Base | Representation |
---|---|
bin | 1100010010011101101110… |
… | …0010001010010110000011 |
3 | 1202211200011120201211111021 |
4 | 3010213123202022112003 |
5 | 3232332223424431431 |
6 | 44423010403330311 |
7 | 2563106660132332 |
oct | 304473342122603 |
9 | 52750146654437 |
10 | 13511355311491 |
11 | 433a151918844 |
12 | 1622711680997 |
13 | 7701665264cb |
14 | 349d4b3a9d19 |
15 | 1866dc141b11 |
hex | c49db88a583 |
13511355311491 has 2 divisors, whose sum is σ = 13511355311492. Its totient is φ = 13511355311490.
The previous prime is 13511355311483. The next prime is 13511355311557. The reversal of 13511355311491 is 19411355311531.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 13511355311491 - 23 = 13511355311483 is a prime.
It is not a weakly prime, because it can be changed into another prime (13511355311191) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6755677655745 + 6755677655746.
It is an arithmetic number, because the mean of its divisors is an integer number (6755677655746).
Almost surely, 213511355311491 is an apocalyptic number.
13511355311491 is a deficient number, since it is larger than the sum of its proper divisors (1).
13511355311491 is an equidigital number, since it uses as much as digits as its factorization.
13511355311491 is an evil number, because the sum of its binary digits is even.
The product of its digits is 121500, while the sum is 43.
The spelling of 13511355311491 in words is "thirteen trillion, five hundred eleven billion, three hundred fifty-five million, three hundred eleven thousand, four hundred ninety-one".
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