## 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 $$83$$.

1. dan815

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2. 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 $\frac{10^{\color{red}n}-1}{9}\equiv0\mod83$

3. anonymous

I suppose this can be simplified a bit... $$n$$ should also satisfy this, no? $10^n\equiv1\mod83$

4. dan815

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

5. dan815

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

6. dan815

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

7. dan815

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

41, 1s

9. anonymous

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

10. dan815

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

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

12. ikram002p

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

13. 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 $$p-1$$

14. Jhannybean

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15. ikram002p

@ganeshie8 are you saying that R_n always composite ?

16. ganeshie8

11 isn't composite

17. ikram002p

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

18. 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

19. ikram002p

got it!

20. ikram002p

:O its awesome isnt it ?

21. ganeshie8

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

22. ikram002p

yeah with even number of digits hmmm

23. ganeshie8

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

24. ganeshie8

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

25. ikram002p

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

26. ikram002p

the 1010101 stuff

27. ikram002p

let me remember how it was like exactly

28. ganeshie8

Ah, right, its easy one sec

29. ganeshie8

$\large R_{2k}=\sum_{k=0}^{\color{red}{2k-1}}10^k=\frac{10^{\color{red}{2k}}-1}{9} = \dfrac{(10^k-1)(10^k+1)}{9}$

30. ganeshie8

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

31. ikram002p

so ovs what l was whole number stays composite :3

32. 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

33. ikram002p

haha sweet !

34. ganeshie8

that also shows : if $$n$$ is composite, then $$R_n$$ is composite

35. ikram002p

correct!