Question

\[x^3+3x^2+3x+1 \\ x^3+1+3x^2+3x \\ (x+1)(x^2-x+1)+3x(x+1) \\ (x+1)(x^2-x+1+3x) \\ (x+1)(x^2+2x+1) \\ (x+1)(x+1)^2 \\ (x+1)^3 \\ \text{ So } 343=1331_6 \\ \] Thought it was weird in base 6 this number had a factorization like this. I get 7 and 7^2... But 6. I wonder why 6 is unique here amongst the non-factors of 343. Or I don't know if it is unique or not. I have just did 343 in base whatever on wolfram all night last night looking for numbers that would give this kinda factorization.

Answer 1

My goal is to somehow find a way to write mnmnmnmnm.... (odd amount of digits and this symmetry all the way through) in some base number x such that the number can factored in a similar manner above.

Answer 2

HI \(\color\magenta\heartsuit\)

Answer 3

Hi!

Answer 4

@Kainui

Answer 5

By the way I do warn you guys I don't know if this can be done.
For some reason I assume it can be.

Answer 6

\[34343=9:14:0_{61} \\ 9x^2+14x+0x \\ x(9x+14) \\ \text{ Like I don't like 61 much because it is a factor of 34343 }\]
It would be cool to find another odd ball like we did above for 343.

Answer 7

If I understand correctly the phenomenon, I wonder if it wasn't because 7=6+1
If we try 6=5+1, we have
\(6^3=216_{10}=1331_5\)

Answer 8

hmm... that's neat

Answer 9

so \[5^3=125=1331_4 \]

Answer 10

It also works for 9:
\(9^3=729=1331_8\) :)

Answer 11

\[n^3=1331_{n-1} \text{ for integer n}\]
I wonder if we can prove that by induction.

Answer 12

n would have to be bigger than ...

Answer 13

Tag @ganeshie8 , he can do anything!

Answer 14

well there is a 3 in there
so n would have to be bigger than or equal to 5

Answer 15

Yep!

Answer 16

Or by mod.

Answer 17

@Marki

Answer 18

hmm im like my brain crashed a bit hahaha :\ somehow

Answer 19

i've seen this ... let me get my memory back

Answer 20

\[\text{ Assume } n^3=1331_{n-1} \\ \text{ for integer } n \ge 5 \\ \text{ We want to show } (n+1)^3=1331_{n}\]
\[n^3+3n^2+3n+1=1331_{n-1}+?????\]
Trying to figure out what to do with the 3n^2+3n+1 part

Answer 21

by the way while I think @rational were you assigned that 343 /34343 / so on thing?

Answer 22

Like was that an assignment ?
And if it was there your teacher ever tell you the way they would have done it?

Answer 23

did* (not there)

Answer 24

i think this is too smart

Answer 25

If it is too smart for you @marki then it is too smart for me :(

Answer 26

You are a genius.

Answer 27

smart = think about = solvable = not out from space = u are smart to get such idea
also dont get such idea's of hmmm :|

Answer 28

\[(1331)_{n-1} = (n-1)^3 + 3(n-1)^2 + 3(n-1) + 1 = n^3\]

Answer 29

neat neat!

Answer 30

@freckles that 34343.. was a very hard problem which i have been working on and off during my free time..

Answer 31

@Marki and @Kainui also know about that problem more than me

Answer 32

Like even digit numbers like are way easy.
http://openstudy.com/users/freckles#/updates/54a9a050e4b054f0c3ba5a87

Answer 33

I been plugging with this symmetry and this way of factoring for a bit.
I only kinda got the different base idea from @ganeshie8.

Answer 34

http://openstudy.com/users/freckles#/updates/54ab17aae4b0b83d9dca4479
When he posted here about it,

Answer 35

freckles we already solved for these cases:-
if number of digits of the form 3n and 3n+1

Answer 36

i cant remember where its been showing all work

Answer 37

for that induction problem?

Answer 38

I think @rational actually just proved it

Answer 39

and I feel like an idiot for not seeing it :p

Answer 40

ugh no

Answer 41

this changing base thingy is giving me some hope for that 34343.. xD
@FibonacciChick666 is good with NT too

Answer 42

@zarkon and @sithgiggles is that how you spell his name (it isn't showing up for me-that sithgiggles guy)

Answer 43

@SithsAndGiggles

Answer 44

know what i'll work in this -.-
i cant run away :P

Answer 45

Would love to help, but I honestly have no clue here

Answer 46

so 6 I think won't help at all for the other numbers
because none of the other numbers in the sequence mentioned by @rational can be written as some number cube

Answer 47

or you know the relationship between 6 and 343 will not help us at all*

Answer 48

ok well I have another way to proceed
...
I wonder if writing a number
like this is : \[mnmnmnmnmnm....=r:0_x\]
is doable
where r is a number between 0 and x (including 0 but not x)
So we have a number in base 10 that we want to write in base x.
And this number in base x only has two digits where the one digit is 0.

Answer 49

for example:
\[34343_{10}=61:0_{563}\]

Answer 50

Here is my guess, the prime factorization of 6 is 2*3 now, 4 is just 2*2 so the digits of 343434.... can be made using only the numbers in the prime factorization of 6

Answer 51

but that is cheating I think

Answer 52

not you @FibonacciChick666
I was talking about myself (and the cheating thing)

Answer 53

lol it's cool

Answer 54

Like I think the (r:0)_x thing works only if you know a factor of the number given
but it would cool to write it in that base number without even assuming it was a factor
but I don't know if that is possible

Answer 55

@freckles here is the main topic
http://openstudy.com/study#/updates/53ecad4be4b01789aba50084

Answer 56

that looks neat :) here are some easy observations about the sequence 34343... :

1)
starting from third term, every other third term is divisible by 3 because the sum of digits is divisible by 3

2)
starting from first term, every other third term is divisible by 7 for another cool reason

Answer 57

yep yep we showed that ealier our problem with
when number of digits is of the form 3n+2 like
34343
34343434343
34343434343434343
..............

Answer 58

but i want to forget this right now and work with bases like what u asked @freckles

Answer 59

I might have to take a break from this because at the moment I feel like I don't know where to go.

Answer 60

I really like this though. People like us working on something new I think. I think is the new.

Answer 61

well i like myself working on this , hmm i dont think other ppl really care :P

Answer 62

Well I like care. I like that you are thinking about it with me.

Answer 63

^_^

Answer 64

\[\large 34343434343_{10}= 3713:0_{9249511}\]

essentially you have rephrased the problem, thats all xD

Answer 65

whats this xD

Answer 66

\[\large 34343434343_{10}= 1:0_{34343434343}\]

Answer 67

?

Answer 68

\[\large 34343434343_{10}= 10_{34343434343}\]

Answer 69

:O

Answer 70

oh no :P

Answer 71

oh God xD
i'll fell down

Answer 72

\[\large (343)_{10} = (7\times 49)_{10} = (70)_{49}\]

Answer 73

>.<

Answer 74

solved omg i still feels like WOW

Answer 75

still not clear ha ?

Answer 76

NO ITS SO CLEAR !

Answer 77

wait but how are you figuring out the base number such that it can be written as a 2 digit number with 1's digit as 0

Answer 78

\[\large 34343434343_{10}= 10_{34343434343} \text{ this one is most interesting }\]

Answer 79

wait

Answer 80

there is no way to find that afaik :P

wolf uncle is helping me

Answer 81

\[34343=10_{34343}\]

Answer 82

so since all of the sequence of the form 10 base 34343...
that means (2*5) in base 3434343...
means all composite :3

Answer 83

so wait wait if they all can be written like this...

Answer 84

oh no

Answer 85

lol its not gona help us

Answer 86

doesn't help because that one digit is 1

Answer 87

;)

Answer 88

so r:0_x
where r is [0,1) U (1,x)

Answer 89

there is no free lunch haha!

Answer 90

hmm ok feeling dull a bit

Answer 91

what does that means ?

Answer 92

I'm beginning to think this approach will not work

Answer 93

brb 5 mnts

Answer 94

But who knows that could be me giving up early

Answer 95

ok ok ok

Answer 96

im quit bored :P