Base | Representation |
---|---|
bin | 1011111011001000000011… |
… | …0110110100111000000111 |
3 | 1201102100020110102102111211 |
4 | 2332302000312310320013 |
5 | 3204300100410342044 |
6 | 43514500251312251 |
7 | 2522123661444133 |
oct | 276620066647007 |
9 | 51370213372454 |
10 | 13110402043399 |
11 | 41a50aa190a81 |
12 | 1578a7314b687 |
13 | 7413c8560a0b |
14 | 3347927823c3 |
15 | 17b071be8634 |
hex | bec80db4e07 |
13110402043399 has 2 divisors, whose sum is σ = 13110402043400. Its totient is φ = 13110402043398.
The previous prime is 13110402043337. The next prime is 13110402043447. The reversal of 13110402043399 is 99334020401131.
13110402043399 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 is an emirp because it is prime and its reverse (99334020401131) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 13110402043399 - 29 = 13110402042887 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13110402043699) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6555201021699 + 6555201021700.
It is an arithmetic number, because the mean of its divisors is an integer number (6555201021700).
Almost surely, 213110402043399 is an apocalyptic number.
13110402043399 is a deficient number, since it is larger than the sum of its proper divisors (1).
13110402043399 is an equidigital number, since it uses as much as digits as its factorization.
13110402043399 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 69984, while the sum is 40.
The spelling of 13110402043399 in words is "thirteen trillion, one hundred ten billion, four hundred two million, forty-three thousand, three hundred ninety-nine".
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