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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

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  • $\begingroup$ feel free to edit the question so that it become more clear and understandable. tnx $\endgroup$
    – Sadegh
    Feb 10, 2016 at 19:00
  • $\begingroup$ Wow. An idea jumped out right away, but it only fit the first two properties. Back to the drawing board... $\endgroup$ Feb 10, 2016 at 19:23
  • $\begingroup$ Share your idea :-) interesting to know $\endgroup$
    – Sadegh
    Feb 10, 2016 at 19:24
  • $\begingroup$ 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. $\endgroup$ Feb 10, 2016 at 19:27
  • 4
    $\begingroup$ Concatenation. So basically what I mean by $ abc $ is $ (100*a)+(10*b)+c $ $\endgroup$
    – Sadegh
    Feb 10, 2016 at 19:31

1 Answer 1

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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.
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  • $\begingroup$ Brute force reveals this to be the only solution to satisfy Properties 1–3. Awesome work with the other properties! $\endgroup$
    – dpwilson
    Feb 10, 2016 at 20:21
  • $\begingroup$ didnt think about simple $ a+d=b+e=c+f=9 $, interesting ! I added it to question $\endgroup$
    – Sadegh
    Feb 10, 2016 at 20:42
  • $\begingroup$ you discovered one of properties: the shifting numbers when multiply it by 1 to 6. about 1/7, you are very near to discover another one. hint: try to make it shifting somehow ;-) $\endgroup$
    – Sadegh
    Feb 10, 2016 at 23:15

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