Base | Representation |
---|---|
bin | 11101010010110110111… |
… | …001011011010111011001 |
3 | 21010110012022122112120022 |
4 | 131102312321123113121 |
5 | 230440323434432022 |
6 | 4140451134240225 |
7 | 265304551612052 |
oct | 35226671332731 |
9 | 7113168575508 |
10 | 2013113202137 |
11 | 706835a14724 |
12 | 2861a3656075 |
13 | 117ab321ca8b |
14 | 6d614285729 |
15 | 37574340c42 |
hex | 1d4b6e5b5d9 |
2013113202137 has 2 divisors, whose sum is σ = 2013113202138. Its totient is φ = 2013113202136.
The previous prime is 2013113202091. The next prime is 2013113202173. The reversal of 2013113202137 is 7312023113102.
Together with next prime (2013113202173) it forms an Ormiston pair, because they use the same digits, order apart.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 1979806587136 + 33306615001 = 1407056^2 + 182501^2 .
It is a cyclic number.
It is not a de Polignac number, because 2013113202137 - 28 = 2013113201881 is a prime.
It is not a weakly prime, because it can be changed into another prime (2013113202337) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1006556601068 + 1006556601069.
It is an arithmetic number, because the mean of its divisors is an integer number (1006556601069).
Almost surely, 22013113202137 is an apocalyptic number.
It is an amenable number.
2013113202137 is a deficient number, since it is larger than the sum of its proper divisors (1).
2013113202137 is an equidigital number, since it uses as much as digits as its factorization.
2013113202137 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 1512, while the sum is 26.
Adding to 2013113202137 its reverse (7312023113102), we get a palindrome (9325136315239).
The spelling of 2013113202137 in words is "two trillion, thirteen billion, one hundred thirteen million, two hundred two thousand, one hundred thirty-seven".
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