Page 1 of 2 12 LastLast
Results 1 to 15 of 19

Thread: 26th digit

  1. #1
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
    Posts
    648

    Smile 26th digit

    N is a 50digit nomber(in the decimal scale). All digits except the 26th digit(from the left) are 1. If N is divisible by 13, find the 26th digit.
    Last edited by malaygoel; Jun 21st 2006 at 06:39 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Feb 2006
    From
    United Kingdom
    Posts
    70
    we can use the facts that
    $\displaystyle 10^6 \equiv 1 \mod 13$
    so $\displaystyle 10^{6k} \equiv 1 \mod 13$
    so $\displaystyle a10^{6k} \equiv a \mod 13$
    also
    $\displaystyle 10^3 \equiv -1 \mod 13$
    then combining this with our other idendity
    $\displaystyle 10^{6k+3} \equiv -1 \mod 13$
    so $\displaystyle a10^{6k+3} \equiv -a \mod 13$

    so our number is
    $\displaystyle N = 11,111,111,111,111,111,111,111,11k$$\displaystyle ,111,111,111,111,111,111,111,111$

    which we can write as..
    $\displaystyle N = 11*10^{48} + 111*10^{45} + ... +$$\displaystyle 11k*10^{24} + ... + 111*10^3 + 111$

    using our idendities from before, we know that for instance
    $\displaystyle 11*10^{48} \equiv 11 \mod 13$
    since 48 = 7*6 so

    $\displaystyle N = 11-111+111-111+111-111+111-111+11k$$\displaystyle -111+111-111+111-111+111-111+111 \mod 13$

    which simplifies to
    $\displaystyle N \equiv 11k - 100 \mod 13$
    so if $\displaystyle N|13$ then $\displaystyle 11k-100 \equiv 0 \mod 13$
    the only value of k where this is true is $\displaystyle k=3 $
    thus this is the answer. $\displaystyle \mbox{the 26th digit is 3.}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
    Posts
    648
    Quote Originally Posted by Aradesh
    we can use the facts that
    $\displaystyle 10^6 \equiv 1 \mod 13$
    so $\displaystyle 10^{6k} \equiv 1 \mod 13$
    so $\displaystyle a10^{6k} \equiv a \mod 13$
    also
    $\displaystyle 10^3 \equiv -1 \mod 13$
    then combining this with our other idendity
    $\displaystyle 10^{6k+3} \equiv -1 \mod 13$
    so $\displaystyle a10^{6k+3} \equiv -a \mod 13$

    so our number is
    $\displaystyle N = 11,111,111,111,111,111,111,111,11k$$\displaystyle ,111,111,111,111,111,111,111,111$

    which we can write as..
    $\displaystyle N = 11*10^{48} + 111*10^{45} + ... +$$\displaystyle 11k*10^{24} + ... + 111*10^3 + 111$

    using our idendities from before, we know that for instance
    $\displaystyle 11*10^{48} \equiv 11 \mod 13$
    since 48 = 7*6 so
    You made amistake.
    48 = 6*8 not 7*6
    never mind.
    I could follow your solution.
    Are there any more methods?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Feb 2006
    From
    United Kingdom
    Posts
    70
    Quote Originally Posted by malaygoel
    You made amistake.
    48 = 6*8 not 7*6
    never mind.
    I could follow your solution.
    Are there any more methods?
    my finger must have slipped.

    this is the only way i know how to do this problem.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
    Posts
    648
    Quote Originally Posted by Aradesh
    my finger must have slipped.

    this is the only way i know how to do this problem.
    I was just wondering could the given number be just divided by 13 and get the answer.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Feb 2006
    From
    United Kingdom
    Posts
    70
    oh, good point. i just checked it on mathematica (a computer program that does big sums) and it gives:

    854700854700854700854701008547008547008547008547
    while other values for our 26th digits don't give a whole number.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by Aradesh
    oh, good point. i just checked it on mathematica (a computer program that does big sums) and it gives:

    854700854700854700854701008547008547008547008547
    while other values for our 26th digits don't give a whole number.
    Of course this is not a proof as M. could have a bug in its BigNum arithmetic

    RonL
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,124
    Thanks
    720
    Awards
    1
    Quote Originally Posted by CaptainBlack
    Of course this is not a proof as M. could have a bug in its BigNum arithmetic

    RonL
    Mathematica have a BUG?? It's not possible!

    -
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
    Posts
    648
    Quote Originally Posted by malaygoel
    I was just wondering could the given number be just divided by 13 and get the answer.
    What I wanted to say(although I have not tried) is that we could substitute the 26th digit as 1 and perform divison(which will be easy due to repeating 1's) and then find find the value of k by guessing the remainder that the 26th digit should give while performing division.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
    Posts
    648
    Quote Originally Posted by malaygoel
    N is a 50digit nomber(in the decimal scale). All digits except the 26th digit(from the left) are 1. If N is divisible by 13, find the 26th digit.
    I think I have found it.
    By performing simple division, I found that 111111 is divisible by 13
    Now,
    $\displaystyle N=11,111,111,111,111,111,111,111,11k,$$\displaystyle 111,111,111,111,111,111,111,111,$
    there are 25 1's before k and 24 1's after k.
    24 1's before k and 24 1's after k are divisible by 13
    We are left with $\displaystyle 1k$ and hence the answer 3.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Excellent solution, Malay!

    I've known for years that $\displaystyle 7 \times 11 \times 13 \:=\:1001$ **

    . . hence: $\displaystyle 111,111\:=\:111 \times 1001 \:=\:111\times 7 \times 11 \times 13$

    I'm embarrassed that none of this occured to me.

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    More trivia . . .

    . . $\displaystyle 10,001 \;=\;73\cdot137$
    . .$\displaystyle 100,001 \;= \;11\cdot9091$
    .$\displaystyle 1,000,001\;=\;101\cdot9901$
    $\displaystyle 10,000,001\;=\;11\cdot909091$


    . . . .$\displaystyle 111\;=\;3\cdot37$
    . . .$\displaystyle 1,111 \;=\;11\cdot101$
    . . $\displaystyle 11,111\;=\;41\cdot271$
    . .$\displaystyle 111,111\;=\;111\cdot1001\;=\;3\cdot7\cdot11\cdot13 \cdot37$
    .$\displaystyle 1,111,111\;=\;239\cdot4649$
    $\displaystyle 11,111,111\;=\;10001\cdot1111\;=\;11\cdot73\cdot10 1\cdot137$

    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
    Posts
    648
    Quote Originally Posted by Soroban
    Excellent solution, Malay!

    I've known for years that $\displaystyle 7 \times 11 \times 13 \:=\:1001$ **

    . . hence: $\displaystyle 111,111\:=\:111 \times 1001 \:=\:111\times 7 \times 11 \times 13$

    I'm embarrassed that none of this occured to me.

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    More trivia . . .

    . . $\displaystyle 10,001 \;=\;73\cdot137$
    . .$\displaystyle 100,001 \;= \;11\cdot9091$
    .$\displaystyle 1,000,001\;=\;101\cdot9901$
    $\displaystyle 10,000,001\;=\;11\cdot909091$


    . . . .$\displaystyle 111\;=\;3\cdot37$
    . . .$\displaystyle 1,111 \;=\;11\cdot101$
    . . $\displaystyle 11,111\;=\;41\cdot271$
    . .$\displaystyle 111,111\;=\;111\cdot1001\;=\;3\cdot7\cdot11\cdot13 \cdot37$
    .$\displaystyle 1,111,111\;=\;239\cdot4649$
    $\displaystyle 11,111,111\;=\;10001\cdot1111\;=\;11\cdot73\cdot10 1\cdot137$

    It seems that 1001,10001,.... are all composite. Can it be proved that $\displaystyle 10^k + 1(k\geq4)$ is composite?
    Can it be proved that 111,1111,.......$\displaystyle \frac{10^k - 1}{9}(k\geq1)$is composite?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by malaygoel
    It seems that 1001,10001,.... are all composite. Can it be proved that $\displaystyle 10^k + 1(k\geq4)$ is composite?
    Can it be proved that 111,1111,.......$\displaystyle \frac{10^k - 1}{9}(k\geq1)$is composite?
    No, because it is not true.

    These numbers are called "repunits" we can however show if a repunit is prime then the number of ones must also be prime.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, ThePerfectHacker!

    Good call!


    These numbers are called "repunits".
    We can however show: if a repunit is prime, then the number of ones must also be prime.

    "Repunits" is an abbreviation for repeated units.

    Even if the numbers of 1's is a prime number, there is no dependable rule.


    Let $\displaystyle 1^{(n)}$ represent a number composed of n 1's.

    $\displaystyle 1^{(3)} \;= (3)(37)$
    $\displaystyle 1^{(5)} \;= (41)(271)$
    $\displaystyle 1^{(7)} \;= (239)(4649)$
    $\displaystyle 1^{(11)} = (21,\!649)(313,\!239)$
    $\displaystyle 1^{(13)} = (53)(79)(265,\!371,\!653)$
    $\displaystyle 1^{(17)} = (2,\!071,\!723)(5,\!363,\!222,\!857)$
    $\displaystyle 1^{(19)} = prime$
    $\displaystyle 1^{(23)} = prime$
    $\displaystyle 1^{(29)} = (3191)(16,\!763)(43,\!037)(62,\!003)(77,\!843,\!83 9,\!397)$
    $\displaystyle 1^{(31)} = composite \text{ (divisible by }2791)$
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
    Posts
    648
    Quote Originally Posted by ThePerfectHacker
    No, because it is not true.

    These numbers are called "repunits" we can however show if a repunit is prime then the number of ones must also be prime.
    I would be happy if you post here the proof of the statement in the quoted work.But the question still remains:Is there any criteria by which we can say that "$\displaystyle \frac{10^k - 1}{9}$ is prime or composite depends on the value of k"?
    Similarly, what can you say about the numbers $\displaystyle 10^k + 1(k\geq4)$.
    Keep Smiling
    Malay
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. The last digit of 7^7^7
    Posted in the Number Theory Forum
    Replies: 7
    Last Post: Nov 5th 2011, 05:59 AM
  2. Digit sum & digit product of number x
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Jan 19th 2011, 08:07 AM
  3. Last digit
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Jun 28th 2009, 02:37 AM
  4. digit nos
    Posted in the Statistics Forum
    Replies: 1
    Last Post: Mar 28th 2009, 08:06 AM
  5. decimal digit as final digit
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: Feb 25th 2008, 07:18 PM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum