ganeshie8
  • ganeshie8
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 \(83\).
Mathematics
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SOLVED
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katieb
  • katieb
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dan815
  • dan815
.
anonymous
  • anonymous
Correction: \[R_n=\sum_{k=0}^{\color{red}{n-1}}10^k=\frac{10^{\color{red}n}-1}{9}\] so find \(n\) such that \[\frac{10^{\color{red}n}-1}{9}\equiv0\mod83\]
anonymous
  • anonymous
I suppose this can be simplified a bit... \(n\) should also satisfy this, no? \[10^n\equiv1\mod83\]

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More answers

dan815
  • dan815
from fermats little theorem we have this bount atleast 10^(82)-1=0 mod 83
dan815
  • dan815
oh we can reduce the ones we have left to check by looking at the divisors of 82 apparently
dan815
  • dan815
acording to this http://prntscr.com/8834ha
dan815
  • dan815
82 = 2*41 therefore all we have left to check is 2,41 10^2-1 =/=0mod83 10^41-1 = 0 mod 83
dan815
  • dan815
41, 1s
anonymous
  • anonymous
I'm getting the same result with Mathematica, checks out
dan815
  • dan815
the 2nd statement that there msut exist some lower exponent d that divides p-1, if p does not divide a looks interesting
ikram002p
  • ikram002p
i has same method finding the first n which satisfy 10^n=1 mod 83
ikram002p
  • ikram002p
so we know 10^82 satisfy, now we check its divisors 1,2,41,82 the end xD
ganeshie8
  • ganeshie8
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 \(p-1\)
Jhannybean
  • Jhannybean
.
ikram002p
  • ikram002p
@ganeshie8 are you saying that R_n always composite ?
ganeshie8
  • ganeshie8
11 isn't composite
ikram002p
  • ikram002p
i know it has some primes im trying to understand what u have said :)
ganeshie8
  • ganeshie8
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
ikram002p
  • ikram002p
got it!
ikram002p
  • ikram002p
:O its awesome isnt it ?
ganeshie8
  • ganeshie8
so baiscally all \(\large R_{p-1}\)'s are composite because they are trivially divisible by \(p\)
ikram002p
  • ikram002p
yeah with even number of digits hmmm
ganeshie8
  • ganeshie8
Oh, all \(\large R_{2k}\) are composite is it ?
ganeshie8
  • ganeshie8
Easy to see that all \(\large R_{3k}\) are composite because they are divisible by \(3\)..
ikram002p
  • ikram002p
yeah for this one, i think we have proved this earlier
ikram002p
  • ikram002p
the 1010101 stuff
ikram002p
  • ikram002p
let me remember how it was like exactly
ganeshie8
  • ganeshie8
Ah, right, its easy one sec
ganeshie8
  • ganeshie8
\[\large R_{2k}=\sum_{k=0}^{\color{red}{2k-1}}10^k=\frac{10^{\color{red}{2k}}-1}{9} = \dfrac{(10^k-1)(10^k+1)}{9}\]
ganeshie8
  • ganeshie8
for \(k\gt 1\), we have \(10^k-1 = 9l\) such that \(l\gt 1\)
ikram002p
  • ikram002p
so ovs what l was whole number stays composite :3
ganeshie8
  • ganeshie8
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
ikram002p
  • ikram002p
haha sweet !
ganeshie8
  • ganeshie8
that also shows : if \(n\) is composite, then \(R_n\) is composite
ikram002p
  • ikram002p
correct!

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