Question

find the period of \(\large 1/83 \) in its decimal expansion

Answer 1

what is period mean here?

Answer 2

is that just any repeating part that might occur in the expansion?

Answer 3

period is the length of the repeating part...

example :
the period of 1/37 is 3 because 1/37 = 0.027027027...

Answer 4

the length of the repeating part

Answer 5

82

Answer 6

This one is going to be pretty tough without any tricks
because it it is suppose to be 41.
:p

Answer 7

how did you get 82

Answer 8

\((10^{41}-1) \pmod {83}=0\)

Answer 9

yes both 41 and 82 are periods.. but 41 is the smallest one
actually i think we can say any multiple of 41 is a period

Answer 10

also \((10^{82}-1) \pmod {83}=0\)

Answer 11

fermat little theorm

Answer 12

interesting, whats that and does that work

Answer 13

and how did u get 41 @myininaya

Answer 14

Ohh I see.. how is that related to period ?

Answer 15

found here

http://hr.userweb.mwn.de/numb/period.html

the length of period of the denominator \(q\) should divide \(10^{k}-1\)

Answer 16

lol I was trying to find a trick how to get 41

Answer 17

but sometimes \(\Large m^{\phi (n)} =m^{\phi(n)/2} \equiv 1 \pmod q\)

Answer 18

\(\Large m^{\phi (n)} =m^{\phi(n)/2} \equiv 1 \pmod n\)

Answer 19

\[(10^{x}-1) \pmod {n}=0\]
so finding x here will give us the period for 1/n
---would this work for all n and x

Answer 20

no u have to check the earlier powers to specially its \(\phi(n)/2\)

Answer 21

n = 0.abcdef abcdef abcdef ...
10^6n=abcdef.abcdef abcdef abcdef ...
--------------------------------------
10^6n-n = abcdef

n(10^6-1) = abcdef

yeah it seem to work in general xD

Answer 22

@mathmath3333 is such a wiz

Answer 23

\(\phi(83) = 82\)
and only factors of \(82\) are \(2\) and \(41\)

so only possible candidates of period are : \(\{2,41,82\}\)

is that how you figured out @mathmath333

Answer 24

i searched "wiz" on google

Answer 25

yes this one

Answer 26

By similar reasoning, the period of 1/7 has to be 6 or any factors of 6 ?

Answer 27

example in \(1/73\) its period of 8
and 72=8*9

there is obvious relation with it

Answer 28

nice nice
for finding the period of \(\large \dfrac{1}{p}\), we can simply try the factors of \(\large p-1\)

Answer 29

\[\frac{1}{43} \\ p=43 \\ p-1=42=6 \cdot 7 \]
And how do you continue from here
I'm still confused

Answer 30

wait so those are the only candidates for the equation given by @mathmath333

Answer 31

\(\Large (10^{\text{try every factor of 42}} -1)\pmod {43} =0\)

Answer 32

the lowest will be the answer

Answer 33

\[(10^x-1) (\mod p)=0 \\ (10^6-1)( \mod 43) \neq 0 \\ (10^7-1)(\mod 43) \neq 0 \\ (10^{21}-1)(\mod 43)=0\]
I think I get it
so the period of 1/43 is 21

Answer 34

yes

Answer 35

thats it i guess
just a side observation.. once we know that \((10^{21}-1)(\mod 43)=0\), below is also trivially true
\[(10^{21x}-1)(\mod 43)=0\]

Answer 36

That will get scarier for numbers with bigger denominators

Answer 37

not if its multiple choice question

Answer 38

completely agree! but thats the best known method we have so far it seems..

Answer 39

Though I guess that is the point in why they choose big numbers in encryption
it runs more difficulty of us using mathematics to get the right key

Answer 40

yeah solving \(10^x \equiv a\pmod {p}\) is a discrete logarithm problem - the most hardest problem for computers... these are used in cryptography .. solve below problem for $100,000,000 :P

http://unsolvedproblems.org/index_files/RSA.htm

Answer 41

That number is crazy and I dare @mathmath333 to take on the challenge and give me half his award :p

Answer 42

i would rather play chess tournaments and gather more

Answer 43

jk

Answer 44

Woah thats not fair, i deserve half and you both can share 25 lol
btw it is given, that number has only two factors

Answer 45

Nothing is fair when it comes to women

Answer 46

I do have one real question on this problem

Answer 47

earlier to find x I was just pluggin in numbers and seeing which gave me remainder 0

Answer 48

I guess that is the best way ?

Answer 49

there is a small trick..

Answer 50

or the best way we know of

Answer 51

do tell :0

Answer 52

primitive roots ?

Answer 53

from the fermat we know that
\[10^{82}\equiv 1 \pmod{83}\]

Answer 54

so \(10^{41}\) has to leave a remainder of \(\pm 1\) because :
\[10^{82}\equiv 1 \pmod{83}\]
\[10^{82}- 1 \equiv 0 \pmod{83}\]
\[(10^{41}- 1)(10^{41}+1) \equiv 0 \pmod{83}\]

Answer 55

if \(10^{41}\) leaves a remainder of \(-1\), then we're sure that the period is \(82\)

Answer 56

that is kinda weird
the period for 1/83 is (83-1)/2
and the period 1/43 is (43-1)/2

Answer 57

the period of 1/97 is 96

Answer 58

can i substitute any primitive root of 83 for 83

Answer 59

But it must be true for some integer n that 1/n has period (n-1)/2
and if so I wonder if we can give the set of numbers n such that is true:

Answer 60

thats too long 96

Answer 61

1/13 has period (13-1)/2

Answer 62

thats an interesting question

primitive roots are also a good idea

Answer 63

is \(n\) a prime in \(1/n\) ?

Answer 64

It seems to be prime so far
I'm just trying to find a set of numbers such that 1/n has period (n-1)/2

Answer 65

but not all primes

Answer 66

1/23 doesn't have period (23-1)/2

Answer 67

Ohhk.. im not even sure if we can say the period function is one-to-one or one-to-many hmm

Answer 68

the (n-1)/2 will not work every time u have to check it

Answer 69

say f(p) gives the period of 1/p in the decimal expansion where p is a prime

it looks like f(p) is one-to-one...idk not so sure..

Answer 70

can we find two primes with same period ?

Answer 71

1/2 and 1/3 have period 1 right?

Answer 72

or can you say 1/2 has period 1

Answer 73

it doesn't repeat

Answer 74

or i mean it terminates

Answer 75

i think the primitive root will not work


http://www.wolframalpha.com/input/?i=10%5E%2896%29+mod+5%3D

and 5 is primitive root of 97

Answer 76

2 is even it wont repeat

Answer 77

2 is also prime

Answer 78

Ohkk.. then there can be more than one prime having the same period

Answer 79

so we should be looking for odd primes

Answer 80

it is the cursed prime

Answer 81

we exclude 2 and 5 as they are factors of 10

Answer 82

lol they are cursed primes only in base 10.. any combination of 2 and 5 in the denominator gives a terminating decimal expansion :

1/(2^a5^b) always terminates

Answer 83

I found primes 13 and 7
and they both seem to have period 6
so it doesn't look lit it is one-to-one

Answer 84

look like*