Find the missing properties of cyclic number

Guess a 6 digits number in the form of $\ abcdef$ which holds Properties #1, #2 and #3. The real challenge is to find 4 other interesting properties of this number:

• Property #1

$\ a+d=b+e=c+f=9$

$\ ab+cd+ef=99$

$\ abc+def=999$

$\ abcd+efab+cdef=9999$

• Property #2

$\ abcdef^2= ghijklmnopq$

$\ ghijk+lmnopq=abcdef$

• Property #3

$\ def^2 - abc^2=fabcde$

$\ efa^2 - bcd^2=abcdef$

$\ fab^2 - cde^2=bcdefa$

• Property #4 (found!)

$\ 1\times abcdef=abcdef$

$\ 2\times abcdef=cdefab$

$\ 3\times abcdef=bcdefa$

$\ 4\times abcdef=efabcd$

$\ 5\times abcdef=fabcde$

$\ 6\times abcdef=defabc$

• Property #5

hint: Improve the equation below to gain cyclic number similar to previous properties (p is a prime number):

$\ \frac1p =0.\overline{abcdef}$ ...

• Property #6

hint: following formula can be starting point for reproducing cyclic number( r and s are two constant numbers that you need to find):

$\ r^1\mod\ s=c$

...

• Property #7

hint: following formula can be starting point for reproducing cyclic number(p is a prime number):

$\ 10=3+(p \times a)$

...

Note: What I mean by $abc$ is concatenation of numbers or mathematicaly speaking, $abc$ is $(100*a)+(10*b)+c$.

Spoiler:

Prime number which produce this cyclic number is p=7

• feel free to edit the question so that it become more clear and understandable. tnx – Woeitg Feb 10 '16 at 19:00
• Wow. An idea jumped out right away, but it only fit the first two properties. Back to the drawing board... – Sean Henderson Feb 10 '16 at 19:23
• Share your idea :-) interesting to know – Woeitg Feb 10 '16 at 19:24
• 333333. Worked great initially. Question, though...When you're stating those properties, are we taking things like abc + def = 999 to be the concatenation of a, b and c or is it the product of the digits? Makes a HUGE difference. – Sean Henderson Feb 10 '16 at 19:27
• Concatenation. So basically what I mean by $abc$ is $(100*a)+(10*b)+c$ – Woeitg Feb 10 '16 at 19:31

Given that $abc+def=999$, we must have $a+d=b+e=c+f=9$ because there's no way to produce a carry anywhere. From property 3, we must have $d>a$, $e>b$, $f>c$, otherwise the numbers on the right would be negative. $$fabcde=def^2-abc^2=(def+abc)(def-abc)=999(def-abc)=999(def+def-999)=999(200d+20e+2f-999)$$ But $$fabcde=1000(fab)+cde=1000(fab)+(999-fab)=999(fab+1)=999(100f+10a+b+1)=999(100f-10d-e+100)$$ So $200d+20e+2f-999=100f-10d-e+100$, which we can rearrange to get $30d+3e=157+14f$. $157+14f$ must be divisible by 3, while $f$ must be at least 5 and at most 9. The only possibility is that $f=7$. Then $abcdef=142857$.
Some other properties of the number $142857$:
• $2\times142857=285714$, which is the same number but shifted over. The same thing happens for 3, 4, 5, or 6 times.
• $7\times142857=999999$.
• If you multiply $142857$ by any positive integer, add the last six digits to the rest, and continue until you have only six digits, the result will be $999999$ or a cyclic shift of $142857$. This results from a combination of the previous two properties.
• $0.\overline{142857}=\frac{1}{7}$, which is part of the explanation for why all of these properties hold.
• didnt think about simple $a+d=b+e=c+f=9$, interesting ! I added it to question – Woeitg Feb 10 '16 at 20:42