# 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.

$$857^2 = 734449$$
$$142^2 = 20164$$
$$857^2 − 142^2 = 714285$$
$$734449 − 20164 = 714285$$