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ganeshie8

  • one year ago

Let \(R_n=111\ldots\text{n 1's}\) be the number with \(n\) ones in decimal number system. Find the smallest \(R_n\) that is divisible by \(83\).

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  1. dan815
    • one year ago
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    .

  2. anonymous
    • one year ago
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    Correction: \[R_n=\sum_{k=0}^{\color{red}{n-1}}10^k=\frac{10^{\color{red}n}-1}{9}\] so find \(n\) such that \[\frac{10^{\color{red}n}-1}{9}\equiv0\mod83\]

  3. anonymous
    • one year ago
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    I suppose this can be simplified a bit... \(n\) should also satisfy this, no? \[10^n\equiv1\mod83\]

  4. dan815
    • one year ago
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    from fermats little theorem we have this bount atleast 10^(82)-1=0 mod 83

  5. dan815
    • one year ago
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    oh we can reduce the ones we have left to check by looking at the divisors of 82 apparently

  6. dan815
    • one year ago
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    acording to this http://prntscr.com/8834ha

  7. dan815
    • one year ago
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    82 = 2*41 therefore all we have left to check is 2,41 10^2-1 =/=0mod83 10^41-1 = 0 mod 83

  8. dan815
    • one year ago
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    41, 1s

  9. anonymous
    • one year ago
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    I'm getting the same result with Mathematica, checks out

  10. dan815
    • one year ago
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    the 2nd statement that there msut exist some lower exponent d that divides p-1, if p does not divide a looks interesting

  11. ikram002p
    • one year ago
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    i has same method finding the first n which satisfy 10^n=1 mod 83

  12. ikram002p
    • one year ago
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    so we know 10^82 satisfy, now we check its divisors 1,2,41,82 the end xD

  13. ganeshie8
    • one year ago
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    Awesome! so it seems, given a prime, \(p\), we can always find the smallest \(R_n\) divisible by \(p\), by looking at the divisors of \(p-1\)

  14. Jhannybean
    • one year ago
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    .

  15. ikram002p
    • one year ago
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    @ganeshie8 are you saying that R_n always composite ?

  16. ganeshie8
    • one year ago
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    11 isn't composite

  17. ikram002p
    • one year ago
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    i know it has some primes im trying to understand what u have said :)

  18. ganeshie8
    • one year ago
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    fix a prime \(p\) and form the set of all \(R_n\)'s that are divisible by \(p\). then we can find the least element in that set using the earlier trick

  19. ikram002p
    • one year ago
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    got it!

  20. ikram002p
    • one year ago
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    :O its awesome isnt it ?

  21. ganeshie8
    • one year ago
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    so baiscally all \(\large R_{p-1}\)'s are composite because they are trivially divisible by \(p\)

  22. ikram002p
    • one year ago
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    yeah with even number of digits hmmm

  23. ganeshie8
    • one year ago
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    Oh, all \(\large R_{2k}\) are composite is it ?

  24. ganeshie8
    • one year ago
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    Easy to see that all \(\large R_{3k}\) are composite because they are divisible by \(3\)..

  25. ikram002p
    • one year ago
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    yeah for this one, i think we have proved this earlier

  26. ikram002p
    • one year ago
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    the 1010101 stuff

  27. ikram002p
    • one year ago
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    let me remember how it was like exactly

  28. ganeshie8
    • one year ago
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    Ah, right, its easy one sec

  29. ganeshie8
    • one year ago
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    \[\large R_{2k}=\sum_{k=0}^{\color{red}{2k-1}}10^k=\frac{10^{\color{red}{2k}}-1}{9} = \dfrac{(10^k-1)(10^k+1)}{9}\]

  30. ganeshie8
    • one year ago
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    for \(k\gt 1\), we have \(10^k-1 = 9l\) such that \(l\gt 1\)

  31. ikram002p
    • one year ago
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    so ovs what l was whole number stays composite :3

  32. ganeshie8
    • one year ago
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    for \(k\gt 1\), \[\large R_{2k}=\sum_{k=0}^{\color{red}{2k-1}}10^k=\frac{10^{\color{red}{2k}}-1}{9} = \dfrac{(10^k-1)(10^k+1)}{9} = l(10^k+1)\] which cannot be a prime as it is product of two factors that are > 1

  33. ikram002p
    • one year ago
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    haha sweet !

  34. ganeshie8
    • one year ago
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    that also shows : if \(n\) is composite, then \(R_n\) is composite

  35. ikram002p
    • one year ago
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    correct!

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