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Math Help - [SOLVED] Show that p is divisible by 641?

  1. #1
    Super Member fardeen_gen's Avatar
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    [SOLVED] Show that p is divisible by 641?

    If p and q are positive integers such that \frac{p}{q} = 1 + \frac{1}{2} - \frac{2}{3} + \frac{1}{4} + \frac{1}{5} - \frac{2}{6} + \frac{1}{7} + \frac{1}{8} - \frac{2}{9} + \mbox{...} + \frac{1}{478} + \frac{1}{479} - \frac{2}{480}, show that p is divisible by 641.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by fardeen_gen View Post
    If p and q are positive integers such that \frac{p}{q} = 1 + \frac{1}{2} - \frac{2}{3} + \frac{1}{4} + \frac{1}{5} - \frac{2}{6} + \frac{1}{7} + \frac{1}{8} - \frac{2}{9} + \mbox{...} + \frac{1}{478} + \frac{1}{479} - \frac{2}{480}, show that p is divisible by 641.
    The last three terms above are not consistent with those before the ellipsis, can you give the expression for the general term in this summation.

    CB
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  3. #3
    Super Member fardeen_gen's Avatar
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    Even I couldn't find the general term. Thats why I posted.
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    The last three terms above are not consistent with those before the ellipsis, can you give the expression for the general term in this summation.

    CB

    Yes they are as 3 divides 480 properly, 3.160 = 480. I believe the series is,

    \displaystyle\sum_{i=1}^{160} 1/(3i-2) +<br />
 \displaystyle\sum_{i=1}^{160} 1/(3i-1) - \displaystyle\sum_{i=1}^{160} 2/3i <br />
= \displaystyle\sum_{i=1}^{160} (1/(3i-2) + 1/(3i-1) - 2/3i) = <br />
\displaystyle\sum_{i=1}^{160} (9i-4)/(3i-2)(3i-1)3i.

    I'm sure that there is a neater what though! I hope that helps.
    Last edited by Swlabr; May 21st 2009 at 11:53 PM. Reason: Bad punctuation
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  5. #5
    Super Member fardeen_gen's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    The last three terms above are not consistent with those before the ellipsis, can you give the expression for the general term in this summation.

    CB
    Quote Originally Posted by Swlabr View Post
    Yes they are as 3 divides 480 properly, 3.160 = 480. I believe the series is,

    \displaystyle\sum_{i=1}^{160} 1/(3i-2) +<br />
\displaystyle\sum_{i=1}^{160} 1/(3i-1) - \displaystyle\sum_{i=1}^{160} 2/3i <br />
= \displaystyle\sum_{i=1}^{160} (1/(3i-2) + 1/(3i-1) - 2/3i) = <br />
\displaystyle\sum_{i=1}^{160} (9i-4)/(3i-2)(3i-1)3i.

    I'm sure that there is a neater what though! I hope that helps.
    We have the series then. Anyone with any ideas on proving the required divisibility?
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    We have the series then. Anyone with any ideas on proving the required divisibility?
    Well, what are your ideas?
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  7. #7
    Super Member fardeen_gen's Avatar
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    641 seems to be completely random to me. I do not see any pattern.
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  8. #8
    Moo
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    Hi,

    I found the same sum as above.

    And as for 641, notice that it is prime. So I guess the result of the summation will have a factor 641 in the numerator.
    That's all I can guess about.

    I tried to compute \int_0^1 \frac{1-x^{480}}{1-x^3} ~dx=\int_0^1 \sum_{k=1}^{160} x^{3k-3} ~dx=\sum_{k=1}^{160}\frac{1}{3k-2}

    But didn't give it enough time (it was late )...
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  9. #9
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    641 seems to be completely random to me. I do not see any pattern.

    641 is prime, as Moo mentioned, so it does not divide the denominator of our fraction, as the denominator is just 1.2.3. ... .480. Thus, if 641 divides the numerator of our product it will divide every such numerator (this is not clear as 3 divides the numerator but also the denominator in our sum).
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  10. #10
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    Quote Originally Posted by fardeen_gen View Post
    641 seems to be completely random to me. I do not see any pattern.
    I have no idea how it would come up in this problem, but 641 doesn't seem random to me: it reminds me immediately the proof by Euler that the Fermat number F_5=2^{2^5}+1 is not prime (thus breaking down Fermat's conjecture). Euler showed that 641 divides F_5. I guess this is not a coincidence, but I may be totally wrong. If it may help anyone...
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  11. #11
    Lord of certain Rings
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    Quote Originally Posted by fardeen_gen View Post
    [SIZE=3]If p and q are positive integers such that \frac{p}{q} = 1 + \frac{1}{2} - \frac{2}{3} + \frac{1}{4} + \frac{1}{5} - \frac{2}{6} + \frac{1}{7} + \frac{1}{8} - \frac{2}{9} + \mbox{...} + \frac{1}{478} + \frac{1}{479} - \frac{2}{480}, show that p is divisible by 641.
    Lets say S = 1 + \frac{1}{2} - \frac{2}{3} + \frac{1}{4} + \frac{1}{5} - \frac{2}{6} + \mbox{...} +  \frac{1}{478} + \frac{1}{479} - \frac{2}{480}

    S  = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}  + \mbox{...} + \frac{1}{478} +  \frac{1}{479} + \frac{1}{480} - 3\left(\frac13 + \frac{1}{6} + \frac{1}{9} +  \mbox{...} +\frac1{480}\right)

    S = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \mbox{...} + \frac{1}{478} + \frac{1}{479} + \frac{1}{480} - \left(1 + \frac{1}{2} + \frac{1}{3} +  \mbox{...} +\frac1{160}\right)

    S = \frac1{161} + \frac{1}{162} + \frac{1}{163} + \frac{1}{164} +  \mbox{...} +\frac{1}{480}

    Now group terms:

    S = \left(\frac1{161} + \frac{1}{480}\right) +\left(\frac1{162} + \frac{1}{479}\right) \mbox{...} + \left(\frac1{320} + \frac{1}{321}\right)

    S = \frac{641}{161 \times 480} +\frac{641}{162 \times 479} + \mbox{...} +\frac{641}{320 \times 321}

    Finally note that 641 will remain on the top always because its prime and all the denominator fellows are less than 641. So done

    Nice problem!
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  12. #12
    Lord of certain Rings
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    Quote Originally Posted by Laurent View Post
    I have no idea how it would come up in this problem, but 641 doesn't seem random to me: it reminds me immediately the proof by Euler that the Fermat number F_5=2^{2^5}+1 is not prime (thus breaking down Fermat's conjecture). Euler showed that 641 divides F_5
    Thats exactly what I thought when he said it looks random. Its a very famous Number Theory anecdote.
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  13. #13
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    Brilliant proof isomorphism o_O
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