Question

Challenge problem , show that for any
n positive even integer

\(\large \gcd(3434,(\frac{10^{6n-6}-1}{99}) )= 1 \)

Answer 1

99 is below the entire thing right ?

Answer 2

yes that gives u
99999999.....9/99=1010101....01

Answer 3

look at the expression in question
it is wrong

Answer 4

how about nw ?

Answer 5

\[\large \gcd\left(3434,~~\frac{10^{6n-6}-1}{99} \right)= 1\]

you want to prove this for n even number right ?

Answer 6

yes

Answer 7

@Marki ,it's ur idea?

Answer 8

means ?

Answer 9

ur doing it for ur main question?

Answer 10

well tbh , this prove nothing t the previous question :P
but its good observation that we always got distinct primes btw those term

Answer 11

seems to be related :D

Answer 12

see , CGD btw them is 1 at each time but there sum is composite which makes me think of this

x N+y M=1

Answer 13

but it sounded more like hmm complicated , if i found something ok for sure i'll share :D
tell now try to prove this one only

Answer 14

I wish i could stay longer and try on this problem but I have to leave soon
I kinda playing with ugly fraction
\[\frac{(10^{n-1}-1)(10^{n+1}-1)(10^{2(n-1)}+10^{n-1}+1)(10^{2(n-1)}-10^{n-1}+1)}{(10-1)(10+1)}\]

Answer 15

sounds interesting !

Answer 16

oops type-0

Answer 17

i hope u to get something ! all the luck

Answer 18

\[\frac{(10^{n-1}-1)(10^{n-1}+1)(10^{2(n-1)}+10^{n-1}+1)(10^{2(n-1)}-10^{n-1}+1)}{(10-1)(10+1)} \]

Answer 19

I want to cancel the factors on bottom
i'm sure we can somehow factor the \[10^{n-1} \pm 1\]

Answer 20

\[3434=(10^2+1)(3 \cdot 10 +4)\]

Answer 21

if n is ever then n-1 is odd
so if I'm not mistaken 10^2+1 will not be a factor of the other number

Answer 22

but i'm just sorta thinking too fast right now because i'm rushed
i must go now

Answer 23

yes u can do

Answer 24

Here is a fancy method :

prime factors of \(\large 3434\) are \( \{2, 7, 101\} \)

so it is sufficient if we show that the number \(\large \frac{10^{12k-6}-1}{99}\) can never have one of those factors

Answer 25

:D

Answer 26

@ganeshie8 wait...

Answer 27

okay :)

Answer 28

i'm right working on that idea...

Answer 29

i think i've proved why it doesn't have the factor 101

Answer 30

this is what ganesh trying to do

Answer 31

what's ur idea about 17?

Answer 32

my idea works for all factors in general but it uses number theory
can i see your method for 101 ? :)

Answer 33

comon , someone type :-|

Answer 34

for all factors? i consider each of them and then try to say that number doesn't have those factors....
this is what i've done so far (not complete)

all of those numbers can be written like this:
\(\frac{ 10^6-1 }{99 } = 10101\) 5 digits
\(\frac{ 10^18-1 }{ 99 }=101010....1 \) 17 digits
\(\frac{ 10^18-1 }{ 99 }=101010....1 \) 29 digits

\(10101 \equiv 1 \ (mod \ 101)\)
well,i'm using calculator to know what would 1010...1 (17 digits) be...

Answer 35

awesome!! it's \( 1010..1 (17 \ digits) \equiv 1 \ (mod \ 101) \) too.

Answer 36

its 1 for any digit ;)

Answer 37

so relax this can be proven , see second comment

Answer 38

yes,i know...now i'm working on 17...what's ur idea?

Answer 39

hmm i cant reveal mine , since i posted this :P
lets see what @ganeshie8 have

Answer 40

but i would like @ganeshie8 to wait a moment..i think i can solve it...

Answer 41

Hint :-
work on mod properties

Answer 42

i'm exactly working on that , i havn't studied n th for 2 weeks and i think i should first have a look at my book XD

Answer 43

hmm...the first one is congruent to 3 (mod 17) and second one is congruent to 1 (mod 17 ) :|

Answer 44

that looks close to what i have :)

Answer 45

what about mod 10 :D

Answer 46

all of them are congruent to 1 @Marki ,does this help?

Answer 47

ok another help
if k=17n
then
k mod 10=7,1,5,9 only

Answer 48

we need this case
so if 17 is a factor then we need this case 17*n=1 mod 10
also n has to have this property
n=3 mod 10

Answer 49

got it! if 17 is a factor of that then it can be written as 17xn and it's congruent to 1,5,7,9 (mod 10) but when is't congruent to 1 then n is congruent to 3 (mod 10) and it cannot ever be happened! is this ur meant?

Answer 50

and it's odd so doesn't have the factor 2...

Answer 51

so the question is solved...@ganeshie8 what was ur method?

Answer 52

why it cant happen ?

Answer 53

a=17n
n=3t + 10
so a=51t + 170

Answer 54

will continue tomorrow,i'm very tired...

Answer 55

sry,
a=17n
n+3t (it can be written as 3t )
--> a = 51t
and \( a \equiv 1 \)(mod 10)
and it happens whenever t = 1,11,111,1111,...
then a would be:
51,561,5661,56661,...

Answer 56

correct?

Answer 57

@ganeshie8 , what do u think?

Answer 58

Looks good to me!

Answer 59

thanks ;) but i'm not very happy as this question was not such a hard question that i spend a lot of time to solve it , i hope to be better at number th ..

Answer 60

here is what i had for proving \(17\) cannot divide \(\large \frac{10^{12x-6}-1}{99}\) :

we want to show that \(\dfrac{10^{12x-6}-1}{99} = 17k\) is not solvable in integers

rearranging we get \(10^{12x-6} = 1+ (99\cdot 17)k\)

which is same as showing below congruence has no solution
\[10^{12x-6} \equiv 1 \pmod{99\cdot 17}\tag{1}\]

Next working out the order of 10 in modulo 99*17 gives 16
so 16 must divide 12x-6 for the congruence to have a solution :
\[12x - 6 \equiv 0 \pmod { 16}\]
\[2x - 1 \equiv 0 \pmod { 4}\]
\[2x - 1 = 4y\]
\[2x - 4y = 1\]

NO solutions as \(2 \not | ~1\) and consequently \(17\) cannot divide \(\large \frac{10^{12x-6}-1}{99}\)

Answer 61

:)

Answer 62

works fine

Answer 63

eh last one of them

Answer 64

no,we showed for 101 and 17 and it obviously is not divisible by 2