Base | Representation |
---|---|
bin | 1100000110110011010101… |
… | …1011010110000110011001 |
3 | 1202010112000110220000120101 |
4 | 3001230311123112012121 |
5 | 3221041412000324121 |
6 | 44150553342055401 |
7 | 2542454663154316 |
oct | 301546533260631 |
9 | 52115013800511 |
10 | 13311000011161 |
11 | 4272188542206 |
12 | 15ab916b7a561 |
13 | 7572b6716768 |
14 | 34038203bd0d |
15 | 1813b28d8491 |
hex | c1b356d6199 |
13311000011161 has 2 divisors, whose sum is σ = 13311000011162. Its totient is φ = 13311000011160.
The previous prime is 13311000011093. The next prime is 13311000011191. The reversal of 13311000011161 is 16111000011331.
13311000011161 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 12887776902025 + 423223109136 = 3589955^2 + 650556^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-13311000011161 is a prime.
It is not a weakly prime, because it can be changed into another prime (13311000011191) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6655500005580 + 6655500005581.
It is an arithmetic number, because the mean of its divisors is an integer number (6655500005581).
Almost surely, 213311000011161 is an apocalyptic number.
It is an amenable number.
13311000011161 is a deficient number, since it is larger than the sum of its proper divisors (1).
13311000011161 is an equidigital number, since it uses as much as digits as its factorization.
13311000011161 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 54, while the sum is 19.
Adding to 13311000011161 its reverse (16111000011331), we get a palindrome (29422000022492).
The spelling of 13311000011161 in words is "thirteen trillion, three hundred eleven billion, eleven thousand, one hundred sixty-one".
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