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

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- dan815

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- 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

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

from fermats little theorem we have this bount atleast 10^(82)-1=0 mod 83

- dan815

oh we can reduce the ones we have left to check by looking at the divisors of 82 apparently

- dan815

acording to this http://prntscr.com/8834ha

- 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

41, 1s

- anonymous

I'm getting the same result with Mathematica, checks out

- 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

i has same method finding the first n which satisfy
10^n=1 mod 83

- ikram002p

so we know
10^82 satisfy, now we check its divisors 1,2,41,82 the end xD

- 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

.

- ikram002p

@ganeshie8 are you saying that R_n always composite ?

- ganeshie8

11 isn't composite

- ikram002p

i know it has some primes im trying to understand what u have said :)

- 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

got it!

- ikram002p

:O its awesome isnt it ?

- ganeshie8

so baiscally all \(\large R_{p-1}\)'s are composite because they are trivially divisible by \(p\)

- ikram002p

yeah with even number of digits hmmm

- ganeshie8

Oh, all \(\large R_{2k}\) are composite is it ?

- ganeshie8

Easy to see that all \(\large R_{3k}\) are composite because they are divisible by \(3\)..

- ikram002p

yeah for this one, i think we have proved this earlier

- ikram002p

the 1010101 stuff

- ikram002p

let me remember how it was like exactly

- ganeshie8

Ah, right, its easy one sec

- 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

for \(k\gt 1\),
we have \(10^k-1 = 9l\) such that \(l\gt 1\)

- ikram002p

so ovs what l was whole number stays composite :3

- 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

haha sweet !

- ganeshie8

that also shows :
if \(n\) is composite, then \(R_n\) is composite

- ikram002p

correct!

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