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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 \(83\).
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Correction: \[R_n=\sum_{k=0}^{\color{red}{n1}}10^k=\frac{10^{\color{red}n}1}{9}\] so find \(n\) such that \[\frac{10^{\color{red}n}1}{9}\equiv0\mod83\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I suppose this can be simplified a bit... \(n\) should also satisfy this, no? \[10^n\equiv1\mod83\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.5from fermats little theorem we have this bount atleast 10^(82)1=0 mod 83

dan815
 one year ago
Best ResponseYou've already chosen the best response.5oh we can reduce the ones we have left to check by looking at the divisors of 82 apparently

dan815
 one year ago
Best ResponseYou've already chosen the best response.5acording to this http://prntscr.com/8834ha

dan815
 one year ago
Best ResponseYou've already chosen the best response.582 = 2*41 therefore all we have left to check is 2,41 10^21 =/=0mod83 10^411 = 0 mod 83

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm getting the same result with Mathematica, checks out

dan815
 one year ago
Best ResponseYou've already chosen the best response.5the 2nd statement that there msut exist some lower exponent d that divides p1, if p does not divide a looks interesting

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2i has same method finding the first n which satisfy 10^n=1 mod 83

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2so we know 10^82 satisfy, now we check its divisors 1,2,41,82 the end xD

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Awesome! so it seems, given a prime, \(p\), we can always find the smallest \(R_n\) divisible by \(p\), by looking at the divisors of \(p1\)

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2@ganeshie8 are you saying that R_n always composite ?

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2i know it has some primes im trying to understand what u have said :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2fix 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
 one year ago
Best ResponseYou've already chosen the best response.2:O its awesome isnt it ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2so baiscally all \(\large R_{p1}\)'s are composite because they are trivially divisible by \(p\)

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2yeah with even number of digits hmmm

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Oh, all \(\large R_{2k}\) are composite is it ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Easy to see that all \(\large R_{3k}\) are composite because they are divisible by \(3\)..

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2yeah for this one, i think we have proved this earlier

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2let me remember how it was like exactly

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Ah, right, its easy one sec

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\large R_{2k}=\sum_{k=0}^{\color{red}{2k1}}10^k=\frac{10^{\color{red}{2k}}1}{9} = \dfrac{(10^k1)(10^k+1)}{9}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2for \(k\gt 1\), we have \(10^k1 = 9l\) such that \(l\gt 1\)

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2so ovs what l was whole number stays composite :3

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2for \(k\gt 1\), \[\large R_{2k}=\sum_{k=0}^{\color{red}{2k1}}10^k=\frac{10^{\color{red}{2k}}1}{9} = \dfrac{(10^k1)(10^k+1)}{9} = l(10^k+1)\] which cannot be a prime as it is product of two factors that are > 1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2that also shows : if \(n\) is composite, then \(R_n\) is composite
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