Base | Representation |
---|---|
bin | 11110010100011001101000… |
… | …000010100001110111011011 |
3 | 122111010111100212002212212221 |
4 | 132110121220002201313123 |
5 | 114434144034320240313 |
6 | 1151333022203244511 |
7 | 40041510166212451 |
oct | 3624315002416733 |
9 | 574114325085787 |
10 | 133343300165083 |
11 | 3953a6377660a8 |
12 | 12b569a9519737 |
13 | 59532b0905738 |
14 | 24cdbc8bc49d1 |
15 | 1063872a69c8d |
hex | 7946680a1ddb |
133343300165083 has 2 divisors, whose sum is σ = 133343300165084. Its totient is φ = 133343300165082.
The previous prime is 133343300165029. The next prime is 133343300165099. The reversal of 133343300165083 is 380561003343331.
It is a strong prime.
It is an emirp because it is prime and its reverse (380561003343331) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 133343300165083 - 229 = 133342763294171 is a prime.
It is not a weakly prime, because it can be changed into another prime (133343300165183) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66671650082541 + 66671650082542.
It is an arithmetic number, because the mean of its divisors is an integer number (66671650082542).
Almost surely, 2133343300165083 is an apocalyptic number.
133343300165083 is a deficient number, since it is larger than the sum of its proper divisors (1).
133343300165083 is an equidigital number, since it uses as much as digits as its factorization.
133343300165083 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 699840, while the sum is 43.
The spelling of 133343300165083 in words is "one hundred thirty-three trillion, three hundred forty-three billion, three hundred million, one hundred sixty-five thousand, eighty-three".
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