    We can now extend this method to generalise for 'volume' in n-dimensional space. Let the n-dimensional volume be denoted by V_n.

    We define our hypothetical region as the set of points satisfying \displaystyle \sum_{i=1}^n \frac{x_i^2}{r_i^2} \le 1. We will also assume that all solutions can be expressed in the form V_n = k \prod_{i=1}^n r_i where k is a positive constant to be found (this assumption can be justified as valid in a number of ways).

    Using our previous method, we note that the boundary can be expressed as \displaystyle \sum_{i=1}^{n-1} \frac{x_i^2}{\left( r_i \sqrt{1-\frac{x_n^2}{r_n^2}} \right)^2} =1 which gives us the following relation.

    Noting that the function we are integrating is even, letting x \mapsto r_n \sqrt{x}, using Beta functions and finally the Gamma representation of the Beta function:

    \displaystyle \begin{aligned} V_n = \int_{-r_{n}}^{r_{n}} k \prod_{i=1}^{n-1} r_i \sqrt{1-\frac{x_n^2}{r_n^2}} \ dx_n = V_{n-1} \int_{-r_{n}}^{r_{n}} \left(1-\frac{x_n^2}{r_n^2} \right)^{\frac{n-1}{2}} \ dx_n = 2 V_{n-1} \int_0^1 (1-x)^{\frac{n-1}{2}} \frac{r_n dx}{2 \sqrt{x}} \end{aligned}
    \displaystyle = r_n V_{n-1} \int_0^1 x^{- \frac{1}{2}} (1-x)^{\frac{n-1}{2}} \ dx = r_n V_{n-1} B \left(\frac{1}{2} , \frac{n+1}{2} \right) = V_{n-1} \frac{r_n \sqrt{\pi} \Gamma \left( \frac{n+1}{2} \right)}{\Gamma \left(1+\frac{n}{2} \right)}

    Noting that the only sensible way to proceed in to let V_0=1 (again, there are various ways we can justify this claim, such as verifying that it leads to the correct values for V_2 and V_3, which it does):

    \displaystyle \Rightarrow V_n = V_0 \prod_{i=1}^n r_i \frac{\sqrt{\pi} \Gamma \left( \frac{n+1}{2} \right)}{\Gamma \left(1+\frac{n}{2} \right)} = \pi^{n/2} \frac{r_1 r_2 \cdots r_n}{\Gamma \left(1+\frac{n}{2} \right)} \ \square

    Woo! This has to be one of the most beautiful results I have discovered independently!

    Is this a well-known result? Is it correct? What you you call V_n?

    Solution 299 (weak corollary)

    Setting r_1=r_2= \cdots r_n=1 gives the required result of \frac{\pi^{n/2}}{\Gamma \left(1+\frac{n}{2} \right)}.
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    (Original post by Jkn)
    Instead of using 'triple integration', I will derive the result in a way that could have been done by an A-level student.
    How likely are we to accept the following?
    Volume has dimension [L]^3. The only dimensioned quantities we have are a,b,c, each of dimension [L].
    Since the volume clearly depends on a, b and c, and is also symmetric on transposing a,b,c, the answer is obviously f abc, where f is a dimensionless function to be determined.
    But since the only dimensionless quantities are constants and ratios of a,b,c (and we know that f should be invariant on cycling a,b,c, so ratios are out), we have f constant.
    Now, a sphere has a=b=c, and volume 4/3 pi a^3. So f=4/3 pi.

    ETA: Yes, I know that f could be \dfrac{(a+b+c)^2}{a^2+b^2+c^2}, for example, but that's too complicated, and it must be the simpler one.
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    (Original post by Jkn)
    Could you verify/falsify this for me for a few values of n greater than or equal to 5 (on Mathematica) just to put my mind at ease?
    Hmm, I'm not sure Mathematica is agreeing for n=5 - with 1000 digits of precision, it claims that the integral is
    Spoiler:
    Show
    2.814748587862548678050445650343 78650553053524527015318777075354 827928\
    30371377245089302614394292704623 80418879979374304513430442428837 653058\
    41624608864703657720269273518028 54473294562709938722054449126147 174140\
    72850162897807163139976305739619 91772662688059938897174929482584 337521\
    46024533297279126519151045044185 28042192530641575921118846740422 514025\
    52880502659077596731252318257658 69581747592075202571315222503434 314109\
    57488248810598252366055487920295 82783810400699542540650976883269 006880\
    95570952708140728142215404454333 62550599588037179595129037214288 176607\
    42051167660894602290092141250264 57206649676739597221299574147120 814876\
    17950725698627131132985739489683 45423998395693930390611697281265 995161\
    24695244147204643630704755698803 98974608587570173500753575778619 442900\
    36035501528596656715108841001854 37843977582718213581082496803166 157351\
    82349156207907687737409162641325 28719697950723862588315707223213 822585\
    29588073383812512325803443310313 53088986130946151406258352788894 008076\
    424627477783602829528*10^1656469

    That is, 2*10^1656469.
    I'm running it with two million digits of precision - hopefully that'll make sure we're on the right order of magnitude. (For reference, your predicted value is 1.19829*10^7)
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    (Original post by Smaug123)
    How likely are we to accept the following?
    Volume has dimension [L]^3. The only dimensioned quantities we have are a,b,c, each of dimension [L].
    Since the volume clearly depends on a, b and c, and is also symmetric on transposing a,b,c, the answer is obviously f abc, where f is a dimensionless function to be determined.
    But since the only dimensionless quantities are constants and ratios of a,b,c (and we know that f should be invariant on cycling a,b,c, so ratios are out), we have f constant.
    Now, a sphere has a=b=c, and volume 4/3 pi a^3. So f=4/3 pi.

    ETA: Yes, I know that f could be \dfrac{(a+b+c)^2}{a^2+b^2+c^2}, for example, but that's too complicated, and it must be the simpler one.
    I considered that but the symmetry condition and a=b=c evaluation is not sufficient to directly deduce a formula for the volume. Also, I'm certain that it constitutes A-level knowledge as it would easily fit in to a C4 or FP3 textbook as an 'extension' exercise. For example, you plot the graph of area against 3rd co-ordinate, etc.. In fact, I am definitely right as this understanding is required for both M3 and M5.

    Oh btw, I may have just generalised the method to n-dimensions. I had to use beta functions but I have reduced it to a collection of series that, if solved, will yield the necessary recurrence relation between n-dimensional 'volume' and 'n+1'-dimensional 'volume' (hypervolume? no idea..)
    (Original post by Smaug123)
    Hmm, I'm not sure Mathematica is agreeing for n=5 - with 1000 digits of precision, it claims that the integral is
    Spoiler:
    Show
    2.814748587862548678050445650343 78650553053524527015318777075354 827928\
    30371377245089302614394292704623 80418879979374304513430442428837 653058\
    41624608864703657720269273518028 54473294562709938722054449126147 174140\
    72850162897807163139976305739619 91772662688059938897174929482584 337521\
    46024533297279126519151045044185 28042192530641575921118846740422 514025\
    52880502659077596731252318257658 69581747592075202571315222503434 314109\
    57488248810598252366055487920295 82783810400699542540650976883269 006880\
    95570952708140728142215404454333 62550599588037179595129037214288 176607\
    42051167660894602290092141250264 57206649676739597221299574147120 814876\
    17950725698627131132985739489683 45423998395693930390611697281265 995161\
    24695244147204643630704755698803 98974608587570173500753575778619 442900\
    36035501528596656715108841001854 37843977582718213581082496803166 157351\
    82349156207907687737409162641325 28719697950723862588315707223213 822585\
    29588073383812512325803443310313 53088986130946151406258352788894 008076\
    424627477783602829528*10^1656469

    That is, 2*10^1656469.
    I'm running it with two million digits of precision - hopefully that'll make sure we're on the right order of magnitude. (For reference, your predicted value is 1.19829*10^7)
    Ah that's what Wolfram gave. Hmm, that is rather annoying! Then again it is worth noting how spookily close that value is to \pi e^e^e^e (or e^e^e^e for that matter).
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    (Original post by Smaug123)
    How likely are we to accept the following?
    Volume has dimension [L]^3. The only dimensioned quantities we have are a,b,c, each of dimension [L].
    Since the volume clearly depends on a, b and c, and is also symmetric on transposing a,b,c, the answer is obviously f abc, where f is a dimensionless function to be determined.
    But since the only dimensionless quantities are constants and ratios of a,b,c (and we know that f should be invariant on cycling a,b,c, so ratios are out), we have f constant.
    Now, a sphere has a=b=c, and volume 4/3 pi a^3. So f=4/3 pi.

    ETA: Yes, I know that f could be \dfrac{(a+b+c)^2}{a^2+b^2+c^2}, for example, but that's too complicated, and it must be the simpler one.
    There surely is an easier way: set n-1 of the n variables equal to , then the extremal solution for the remaining kth variable will be equal to that dimension's semi-major/minor/in-between axis, r_k. Thus the unit hypersphere was stretched in that dimension by \frac{r_k}{1}. The stretch multiplies the volume by a factor that is independent of the other dimensions' stretching, and the volume of the unit hypersphere is \frac{\pi^{n/2}}{ \Gamma \left(1+\frac{n}{2} \right)} (Problem 299* : prove this), so the volume of the stretched hypersphere is
    \displaystyle \frac{\pi^{n/2}}{ \Gamma \left(1+\frac{n}{2} \right)} \prod_{k=1}^nr_k

    This is probably what you meant or had in mind, but I wrote it here for explicitness.
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    (Original post by henpen)
    This is probably what you meant or had in mind, but I wrote it here for explicitness.
    No, I wasn't being entirely serious - I was doing to the volume of an ellipsoid what we did in lectures to Kepler's Laws
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    (Original post by Smaug123)
    No, I wasn't being entirely serious - I was doing to the volume of an ellipsoid what we did in lectures to Kepler's Laws
    It would be enough for a physicist. Physicists' proofs are quite nice in a way entirely different to mathematical ones.
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    (Original post by henpen)
    It would be enough for a physicist. Physicists' proofs are quite nice in a way entirely different to mathematical ones.
    If you come from the "don't worry, be happy" school of mathematical rigour, as one of my other lecturers said…
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    (Original post by henpen)
    There surely is an easier way: set n-1 of the n variables equal to , then the extremal solution for the remaining kth variable will be equal to that dimension's semi-major/minor/in-between axis, r_k. Thus the unit hypersphere was stretched in that dimension by \frac{r_k}{1}. The stretch multiplies the volume by a factor that is independent of the other dimensions' stretching, and the volume of the unit hypersphere is \frac{\pi^{n/2}}{ \Gamma \left(1+\frac{n}{2} \right)} (Problem 299* : prove this), so the volume of the stretched hypersphere is
    \displaystyle \frac{\pi^{n/2}}{ \Gamma \left(1+\frac{n}{2} \right)} \prod_{k=1}^nr_k

    This is probably what you meant or had in mind, but I wrote it here for explicitness.
    OMG! This follows as a corollary from what I just typed up! (I'll edit it on as solution 299)

    So satisfying to rediscover something! Do you know when my result was established and by who?

    Edit: How did you get the result for a the generalized ellipse btw? I can't find stuff about it anywhere online (did you read it somewhere and, if so, is the original derivation the same as mine?)

    Problem 300
    */***

    Working within the base-10 number system, find all 9-digit numbers containing all digits except zero (a zeroless Pandigital number) such that the first n digits form a number that is divisible by n for all n.

    [For example, 123456789 is of the correct form. We also note that 1 is divisible by 1, 12 is divisible by 2, 123 is divisible by 3 though, as 1234 is not divisible by 4, this number fails to meet the requirements.]

    Can you generalise interestingly to number systems with other bases and/or in any other way? (bonus-only)
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    Solution 300

    Let  abcdefghi be the number such that a is the first digit, b is the second and so forth.

    We have, from basic divisibility criteria the following set of equations:

     \Rightarrow (1) b \equiv 0 \ mod \ 2



\Rightarrow (2) a+b+c \equiv 0 \ mod \ 3



\Rightarrow (3) cd \equiv 0 \ mod \ 4, d \equiv 0 \ mod \ 2 



\Rightarrow(4) e = 5



\Rightarrow (5)  f \equiv 2 \ mod \ 2; a+b+c+d+e+ f \equiv 1 \ mod \ 3 \Rightarrow d+f \equiv 1 \ mod \ 3



\Rightarrow (6) a + 5b+ 4c + 6g + 2e + 3f + g \equiv 0 \ mod \ 7



\Rightarrow (7) fgh \equiv 0 \ mod \ 8

    From, Equations (1), (3), (5), (7), b, d, f, h are even and the rest are odd. e is clearly equal to 5, so the rest of the numbers must be either 1, 3, 7, or 9.

    From (3), and the fact that c is odd, we have  d = 2, 6 .

    As d and f are even, we have  d+f \equiv 0 \ mod \ 2, d+f \equiv 1 \ mod \ 3 .

    \Rightarrow d+f = 4, 10, 16 \Rightarrow d+f = 10 as 4 is too small and 16 is too big to be the sum of distinct even numbers under 10.

    This, coupled with the fact that d is 2 or 6 gives us,  (d,f) = (2,8), (6,4) .

    From (7) and the fact that  f = 4, 8 we get  gh \equiv 0 \ mod \ 8 .

    But, as g is odd, we have  h= 2, 6 . In fact, we have,  (g, h) = (1, 6), (3, 2), (7, 2), (9, 6) .

    This leaves b=2, when h = 6 and b = 6 when h = 2.

    So, as far as the configuration of even integers is concerned we have,

     (b, d, f, h) = (4, 2, 8, 6), (8, 6, 4, 2)

    Now, note that from (3),  b = 4 \Rightarrow a+c \equiv 2 \ mod \ 3 and  b = 8 \Rightarrow a+c \equiv 1 \ mod \ 3 .

    Collating the  (g,h) pairs with the appropriate sets of even integers, we get the following:

     (b, d, f, g, h) = (4, 2, 8, 1, 6), (4, 2, 8, 9, 6), (8, 6, 4, 3, 2), (8, 6, 4, 7, 2) .

    Now, finally, look at (6). This and (2) will help us fix a and c and conclude.

    Noting that  3(f+d) = 30, 2e = 10 , we have,

     a+4c + (5b + 6d + 10 + 3f + g) \equiv 0 \ mod \ 7



\Rightarrow a+4c \equiv 2 - (5b+3d+g) \mod \  7

    Now, remember that, when, (3),  b = 4 \Rightarrow a+c \equiv 2 \ mod \ 3 and  b = 8 \Rightarrow a+c \equiv 1 \ mod \ 3 .

    So, notice that,

     (b, d, f, g, h) = (4, 2, 8, 1, 6) \Rightarrow a+ 4c \equiv 3 \ mod \ 7, a+c \equiv 2 \ mod \ 3



 (b, d, f, g, h)  =  (4, 2, 8, 9, 6) \Rightarrow a+ 4c \equiv 2 \ mod \ 7,  a+c \equiv 2 \ mod \ 3



 (b, d, f, g, h)  = (8, 6, 4, 3, 2) \Rightarrow  a+ 4c \equiv 4 \ mod \ 7,  a+c \equiv 1 \ mod \ 3



 (b, d, f, g, h)  = (8, 6, 4, 7, 2) \Rightarrow  a+ 4c \equiv 0 \ mod \ 7,  a+c \equiv 1 \ mod \ 3

    So, we have 4 sets of congruences. Notice that a and c are odd, so, we only have the integers  \left [ 1, 3, 7, 9 \right ] to choose from.

    But, note that for each set of congruences, one of these has already been used up by  g . So, one can only consider the other 3, narrowing down the bunch of solutions.

    In fact, after some computation, it turns out that only the last set of equations is solvable with  a= 3, c= 1 .

    So, that leaves  i = 9 as the only thing left.

    And gives us with just such an integer we want:

     abcdefghi = 381654729 .

    A quick check reveals this to be acceptable.
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    (Original post by Jkn)
    Edit: How did you get the result for a the generalized ellipse btw? I can't find stuff about it anywhere online (did you read it somewhere and, if so, is the original derivation the same as mine?
    )
    I just came across this. There's an off reference in the section titled "Integration Problems." Read the last paragraph.

    http://en.wikipedia.org/wiki/Gamma_function
 
 
 
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