142857


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.

\(857^2 = 734449\)
\(142^2 = 20164\)
\(857^2 − 142^2 = 714285\)
\(734449 − 20164 = 714285\)