In different bases:
| Binary (base 2) | \(100010111000001001_2\) |
| Ternary (base 3) | \(21020222000_3\) |
| Quaternary (base 4) | \(202320021_4\) |
| Quinary (base 5) | \(14032412_5\) |
| Senary (base 6) | \(3021213_6\) |
| Octal (base 8) | \(427011_8\) |
| Decimal (base 10) | \(142857_{10}\) |
| Duodecimal (base 12) | \(\rm{6A809_{12}}\) |
| Hexadecimal (base 16) | \(\rm{22E09_{16}}\) |
| Vigesimal (base 20) | \(\rm{HH2H_{20}}\) |
| Base 36 | \(\rm{3289_{36}}\) |
Other miscellaneous information:
| Prime factorization | \(3^3 × 11 × 13 × 37\) |
| Divisors | \(1, 3, 9, 11, 13, 27, 33, 37, 39, 99, 111, 117, 143, 297, 333, 351, 407,\\ 429, 481, 999, 1221, 1287, 1443, 3663, 3861, 4329, 5291, 10989, 12987,\\ 15873, 47619, 142857\) |
| Roman numeral | \(\rm{\overline{CXL}MMDCCCLVII}\) (vinculum system) |
If it is multiplied by 1, 2, 3, 4, 5, or 6, the answer will be a cyclic permutation of itself.
\(142857 × 1 = 142857\)
\(142857 × 2 = 285714\)
\(142857 × 3 = 428571\)
\(142857 × 4 = 571428\)
\(142857 × 5 = 714285\)
\(142857 × 6 = 857142\)
\(142857 × 7 = 999999 \space\longleftarrow\) Doesn’t work.
Those numbers are the six repeating digits of \(\frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}\), and \(\frac{6}{7}\).
\(1/7 = 0.\overline{142857}\)
\(2/7 = 0.\overline{285714}\)
\(3/7 = 0.\overline{428571}\)
\(4/7 = 0.\overline{571428}\)
\(5/7 = 0.\overline{714285}\)
\(6/7 = 0.\overline{857142}\)
\(7/7 = 0.\overline{999999} = 1\)
\(8/7 = 1.\overline{142857} \space\longleftarrow\) The cycle starts repeating.
\(9/7 = 1.\overline{285714}\)
\(…\)
If you square the last three digits and subtract the square of the first three digits, you also get back a cyclic permutation of the number.
\(857^2 = 734449\)
\(142^2 = 20164\)
\(857^2 − 142^2 = 714285\)
\(734449 − 20164 = 714285\)